NEET - PHYSICS ELECTROSTATICS

12th PHYSICS MODULE - 1

The resultant field is      . For Q.6 What is the difference between dielectric strength
E E0 EP and the dielectric constant?

homogeneous and isotropic dielectrics, the Q.7 Explain why a water molecule is permanently
 polarized. What type of molecule has no
direction of EP is opposite to the direction of permanent polarization?
 
E0 . The resultant field E is in the same direction 1.22 CAPACITORS AND
CAPACITANCE
 but magnitude is
as the applied field E0 its * Capacitance is one of the basic concepts behind
reduced. We can write   E0 electronics, and it is widely used, as seen by the
E K number of capacitors that are used in electronic
circuits. Second toresistors,capacitorsarethenext
where K is a constant for given dielectric which

has a value greater than one. This constant K is

called the dielectric constant or relative

permittivity of the dielectric.

Checkup 8 * most used component in the electronics industry.
Capacitors find uses in all types of circuit from
Q.1 Can there be an electric field within the material logic circuits, to power supplies and radio
frequency circuits to audio one. In addition to
of a conductor in electrostatic equilibrium?

Q.2 A point charge Q is at the center of an uncharged this there are many types of capacitor, but despite
their differences, they all rely on the basic
spherical conducting shell. How much charge is concepts of capacitance.
Capacitance is effectively the ability to store
on the inner surface of the shell? The outer * charge. In its simplest form a capacitor consists
surface? of two parallel plates. It is found that when a
battery or any other voltage source is connected
(A) 0; 0 (B) Q; 0

(C) –Q; 0 (D) –Q; Q

Q.3 A person is placed in a large hollow metallic to the two plates a current flows for a short time
sphere that is insulated from ground. If a large and one plate receives an excess of electrons,
charge is placed on the sphere, will the person while the other has too few. In this way one plate,
be harmed upon touching the inside of the the one with the excess of electrons becomes
sphere? Explain what will happen if the person negatively charge, while the other becomes
also has an initial charge whose sign is opposite * positively charged.
that of the charge on the sphere. If the battery is removed the capacitor will retain
its charge. However if a resistor is placed across

Q.4 Two identical conducting spheres each having a the plates, a current will flow until the capacitor

radius of 0.500 cm are connected by a light  becomes discharged.

2.00m-long conducting wire.Acharge of 60.0µC

is placed on one of the conductors.Assume that

the surface distribution of charge on each sphere

is uniform. Determine the tension in the wire.

Q.5 A conducting spherical shell of inner radius a and Figure : Any two conductors insulated
outer radius b carries a net charge Q. A point from one another form a capacitor.
charge q is placed at the center of this shell.
Determine the surface charge density on (a) the
inner surface of the shell and (b) the outer surface
of the shell.

53 ELECTROSTATICS

12th PHYSICS MODULE - 1

* Any two conductors insulated from one another Thus, 1 farad = 1 coulomb
form a capacitor. 1 volt

Capacity Since farad is a large unit, its submultiples are
used for practical purposes.
* It can be easily shown that the change in potential
of a conductor, due to charge it acquires, 1.23 PARALLEL PLATE
depends not only upon the magnitude of the CAPACITOR (PPC)
charge but also upon its size and shape, and its
surroundings. Thus different conductors, when * A PPC consists of two equal parallel metal plates
in conducting communication, although at the facing each other.
same potential, have different capacities for
holding a charge of electricity. * The plates may be square, rectangular or circular
in shape.
* If the conditions of a conductor remain the same,
it can be shown that the potential is proportional 
to the charge. The ratio of the charge to the
potential to which it is raised, is called the AA
capacity or capacitance of the conductor; and is
constant under the given conditions. –q + + ++ ++ ++
Thus if V is the potential (with respect to earth) + +
to which a body is raised due to a charge Q on +
it, then V  Q,
B –– +q
Q = constant, say, C = capacity of the –– E
V ++ – – Va
++ ++ d

Vb

conductor. If V = 1, Q = C.

* The capacity or capacitance of a conductor may * E be the electric field between the plates at any

be defined as the charge to be given to the point.

conductor to raise its potential by unity. * Surface charge density,   q ... (1)
A
* If there are a number of charged conductors near

each other, the potential of each is determined From Gauss law  E .ds  q
0
not only by its own charge but also by the

magnitude and sign of charges (induced or

independent) on the other conductors. E  q  
A0 0
* For example, a conductor is charged positively, ... (2)

and a negatively charged conductor is brought The relation between electric field and potential

near it, the potential of the first is lowered and its

capacity is increased. Similarly the capacity of is E  V ... (3)
d
second conductor is increased in this condition.

* E   dV ; V   E dr ; C  Q  Q From eq. (2) and (3),
dr V E
 dr q  V ; q  A0 ... (4)
A0 d V d
We often ignore the minus sign. Thus if the value
of E is known as a function of r, the denominator But from principle of capacitor, q = CV ... (5)
is integrable.
* The SI or practical unit of capacity is a farad From eq. (4) and (5), C  A0
which is the capacity of a body whose potential d
is raised by 1 volt by a charge of 1 coulomb. A0
For air medium capacitance is C0  d

54 ELECTROSTATICS

12th PHYSICS MODULE - 1

If the capacitor is completelyfilled with dielectric D   0  0K
K then capacitance E    
 p P
A AK0  A0    0E 
C d  d  K d  KC0 D  0K E and  D   0 (K 1) E

   This gives for the electric susceptibility e defined
0  as e =0 (K–1).
 C = KC0 r  K 
Note : Dielectric strength : For any given plate

K  C  Cmedium separation, there is a maximum electric field that
C0 Cair
or can be produced in the dielectric before it breaks

downandbegins toconduct thismaximum electric

field is called the dielectric strength.

1.24 ELECTRIC FIELD When designing circuits, one always needs to
IN DIELECTRIC
insure that the electric field generated by the

stored charge in the capacitor does not exceed

* Let Ere0subletatnhtefaieplpdliisedEfielEd0dueEtop polarisation. the dielectric strength of the dielectric material.
The .
If this occurs, the capacitor will short circuit (and

For homogeneous and isotropic dielectric, the sometimes blow up).
 Air has a dielectric strength of 3 × 106 V/m.

direction of Ep is opposite to the direction of 1.25 SPHERICAL
 CAPACITOR
E0 . So, resultant field is E = E0 – EP.

(i) Sperical conductor : When a charge Q is given
to a isolated spherical conductor then its potential

–q +q rises. V= 1 Q
40 R

Figure : The surface charges on the C  Q  40R
dielectric reduce the electric field V
inside the dielectric.
If conductor is placed in a medium then
Cmedium = 4R or Cmedium = 40rR
 * The capacitance of the earth, assuming it as
Field due to polarisation is  * isolated conducting sphere of radius 6400 km is
P (ii) 711 µF.
  D0P0dEs=0PP Q;SfroEee,     p =   0 Outer sphere is earthed :
EP E0 E E0 = When a charge Q is given to inner sphere it is
ds Qfree uniformly distributed on its surfaceAcharge –Q
0E 0 (0E   
P)

or

Where D is electric displacement. is induced on inner surface of outer sphere.

The significance of D: In vacuum, E is related
to the free charge density . When a dielectric –Q

medium is present, the corresponding role is +Q R1


taken up by D. For a dielectric medium, it is D

not E that is directlyrelated to free charge density R2

. The ratio of the magnitudes of  and  is
D E

55 ELECTROSTATICS

12th PHYSICS MODULE - 1

The charge +Q induced on outer surface of outer 1.26 ENERGY STORED
sphere flows to earth as it is grounded. IN A CAPACITOR
E = 0 for r < Rl and E = 0 for r > R2
Potential of inner sphere

V1 = Q + Q  Q  R2  R1  When a conductor is charged its potential
40 R1 40 R 2 40  R1R 2  changes from 0 to V and work is done against
repulsion between charge stored on the
As outer surface is earthed so potential V2 = 0 conductor and charge coming from the charging
Potential difference between plates body. This work is stored as electrostatic

V  V1  V2  Q  R 2  R1  potential energy.
40  R1R 2 
U  W  1  q . dq  U  q2  1 qV  1 CV2
C 2C 2 2
 Q  4 0 R1R 2
C V R2  R1 (in air or vacuum} The energy is stored in the electric field between
the plates of the capacitor.
In presence of medium between plate The energy stored per unit volume in the electric

C  4  r 0 R1R 2 field between the plates is called energy density.
R2  R1
  1 0E2  1 2
(iii) Inner sphere is earthed 2 2 0

Here the system is equivalent to a spherical Work done by battery :
capacitor of inner and outer radii R1 and R2 * W = Charge that flow through battery
respectively and a spherical conductor of radius
× battery EMF
R2 in parallel. –Q Whenever there is a charging of capacitor. Work
done by battery, in part is stored as electrostatic

+Q R1 energy in between capacitor plate and remaining
is dissipated as heat due to charge flow through

R2 connecting wires.

In general, Heat produced is given by

H = (Work done by battery

This is because charge Q given to outer sphere – Energy stored in capacitor)

distributes in such a way that for the outer sphere

charge on the inner side is R1 Q and charge on 1.27 REDISTRIBUTION
R2 OF CHARGES

the outer side is Q  R1 Q  (R2  R1) Q (When two charged conductors are mutually
R2 R2 connected)
When two charged conductors are connected
So total capacity of the system, by a conducting wire then charge flows from a
conductor at higher potential to that at lower
C  4 0 R1R 2  4 0 R2 potential. This flow of charge stops when the
R2  R1 potential of two conductors became equal.
Let the amounts of charges after the conductors
C  40R 2 are connected are Ql' and Q2' respectively and
2 potential is V then
R2  R1

56 ELECTROSTATICS

Q1 Q2 Q1' Q2' 12th PHYSICS MODULE - 1
C1 C2 C1 C2
q1 + q2 = q
V1 R1 V2 R2 V R1 R2 q1 = C1V & q2 = C2V
V
(where V is common potential)

Common potential V  C1 q
According to law of Conservation of charge  C2
Qbefore connection = Qafter connection
Case 2 : Positive plate of capacitor C1 connected
C1V1 + C2V2 = C1V + C2V to positive plate of another charged capacitor
Common potential after connection C2.

V  C1V1  C2V2 Q1 ++++ –––– C1  q1 ++++ –––– C1
C1  C2

Charges after connection Q2 ++++ –––– C2 q2 ++++ –––– C2

Q1' = C1V =  Q1  Q2    C1  Q Q1 + Q2 = q1 + q2
C1  C2   C1  C2  C1V1 + C2V2 = C1V + C2V
C1 

(Q : Total charge on system)

Q2' = C2V =  Q1  Q2    C2  Q V C1V1  C2V2  Q1  Q2
C1  C2   C1  C2  C1  C2 C1  C2
C2 

Ratio of the charges after redistribution (Q1 & Q2 initial charge on capacitor, V final
Q1 common potential)
Q2  C1V  R1
C2V R2
(in case of spherical conductors) Case 3 : Positive plate of C1 connected to
negative plate of C2.
When charge flows through the conducting wire

then energy is lost mainly on account of Joule Q1++++ –––– C1 q1 ++++ ––––

effect, electrical energy is converted into heat

energy, so change in energy of this system, 
U = Uf – Ui
Q2 –––– ++++ C2 q2 –––– ++++
  1 C1V2  1 C2V2    1 C1V12  1 C2V22 
2 2 2 2
Assuming Q1 > Q2 ,
U   1  C1C2  (V1  V2 )2 Q1 – Q2 = q1 + q2 = C1V + C2V
2  C1  C2 
V Q1  Q2  C1V1  C2V2
C1  C2 C1  C2
Here negative sign indicates that energy of the
system decreases in the process.

In parallel plate capacitor we can consider 3 cases 1.28 COMBINATION OF
CAPACITORS
Case 1 : Charged capacitor (say C1) connected
to uncharged capacitor (say C2) Combination of capacitors by a single equivalent
q1 ++++ –––– C1 capacitor is called ‘grouping of capacitors’.
q C1
(a) Capacitor in Parallel

* The potential difference a cross each capacitor
uncharged q2 ++++ –––– C2 is same V1 = V2 = V
C2

57 ELECTROSTATICS

12th PHYSICS MODULE - 1

* The potential differences across capacitor plates

C1 are different V1  Q , V2  Q
+– C1 C2
V1 = V2 = V
C2 C1 Ceq = C1 + C2 * The voltage across the equivalent capacitor is
+–
Q1 V1 = V1 + V2

C2  V = Q/C  1  1  1
Q2 C C1 C2

* The reciprocal of the net capacitance equal to

+– +– +– the sum of the reciprocals of capacitances of
(a) V V
(b) (c) individual capacitors.
V
* The total energy stored in the series combination

Figure : (a) A parallel combination of two capacitors in an electric circuit in which the of capacitors is
potential difference across the battery terminals is V. (b) The circuit diagram for the
parallel combination. (c) The equivalent capacitance is Ceq = C1 + C2. U = U1 + U2 = 1 QV = ( 1 QV1 + 1 QV2 )
2 2 2
* The charge on different capacitors are different
Q1 = C1 V . Q2 = C2 V * If n equal capacitance, each C, are connected in

* The charge on equivalent capacitor plate is series than, equivalent capacitance is C/n.
Q = Q1 + Q2 and Q = CV  C = C1 + C2
V1 : V2 : : C1 : C2
* The total energystored in theparallel combination * To get minimum capacity capacitor must be

connected in series and the resultant capacity is

is U = U1 + U2 = 1 CV2. lesser than the least individual.
2

* If n equal capacitances, each C, are connected 1.29 METHODS OF
FINDING EQUIVALENT
in parallel, then the equivalent capacitance is nC. CAPACITANCE

Q1 : Q2 : : C1 : C2 We know that in series,
* To get maximum capacity capacitors must be

connected in parallel and the resultant capacity

is greater than largest individual. 1  1 1  ........  1 and in parallel,
Ceq C1 C2 Cn

(b) Capacitor in Series Ceq = C1 + C2 + ...... + Cn
Sometimes there are circuits in which capacitors
* The charges on the capacitor plates are same are in mixed grouping. To find Ceq for such
 Q1 = Q2 = Q circuits few methods are suggested here which
help you in finding Ceq.

1. Method of same potential

Figure : (a) A series combination of two capacitors. The charges on Give any arbitrary potentials (V1, V2 ........ etc.)
the two capacitors are the same. (b) The circuit diagram for the series combination. to all terminals of capacitors. But notice that the
(c) The equivalent capacitance can be calculated from the relationship points connected directly by a conducting wire
will have at the same potential. The capacitors
having the same P.D are in parallel. Make a table
corresponding to the figure. Now corresponding
to this table a simplified figure can be formed
and from this figure Ceq can be calculated.

58 ELECTROSTATICS

12th PHYSICS MODULE - 1

EXAMPLE 36 A,1,3 C1
C2 2,4,B
Find the equivalent capacitance between points C3
A and B shown in figure.

AC CC C C CB Ceq = C1 + C2 + C3

SOLUTION: EXAMPLE 38
Number the junction as shown
Find the equivalent capacitance between points
A and B.

1 AC CB

AC C 2 C 3C C 4 CB C
C
VA = V2
V1 = V3 = V4 SOLUTION:
Redraw the network as Circuit can be redrawn as

CC

A,2 1,3,4 B A CB

Now, 3C and C are in series and their equivalent C

capacitance is, Ceq = (3C) (C)  3 C  Parallel combination, Ceq = C + C = 2C
3C  C 4 Series combination

EXAMPLE 37 1  1  1  3 So, Ceq  2 C
Find the equivalent capacitance betweenA& B. Ceq C 2C 2C 3

Parallel combination, Ceq  2 C  C  5 C
3 3

A C1 C2 C3 B 2. Infinite series problems

SOLUTION: This consistsofan infiniteseriesofidentical loops.
Number the junction as shown To find Ceq of such a series first we consider by
ourself a value (say x) of Ceq. Then we break
2 4 the chain in such a manner that only one loop is
C1 left with us and in place of the remaining portion
A1 C2 3 C3 B we connect a capacitor x. Then we find the Ceq
and put it equal to x. With this we get a quadratic
VA = V1 = V3 equation in x. By solving this equation we can
V2 = V4 = VB find the desired value of x.
Redraw the network

59 ELECTROSTATICS

EXAMPLE 39 12th PHYSICS MODULE - 1
Find the effective capacitance betweenAand B.
In such cases some junctions are unnecessarily
A C1 C1 C1 made. Even if we remove that junction there is
no difference in the remaining circuit or charge
distribution. But after removing the junction, the
problem becomes very simple.

C2 C2 C2 To infinite EXAMPLE 40
Find the equivalent capacitance between points
B A and B.

SOLUTION: C
Suppose the effective capacitance betweenAand
C C CC
B is CR. Since the network is infinite, even if we
remove one repeating unit from the chain AC CB

remaining network would still have infinite cells, SOLUTION:

i.e., effective capacitance between DE would also

be CR. C

A C1 D A D C1 CC CC

CR C2 CR CR C0 C2 C0 A B
CC

B (A) E B E (B) 1  1  2 C1 = C
C C C 2
Inotherwords thegiveninfinitechainis equivalent

to capacity C1 in series with combination of C2 C C
and CR in parallel as shown in Fig. (A). So B
C/2
CR = C1 S [C2 + CR]
C
i.e., CR  C1 (C2  CR )
C1  C2  CR A
CC

or CR2 + C2CR – C1C2 = 0 C  C  3C C2 = 3C
2 2 2

i.e., CR  1 C2  C22  4C1C2  3C/2
2
C C
And as capacitance cannot be negative, only  C/2 B
permissible value of CR is :
A

CR  C2  1  4 C1   1 1  1  2  8 C3 = 3C
2 C2  C C 3C 3C 8

However, if C1 = C2 = C; CR = [5 – 1]C/2 3C/8

3. Connection removal method  B

This method is useful when the circuit diagram is A C/2
symmetric except for the fact that the input and
output are reversed. That is the flow of charge is C  3C  7C Ceq = 7C
a mirror image between input and output above 2 8 8 8
a particular axis.

60 ELECTROSTATICS

4. Balanced Wheatstone's Bridge 12th PHYSICS MODULE - 1

 Similarly, a star network can be transformed to
a delta network according to the formula.
VE
C1 C3 Ca  C1C3 , Cb  C1C2 ,
C1  C2  C3 C1  C2  C3

A C5 B Cc  C1 C2C3
 C2  C3
C2 C4

VD EXAMPLE 41
Find the equivalent capacitance across AB.
If C1  C3 , bridge is said to be balanced and
C2 C4 D
C 2C
in that case.
VE = VD or VE – VD = 0 or VED = 0 A CB
i.e., no charge is stored in C5. Hence, it can be
removed from the circuit. 2C C
E
CAB = (C1S C3) || (C2S C4)
SOLUTION:
CAB = C1C3  C2C4 Using Star-Delta conversion
C1  C3 C2  C4
D
5. Unbalanced Wheatstone's Bridge C 2C D 2C
5C/2
For solving unbalancedWheatstone’s bridge use B
Star-delta conversion method. A C  5C C
A delta network is transformation to a star-
network according to the formula BA

2C C 5C
E E

C1 C2 10C/9

Ca 5C
Cb

Cc AB
C3
This implies that 5C/6
 C1C2  C2C3  C3C1
Ca C3 , (105C / 54)  5C 7C
Ceff  CAB  (375C / 54)  5

Cb  C1C2  C2C3  C2C1
C1
6. By distributing charge
Cc  C1C2  C2C3  C3C1
C2 In this method assume a main charge q. Distribute
it in different capacitors as q1, q2 .... etc. Using
C1 Ca Cb Kirchhoff's laws (Refer Current electricity
C2 Cc chapter) find q1, q2 .... etc. in terms of q. Then
find the potential difference between starting and
C3 end point through any path and equate it with
q/Cnet. By doing so we can calculate Cnet.

61 ELECTROSTATICS

12th PHYSICS MODULE - 1

EXAMPLE 42 EXAMPLE 43

Twelve capacitors, each having a capacitance C, If a capacitor is formed by n plates with alternate
are connected to form a cube. Find the equivalent plates connected together, and separation
capacitance between the diagonally opposite between successive plates is d, then find net
corners such as A and B. capacitance.

B SOLUTION:

C  (n 1) 0A
d

D EXAMPLE 44

AC Find the equivalent capacitance betweenAand
B.
SOLUTION:
Connect a battery between point A and B. Let A 1
say charge 6Q flows from battery. 2
Due to symmetry charge divides as shown in fig. SOLUTION: 3
4
B
1 12
Ceq A2
2Q 32
3 AB
4
34
 C' = 3C0
2Q Q F
2Q Q

A 2Q E 6Q V
6Q
EXAMPLE 45

V Find the equivalent capacitance betweenAand
B.
From Kirchhoff’s rule for loopABFBA
1
2
2Q  Q  2Q  V  0 ....... (1) A 3 B
C C C + 54 –

From equivalent capacitance concept SOLUTION:

6Q = CeqV ....... (2) 1 12
From (1) and (2)
2 B+ –
5Q  6Q ; Ceq  6C A 3 –A B
C Ceq 5 + 54
54

Ceq between A and C = 12 C C' = 2C0
7
EXAMPLE 46
4
Ceq between A and D = 3 C Find the equivalent capacitance betweenAand

1 A
2 B
B. 3
4

62 ELECTROSTATICS

12th PHYSICS MODULE - 1

SOLUTION: 21  C  q  0A 1 
23 V t 1 r 
1 A d 
2 BA 3 4B
3
4

 C  2 C0  C  0A 1  ..........(1)
3 t 1 r 
d 

1.30 CAPACITY OF If medium is fully present between the space.
DIFFERENT  t=d
CONFIGURATION
Now from equation (1), Cmedium = 0rA
d
0A
In case of parallel plate capacitor C  d If capacitor is partiallyfilled bya conducting slab

If capacitor is partially filled with dielectric of of thickness (t< d).
thickness t (t < d)  r =  for conductor

If no slab is introduced between the plates of the C  0A 1  ; C  0A
t 1  (d  t)
 d 
capacitor, then a field E0 given by E0 = 0 ,

exists in a space d.

A 1.31 FORMING OF
CAPACITOR
E0 K=r E0
Distance division
t
Case 1 :
On inserting the slab of thickness t, a field
d

E= E0 exists inside the slab of thickness t and K1 K2
r d1 d2
C1 C2
a field E0 exists in remaining space (d – t).
If V is total potential then (i) Distance is divided and area remains same.
(ii) Capacitors are in series.
V = E0(d – t) + E t

 V  E0 d  t   E  t  C  K0A , C1  K10A , C2  K20A
 E0  d d1 d2

E0 These two in series
E
  r = Dielectric constant 1  1  1  d1  d2
C C1 C2 K10A K20A
 d  d 
 V  0  t  t  q  t  t  1  1  d1  d2 
r A0 r C 0 A K1 K2

63 ELECTROSTATICS

12th PHYSICS MODULE - 1

 C  0A  K1K2  If d1 = d2 = d3 = d
K2d1  K1d2 3

If d1 = d2 = d C  0A  K1K 2 3K1K2K3 K3K1 
2 d   K2K3  

 C  0A  2K1K2   K eq  K1K 2 3K1K 2 K 3 K3K1
d K1  K2  K2K3 

 Keq  2K1K2 Area division
K1  K2
Case 3 :

Case 2 : Similarly if three dielectric slabs are d
arranged as shown in figure.
AK11 C1
d
A AK22 B A
B

K1 K2 K3 C2
d1 d2 d3
(i) Area is divided and distance remains same.
(ii) Capacitors are in parallel.

C1  K10A1 , C2  K20A2
d d

C1 and C2 are in parallel

C1 C2 C3 C = C1 + C2 = K10A1  K20A2
d d

C1  K10 A , C2  K20A , C3  K30A C  0 (K1A1  K2A2)
d1 d2 d3 d

They are in series A  0A  K1  K2 
2 d 2
1  1  1  1 If A1 = A2 = ; C
C C1 C2 C3
K1  K2
d1 d2 d3  Keq  2
 K10A  K20A  K30A

Similarly for three dielectrics

 1  1  d1  d2  d3  K1 A1 C1
C 0 A  K1 K2 K3 
K2 A2 C2
A B A B

1  1  d1K2K3  d2K1K3  d3K1K2  K3 A3 C3
C 0 A K1K2K3
C = C1 + C2 + C3

C  0 A  d1K 2 K 3  K1K2K3  d3K1K2  = K10A1  K20A2  K30A3
d2K1K3 d d d

64 ELECTROSTATICS

C  0 (K1A1  K2A2  K3A3) 12th PHYSICS MODULE - 1
d
A 1.32 VARIATION IN PPC
If A1 = A2 = A3 = 3 UNDER TWO
CONDITIONS
C  0A  K1  K2  K3 
d 3 * If nothing is mentioned then assume battery is

 Keq  K1  K2  K3 disconnected and Q is constant.
3 * A parallel platecapacitor is connected to battery

(V = const.) and a slab of dielectric constant r
is inserted between the plates then total energy
given by battery is divided into two parts :
(i) Half is used to insert the slab (work is done by
field)
(ii) Half is stored in form of electrostatic potential
energy.

Battery disconnected (Q is constant)

Change Q V= Q E= Q C= 0rA U= Q2
C 0rA d 2C
Medium is filled Unchange
Distance Decrease Unchange V  1 U  1
Area Increase Unchange C 2C

Decrease Decrease Increase Decrease
Unchange Increase
Decrease Decrease Increase Decrease

Decrease Decrease

Battery still connected (V is constant)

Change Q = CV V constant E= V C= 0rA U  1 CV2
d d 2
Medium is filled
Distance Decrease QC E  1 UC
Area Increase d
Increase Increase
Increase Unchange Unchange Increase Increase
Increase Unchange Increase Increase
Unchange Increase Increase

Unchange

Steps to Solve Question :

(1) Decide about capacity 1.33 VAN DE GRAAFF
(2) Decide what is constant (Q or V) GENERATOR

[If not given then assume Q is constant]
(3) Remaining parameters are decided by using * It is a machine capable of building up potential
difference of a few million volts, and fields close
formula according to constant parameter.
to the breakdown field of air which is about
3 × 106 V/m.

65 ELECTROSTATICS

* Principle : It is based on thefollowingprinciples: 12th PHYSICS MODULE - 1
(1) Action of sharp points: Charges are leaked from
The uncharged belt again collects the positive
pointed ends of charged conductors. This creates charge and the process continues. Thus, the
an electric wind (as moving air is ionized) which positive charge goes on accumulating and hence
moves away from the conductor. the potential goes on increasing.

+ Metal dome + Checkup 9
+ +
+ Q.1 The plates of a capacitor are connected to a
+ + Belt + battery. What happens to the charge on the plates
B + if the connecting wires are removed from the
+ P battery? What happens to the charge if the wires
are removed from the battery and connected to
++ each other?

+ Q.2 A farad is a very large unit of capacitance.
+ Calculate the length of one side of a square, air-
+ filled capacitor that has a capacitance of 1 F and
+ a plate separation of 1 m.
+
+ Q.3 The sum of the charges on both plates of a
+ capacitor is zero. What does a capacitor store?
+
+ Q.4 Evaluate the equivalent capacitance of the
+ configuration shown in Figure.All the capacitors
+ are identical, and each has capacitance C.
++
+
A

Ground Insulator

Figure : Diagram of a Van de Graaff Q.5 Find the equivalent capacitance between points
generator. Charge is transferred to a and b for the group of capacitors connected as
the metal dome at the top by means shown in figure. Take C1 = 5.00 µF, C2 =10.0µF,
of a moving belt. The charge is and C3 = 2.00 µF.
deposited on the belt at point A and
transferred to the hollow conductor
at point B.

(2) The property that the charge given to a hollow
conductor is transferred to the outer surface and
is distributed uniformly on it.

Working : Apulleydrives an insulating belt by a ELECTROSTATICS
sharplypointed metal comb which has been given
a positive charge by a power supply. Electrons
are removed from the belt, leaving it positively
charged.Asimilar comb at the top allows the net
positive charge to spread to the dome.

66

12th PHYSICS MODULE - 1

Q.6 A parallel-plate capacitor is charged and then IMPORTANT POINTS
disconnected from a battery. By what fraction
does the stored energy change (increase or Charge is quantized, i.e., q = ne with
decrease) when the plate separation is doubled? n = 1, 2, ..... (e = 1.602 × 10–19 coulomb).

Q.7 A parallel-plate capacitor is constructed byfilling * Coulomb's law : F  q1q2
the space between two square plates with blocks 4 0 r R 2
of three dielectric materials, as in Figure . You *
may assume that  >> d. Find an expression for r : relative permittivity or dielectric constant of
the capacitance of the device in terms of the plate the medium in which charges have been
area A and d, K1, K2, and K3.
submerged. 1  9 109 Nm2 / (coulomb)2 .
 4

0
K 2 d/2 Permittivity of a medium = 0 r
d K1 K3
q
* Electric field due to a point charge E  40r2

/2 * Electric field on the axis of a uniformly charged

Q.8 Many computer keyboard buttons are ring of radius R at a distance x from the centre is
constructed of capacitors, as shown in Figure.
When a key is pushed down, the soft insulator E Qx , (where Q is total
between the movable plate and the fixed plate is 40 (R2  x2 )3/2 R
compressed. When the key is pressed, the
capacitance (a) increases, (b) decreases, or (c) * charge on the ring). E is maximum at x = ± 2 .
changes in a way that we cannot determine
because the complicated electric circuit Eplec2traicqfaietlda on the axis of a dipole of moment
connected to the keyboard button may cause a distance R from the centre is
change in V. 2pR
  40 (R2  a2 )2 .
E

If R >> a then   2p .
E 40R3

Key * Electric field on the equatorial line of the dipole
*
Movable at a distance R from the centre is
plate p
Soft   40 (R2  a2 )3/2 .
insulator E
Fixed
plate If R >> a then   p .
E 40R3

At a general far-off point p(R, ) from the axis of
a short dipole the electric field makes an angle 

 tan   tan  .
with R 2

67 ELECTROSTATICS

12th PHYSICS MODULE - 1

* Torque  experienced by a short dipole kept in * Potential due to a point charge V  Q
 40r
uniform external electric field E is
 p cos 
  p  E  pE sin . * Potential due to an electric dipole V  40r2

* If I is the moment of inertia of the short dipole

then the time period of its SHM in the electric * Potential due to a system of charges

field is T = 2 I . V  1  q1  q2  q3  ...  qn 
PE 40  r1 r2 r3 rn 

* Force between two short electric dipoles is * Charge density and intensity of electric field at
* proportional to (1/R4), where R is the separation pointed ends is more while electric potential is
* * same as that of other points.
between two short dipoles. * The electric potential due to a unipolar charge
1/0 number of electric lines of forces originate V  1/x , due to dipolar charge V  1/x2.
from unit positive charge. Potential due to continuous charge distribution
  S
Electric flux E  E  EScos . Area vector

 V   dV  1  dq
S is perpendicular to the surface area. 40 r

* Gauss’s theorem says    ds  Q . Here  * Potential due to a charged ring of radius R on its
E 0 E

is the electric field due to all the charges inside as axis V  1 Q
40 R2  x2
well as outside the Gaussian surface.
* Electric field due to infinitely long charged wire * Potential due to a charged conducting shell or
of linear charge density  at a perpendicular sphere of radius R

distance R is E   . (a) For outside point V Q (r > R)
20R 40r

* Electric field due to conducting sheet of charge (b) For inside point or on surface

density  is  and due to nonconducting sheet V  Q (r  R)
0 40R

is  . * Potential due to uniformly distributed charge in a
20 spherical volume of radius R

* Field due to a uniformly charged thin spherical (a) For outside point V  Q (r  R)
(b) For inside point 40r
shell ofradius R is E = Q foroutside points
4 0r 2
Q
and zero inside. r is distance from the centre of V  80R3 (3R 2  r2) (r  R)
shell and r > R.

* Field due to a charge uniformly distributed in a * Relation between E and V

spherical volume is E = pr for inside points E   V   ˆiE x  ˆjE y  kˆ E z )
30 r (E

and E Q for outside point. Where, Ex   V , Ey   V , Ez   V
40r2 x y z

68 ELECTROSTATICS

12th PHYSICS MODULE - 1

* Electric field at the surface of a charged Q2
* Force between parallel plates is F = 2A0
conductor is E =  , where  is local surface *
0 *
Capacitance of parallel plate capacitor with
charge density.
Potential Energy between two charges separated dielectric of thickness t filled in,

by R, U  Q1Q2 C  0A t
40R  t)  k
(d

* Potential Energy among group of three charges * Capacitance of cylindrical capacitor

U  1  Q1Q2  Q2Q3  Q1Q3  C  20KL
40  r12 r23 r13  loge (R2 / R1)

(put ± sign for charges) * Spherical capacitor C   40K 
(formula can be extended for more charges)  
* Work done by external agent in moving charge 1  1
q from point 1 to 2 is W12 = q (V2 – V1), R1 R2

where V1, V2 are potential at point 1 and 2 due * Equivalent capacitance in series
to charges other than q.
* Work done by Electric field is equal and opposite 1  1  1  1  ....  1 ,
Ceq C1 C2 C3 Cn
to work done by external agent.

* Self potential energy of a uniformly charged in parallel, Ceq = C1 + C2 + C3 + ...... + Cn

 Q2 * Energy stored in a capacitor
80R
conducting sphere of radius R is U .  Q2  1 CV2  QV
2C 2 2
* Potential energy of an electric dipole kept in U

uniform electric field U = p    pE cos  . * Energy density u = 1 0E2
E * 2

* Work done by external agent in rotating the Common potential
* dipole from 1 and 2 is W = pE (cos 1– cos2).
Charge position (a) When plates of similar nature connected
*
Cube centre edge, ===qqq=///82q0./004.. 0. together = C1V1  C2V2
Face centre C1  C2
At corner
At centre of (b) When plates of opposite nature connected

Capacitance of an isolated body is given by together = C1V1  C2V2
C1  C2

C= Q (Q: total charge on the body; V is the * Loss of energy in charge sharing
V (a) When plates of similar nature are connected

potential of the body) together U  1 C1C2 (V1  V2 )2
2 (C1  C2 )
* Capacitance of isolated sphere C = Q = 40R
V (b) When plates of opposite nature are

Q  0A connected together
V d
* Capacitance of PPC, C= U  1 C1C2 (V1  V2 )2
2 (C1  C2 )
(Here Q = charge on one of the plates;

V = potential difference between the plates)

69 ELECTROSTATICS

12th PHYSICS MODULE - 1

* Magnitude of induced charge Qp  Q 1  1  E  9  109  (20 106 ) ( 2ˆi  ˆj  kˆ )
K 8

where Q is charge on the capacitor plate.  22.5 103 ( 2ˆi  ˆj  kˆ ) N / C
 0E , 
* Polarization vector P  where E is net

electric field inside the dielectric and e is electric EXAMPLE 2
susceptibility of the dielectric medium.
e 00E(K Two point charges of 2µC and 8µC are placed
* D = – 1)
*   is electric displacement and 12cm. apart. Find the position of the point where
P, D the electric field intensity will be zero.

equal to surface charge density on the capacitor SOLUTION:

plate. 2µC 12cm 8µC
+P +
* If n identical charged liquid drops are combined x

to form a big drop then Let at point P electric field intensity due to both

S.N. Quantity For each For big the charge is zero, then
charged drop
1. Radius small drop 1 (2 106 )  (8 106 ) 1
2. Charge R = n1/3r 4 0 x 2 40 (12  x)2
3. Capacitance r Q = nq
4. Potential q C' = n1/3C (12 – x)2 = 4x2 12 – x = 2x x = 4 cm.
5. Energy C V' = n2/3V
6. Surface V U' = n5/3U EXAMPLE 3
U  ' = n1/3
charge  Calculate the total electric flux through the pa-
density raboloidal surface due to a uniform electric field
of magnitude E0 in the direction shown in Figure.

r
d

SOLVED EXAMPLES E0

EXAMPLE 1 SOLUTION:
The flux through the curved surface is equal to
Find out electric field intensity at point the flux through the flat circle, E0 r2.
A (0, 1m, 2m) due to a point charge –20µC
EXAMPLE 4
situated at point B ( 2m, 0, 1m) .
SOLUTION: Findtheelectric field insideaspherewhich carries
a charge densityproportional to the distance from
E  KQ r  KQ rˆ the origin  =  r (is a constant).
| r |3 | r |2 SOLUTION:
We can consider all the charge inside the sphere
 r  P.V. of A – P.V. of B to be concentrated on the center of sphere.
Consider an elementary shell of radius x and
 ( 2ˆi  ˆj  kˆ ) (P.V. = position vector) thickness dx.

| r |  ( 2)2  (1)2  (1)2  2  E k 4x2dx (x) r 2
k dq r2 40
 r2  

70 ELECTROSTATICS

12th PHYSICS MODULE - 1

EXAMPLE 5 EXAMPLE 7 +
Electric charge is uniformly +
A point charge Q = 5.00 µC is located at the +
center of a cube of edge L = 0.100 m. In addi-
tion, six other identical point charges having distributed along a long
q = –1.00 µC are positioned symmetrically
around Q as shown in Figure. Determine the straight wire of radius 1mm. 1m
electric flux through one face of the cube.
The charge per cm length of

the wire is Q coulomb. +

Another cylindrical surface of + 50cm
radius 50 cm and length 1m
sym+metrically

L encloses the wire as shown in the figure. F i n d

q thetotalelectricflux passingthroughthecylindrical
qq
surface.
q Qq SOLUTION:
L Charge enclosed by cylindrical surface (length

q 100 cm) is Qenc = 100Q.
By applying Gauss’s law,

L   1 (Qenc. )  1 (100Q)
0 0
SOLUTION:
The total charge is Q – 6 | q |. The total outward EXAMPLE 8

Q6|q| A free pith ball of 8 gram carries a positive charge
flux from the cube is 0 , of which one- of 5 × 10–8 C. What must be the nature and

sixth goes through each face: magnitude of charge that should be given to a

(E )one face  Q 6| q | second pith ball fixed 5cm, vertically below the
60
former pith ball so that the upper pith ball is

(5.00  6.00) 106 CNm2 stationary.
6  8.85 1012 C2
 SOLUTION:
Let a charge q is given to the upper pith ball.

= –18.8 kN.m2/C So that the first pith ball

remain stationary q, should F

EXAMPLE 6 be positive, a as shown in the q0 +
figure. Further mg

A long string with F = mg +q

charge per unit length  q  5 108C = 8 × 10–3 × 9.8 N
 on it passes through 40 (5 102 )2 m2
a cube of side a. Find

the minimum flux a  q  8 103  9.8  25 104 C
through the cube. (5 108)  (9 109 )
SOLUTION:
The minimum length which can fit in the cube is q = 43.55 × 10–8 C = 4.355 × 10–7 C

a. Hence,  q  a
0 0

71 ELECTROSTATICS

EXAMPLE 9 12th PHYSICS MODULE - 1

A parallel plate capacitor of capacitance C is SOLUTION:
charged to a potential V. It is then connected to The flux entering the closed surface equals the
another uncharged capacitor having the same flux exiting the surface. The flux entering the left
capacitance. Find out the ratio of the energy
stored in the combined system to that stored side of the cone is E   E  dA  ERh . This is
initially in the single capacitor.
SOLUTION: the same as the flux that exits the right side of the
cone. Note that for a uniform field only the cross
sectional area matters, not shape.

UInitial = 1 CV2 0 1 CV2 EXAMPLE 12
2 2
An electron is moving round the nucleus of a
After connecting common potential,
hydrogen atom in a circular orbit of radius r. Find
C1V1  C2V2 V 
Vcommon  C1  C2  2 the coulomb force F between the two.

  1  V  2  1  V  2  1 CV2 (Where k  1 )
2 2 2 2 4 40
Ufinal C C

SOLUTION:

Ufinal 1 CV 2   k e2   e2 r  rˆ  r 
Uinitial 4 F r2 r3 r
  1 1: 2 rˆ k.
2
CV 2

EXAMPLE 10 EXAMPLE 13 y
– -q3
Total charge on a sphere of radius 10 cm is 1µC. Three charges –q1, )
Find themaximum electric field dueto the sphere. +q2 & –q3 are placed a
SOLUTION: as shown in figure. b
The electric field due to a charged sphere is Find the x-component –
maximum at its surface. Thus of the force on –q1. -q1 +x
SOLUTION: +q2

Emax = kq  9  109  10–6 = 9 × 105 N/C
R2 100  10–4
– –q3

EXAMPLE 11  F12 q2
––q1 F13
A cone with base radius R and height h is located  +
on a horizontal table.Ahorizontal uniform field E
penetrates the cone, as shown in Figure. x component of force on
Determine the electric flux that enters the left- q1 = F12 + F13 sin 
hand side of the cone.
 kq1q 2  kq1q3 sin  .
b2 a2

72 ELECTROSTATICS

EXAMPLE 14 12th PHYSICS MODULE - 1

Consider two hollow 4Q SOLUTION:
(a) Every object has the same volume,
concentric spheres, S1 and 2Q S2 V = 8 (0.030 0 m)3 = 2.16 × 10–4 m3.
S2, enclosing charges 2Q S1 Q = V = (400 × 10–9) × (2.16 × 10–4)
and 4Q respectively as = 8.64 × 10–11 C
shown in the figure (b) We must count the 9.00cm2 squares
painted with charge:
(i) Find out the ratio of the electric flux through (i) 6 × 4 = 24 squares
Q = A = (15.0 × 10–9)×24.0×(9.00×10–4)
them. (ii) How will the electric flux through the = 3.24 × 10–10 C
(ii) 34 squares exposed
scpohnsetraentS1r change if a medium of dielectric Q = A = (15.0 × 10–9 )×34.0×(9.00 × 10–4 )
is introduced in the space inside S1 in = 4.59 × 10–10 C
(iii) 34 squares
place of air ? Deduce the necessary expression. Q = A = (15.0 × 10–9) × 34.0× (9.00 × 10–4)
SOLUTION: = 4.59 × 10–10 C
(i) According to Gauss theorem, (iv) 32 squares
Q = A = (15.0 × 10–9) × 32.0× (9.00 × 10–4)
net  q  q ; s1  2Q  1 = 4.32 × 10–10 C
0r s2 2Q  4Q 3 (c) (i) Total edge length:  = 24 × (0.0300 m)
Q =   = (80.0 × 10–12) × 24 × (0.0300m)
(ii) If medium is filled in S1 then = 5.76 × 10–11 C
q (ii) Q =   = (80.0 × 10–12) × 44× (0.0300 )
s1  0r  2Q = 1.06 × 10–10 C
0r (iii) Q =   = (80.0 × 10–12) × 64× (0.0300)
= 1.54 × 10–10 C
EXAMPLE 15 (iv) Q =   = (80.0 × 10–12) × 40× (0.0300m)
= 0.960 × 10–10 C
Eight solid plastic cubes, each 3.00 cm on each
edge, are glued together to form each one of the EXAMPLE 16
objects (i, ii, iii, and iv) shown in Figure.
(a) Assuming each object carries charge with Figure shows the electric field lines for two point
uniformdensity400 nC/m3 throughoutits volume, charges separated by a small distance.
find the charge of each object. (b)Assuming each (a) Determine the ratio q1/q2. (b) What are the
object carries charge with uniform density signs of q1 and q2?
15.0nC/m2 everywhere on its exposed surface,
find the charge on each object. (c) Assuming
charge is placed only on the edges where
perpendicularsurfaces meet,withuniform density
80.0 pC/m, find the charge of each object.

q2
q1

SOLUTION:

(i) (ii) (iii) (iv) (a) q1  6   1
q2 18 3

(b) q1 is negative, q2 is positive

73 ELECTROSTATICS

12th PHYSICS MODULE - 1

EXAMPLE 17 y
C(–3,4) E B (4,4)
A charge of 10µC is kept at the origin of X–Y
coordinate system. Find the potential difference

in voltsbetween two points (a,0)&(a/ 2 ,a/ 2 ). A (4,1)

SOLUTION: xx
The distance of both these points from the origin
is a. Hence potential at these points will be equal SOLUTION:
and potential difference will be zero. Potential difference between Aand B is zero. A
&Bsame potential point (lineABis perpendicular
V= kq  kq  0 to electric field). E = 300 NC–1 ;
a a VC – VB = E (BC) = 300 × 7 = 2100 volt

EXAMPLE 18

Three charges –q, +q and +q are situated in X- EXAMPLE 21

Y plane at points (0, –a), (0, 0) and (0, a) In a region the electric field intensity E is given
byE = 100/x2 wherex in metre. Find the potential
respectively. Find the potential at a point distant difference between the points at x = 10 m and
r (r>a) in a direction making an angle fromY-axis. x = 20 m
SOLUTION: SOLUTION:
+q +

V= kq.2a cos   kq r
r2 r
+   100 xˆ
E x2
+q

= kq  2a cos   1 –q –
r r
 VB – VA = 20 E.da 20 100
=– 10 x2 dx
10
EXAMPLE 19
= + 100  1 1200 = 100  1 – 1  =– 5V
Two point charge of 3.2 × 10–19 C and x 20 10
–3.2 × 10–19 C are separated from each other
by 2.4 × 10–10 m. The dipole is situated in a |VB – VA| = 5V
uniform electric field of intensity 4 × 105 V/m.
Calculate potential energyof dipole ininitial state. EXAMPLE 22
SOLUTION:
Potential energyof the dipole in the electric field A charge of +10µC is given to a hollow metallic
E is U = – pE cos  sphere of radius 0.1m. Find the potential at the
I=n–eq(7u.i6li8br×iu1m0–1=4)0×(4U.00×=1–0p5E) (i) outer surface and (ii) centre of the sphere.
= – 3.07 × 10–23 J SOLUTION:
(i) Potential on the surface of the sphere,

EXAMPLE 20 V  1 . q
40 r

A uniform electric field E of 300 NC–1 is But q = + 10µC = 10–5 C, r = 0.1m

directed along PQ. A, B and C are three points   9 109 105 = 9 × 105 V
0.1
in the field having x and ycoordinates (in metres) V

as shown in figure. Calculate potential difference (ii) Potential at the centre of the sphere
= Potential on its surface = 9 × 105 V
between the points : (i)Aand B and (ii) B and C.

74 ELECTROSTATICS

12th PHYSICS MODULE - 1

EXAMPLE 23 EXAMPLE 25

A parallel plate capacitor with air between the An infinite number of capacitors of capacitance
plates has a capacitance of 8 pF. Calculate the C, 4C, 16C ........... are connected in series
capacitance if the distance between the plates is then what will be their resultant capacitance ?
reduced by half and the space between them is SOLUTION:
filled with a substance of dielectric constant. Let the equivalent capacitance of the combination
(r = 6) = Ceq
SOLUTION:
Capacity of parallel plate capacitor 1  1  1 1  ......
Ceq. C 4C 16C
r0A air r 0A  8 1012
C  d (For = 1). ; d 1  1 1 ...... 1
4 16 C
If d  d/2 and r  6 then new capacitance    (this is G.P. series)

C' = 6 × 0A = 12 0A = 12 × 8 pF = 96 pF [S = a r ; first term a = 1, common ratio r = 1/4]
d/2 d 1

EXAMPLE 24 1  1 1  1  Ceq = 3 C
Ceq  4 C 4
Two identical metallic blocks resting on a 1
frictionless horizontal surface are connected by
a light metallic spring having a spring constant k EXAMPLE 26
as shown in Figure a and an unstretched length
Li. A total charge Q is slowly placed on the Obtain the equivalent capacitance for the
system, causing the spring to stretch to an following network.
equilibrium length L, as shown in Figure b.
Determine the value of Q, assuming that all the 100pF
charge resides on the blocks and modeling the
blocks as point charges. C1

200pF 200pF

C2 +
C3 300V
–
C4

100pF

SOLUTION: For 300V supply, determine the charge and
Charge Q/2 resides on each block, which repel voltage across each capacitor.
as point charges: SOLUTION:
C2 and C3 in series with C1 in parallel

C5  C2C3  C1
C2  C3

F  ke (Q / 2) (Q / 2)  k (L  Li ) Let 200pF = C5 200 100
L2 300
 Cequivalent   200 pF ....... (1)
3
k (L  Li )  q = Ceq × V
Solving for Q, Q  2L ke
200
3  300  2  104 pC

75 ELECTROSTATICS

12th PHYSICS MODULE - 1

10µF

C5 = 200pF 300V AB

C

C4 = 100pF 10µF

Across C4, q = 2 × 104 pC ; SOLUTION:

 q  2 104  1  1  1  1  1  1  43
C4 100 20 C 15 C 15 20 60
V 200V
1 1
Across C5, V = 100V ; C  60 , C = 60µF

Across C1, V = 100V 104 pC EXAMPLE 29
q = CV = 100 × 100 =
Calculate the capacitance of the capacitor C in
 Now, across C2 and C3, V = 100V the figure.The equivalent capacitance of the
Across C2, V = 50V & C3, combination between P and Q is 30µF.
V = 50V [C2 = C3]
 Across C2 & C3 20µF
Q = CV = 200 × 50 = 103 pC

EXAMPLE 27

A charge of 10–9C is placed on each of the 64 P C R 20µF S Q

identical drops of radius 2cm. They are then 20µF
combined to form a biggerdrop.Find its potential.
SOLUTION: SOLUTION:
From total charge conservation charge on bigger
drop, Q = nq 1  1  1 ; 1  1  1  1
From mass conservation radius of bigger drop 30 C 60 C 30 60 60

4 R 3  n 4 r3 C = 60µF
3 3
R, ; R  n1/3r

Potential of bigger drop EXAMPLE 30

V= KQ  knq  n2/3v small drop Consider a closed triangular box resting within a
R n1/3r horizontal electric field of magnitude
E = 7.80 × 104 N/C as shown in Figure.
V  n2/3v Calculate the electric flux through (a) the vertical
rectangular surface, (b) the slanted surface, and
 V  (64)2/3  9  109  109 (c) the entire surface of the box.
 102
= 7.2  103 V (2 )

30 cm

E

EXAMPLE 28 10 cm 60°

Calculate the capacitance of the capacitor C in
the figure, if the equivalent capcacitance of the
combination between Aand B is 15µF.

76 ELECTROSTATICS

SOLUTION: 12th PHYSICS MODULE - 1
(a) A' = (10.0 cm) (30.0 cm)
A' = 300 cm2 = 0.0300 m2 EXAMPLE 32
(b) EEEE,,,,AAAA'''=====E(E–(77A2.A.8.83c'00oc4×os×ks1N100.4m4))2((/AC0.)03c0o0s)6c0o.s0°180° Four balls, each with mass m, are connected by
four nonconducting strings to form a square with
A = (30.0 cm) (w) = (30.0 cm)  c1o0s.060c.m0  side a, as shown in Figure. The assembly is
 placed on a horizontal nonconducting frictionless
surface. Balls 1 and 2 each have charge q, and
= 600 cm2 = 0.0600 m2 balls 3 and 4 are uncharged. Find the maximum
E,A (7.80 × 104) (0.0600) speed of balls 1 and 2 after the string connecting
= +2.34 kN.m2/C cos 60.0° them is cut.
=
12

(c) The bottom and the two triangular sides all
liEe,ptaortaalll=el–to2.E3,4so+2E.3=40+fo0r
each of these.
+0+0=0.

EXAMPLE 31

Two charges ±10 µC are placed 5.0 mm apart. 34
Determine the electric field at a point P on the
axis of the dipole 15 cm away from its centre O SOLUTION:
on the side of the positive charge. Take the illustration presented with the problem
as an initial picture.
A OB
P vv
– +
12
–10 µC +10µC + 3 CM 4 +
vv
SOLUTION:
The electric field at a distance r from the centre No external horizontal forces act on the set of
on the axis of the dipole has a magnitude four balls, so its center of mass stays fixed at the
location of the center of the square. As the
E  2p (r/a >> 1) charged balls 1 and 2 swing out and away from
40r3 each other, balls 3 and 4 move up with equal y-
components of velocity. The maximum kinetic
where p = 2aq is the magnitude of the dipole energy point is illustrated. System energy is
conserved:
moment. The direction of electric field on the di-

pole axis is always along the direction of the di-

pole moment vector (i.e., from –q to q). keq2 keq2
Here, p =10–5 C × 5 × 10–3 m = 5 × 10–8 C m a 3a

E  2  5 108Cm   1 mv2  1 mv2  1 mv2  1 mv2
(8.854 1012 C2N1m2 ) 2 2 2 2
4

 (15)3 1 m3 2keq2  2mv2  v  keq2
 106 3a 3am

= 2.6 × 105 N C–1.

77 ELECTROSTATICS

12th PHYSICS MODULE - 1

EXAMPLE 33 EXAMPLE 34

Consider two thin, conducting, spherical shells A variable air capacitor used in a radio tuning
as shown in Figure. The inner shell has a radius circuit is made of N semicircular plates each of
r1 = 15.0 cm and a charge of 10.0 nC. The outer radius R and positioned a distance d from its
shell has a radius r2 = 30.0 cm and a charge of neighbors, to which it is electrically connected.
–15.0 nC. Find (a) the electric field E and As shown in Figure, a second identical set of
(b) the electric potential V in regions A, B, and plates is enmeshed with its plates halfway
C, with V = 0 at r = . between those of the first set. The second set
can rotate as a unit. Determine the capacitance
r1 as a function of the angle of rotation , where
A = 0 corresponds to the maximum capacitance.

r2

B

C d

SOLUTION: R
(a) From Gauss’s law,

EA = 0 (no charge within)

EB  ke qA   89.9  V/m SOLUTION:
r2 r2 With  = , the plates are out of mesh and the
overlap area is zero. With  = 0, the overlap area
EC  ke (qA  qB)    45.0  V / m is that of a semi-circle, R2/2. By proportion,
r r2 the effective area of a single sheet of charge is

(b) VC  ke (qA  qB) d
r

 (8.99  109 ) (5.00  109 )    45r.0  V (  ) R2 .
r 2

At r2, V   45.0  150V When there are two plates in each comb, the
0.300 number of adjoining sheets of positive and
negative charge is 3, as shown in the sketch.
Inside r2, r 89.9 When there are N plates on each comb, the
VB  150V  r2 r2 dr number of parallel capacitors is 2N – 1 and the
total capacitance is
 150  89.9  1  0.3100 
r 0Aeffective
C = (2N – 1) distance
  450  89r.9 V
 (2N  1) 0 (  ) R2 / 2
d/2
At r1, V  450  89.9  150 V
0.150  1) 0 (  ) R2
 (2N d
VA = +150V

78 ELECTROSTATICS

EXAMPLE 35 12th PHYSICS MODULE - 1

The circuit in Figure consists of two identical EXAMPLE 36
parallel metal plates connected byidentical metal
springs to a 100-V battery. With the switch open, eFaocuhropfaarraelale7l.5m0ecmtal2,palraetesespPa1ra, tPed2,sPuc3c, easnsdivPel4y,
the plates are uncharged, are separated by a by a distance d = 1.19 mm, as shown in Figure.
distance d = 8.00 mm, and have a capacitance P1 is connected to the negative terminal of a
C = 2.00 µF. When the switch is closed, the battery, and P2 to the positive terminal. The
distance between the plates decreases bya factor batterymaintains a potential difference of 12.0V.
of 0.500. (a) How much charge collects on each (a) If P3 is connected to the negative terminal,
plate and (b) what is the spring constant for each what is the capacitance of the three-plate system
spring? P1P2P3? (b) What is the charge on P2? (c) If P4
is now connected to the positive terminal of the
d battery, what is the capacitance of the four plate
system P1P2P3P4? (d) What is the charge on P4?
kk

P1 P2 P3 P4

12.0 V

S

SOLUTION: ddd
With switch closed, distance d' = 0.500d and

capacitance C  0A  20A  2C SOLUTION:
d d Each face of P2 carries charge, so the three-plate
(a) Q = C' (V) = 2C (V) system is equivalent to

= 2 (2.00 × 10–6 F) (100 V) = 400 µC P1 P2 P2
P3
(b) The force stretching out one spring is

F Q2  4C2 (V)2  2C2 (V)2 Each capacitor by itself has capacitance
20A 20A (0A / d) d
 K0A  1 (8.85 1012 C2 ) 7.5 104 m2
 2C (V)2 C d N.m2 1.19 103 m
d
= 5.58 pF
One spring stretches by distance x = d/4,
Then equivalent capacitance

k  F  2C (V)2  4   8C (V)2 = 5.58 + 5.58 = 11.2 pF .
x d d d2 (b) Q = CV + CV = 11.2 × 10–12 F (12V)
= 134 pC

 8 (2.00 106 F) (100V)2 (c) Now P3 has charge on two surfaces and in
(8.00 103m)2 effect three capacitors are in parallel:

C = 3 (5.58 pF) = 16.7 pF .

= 2.50 kN/m (d) QOn=lyConeVfa=ce5o.5f8P4×c1ar0r–ie12s charge:
F (12 V)

= 66.9 pC

79 ELECTROSTATICS

12th PHYSICS MODULE - 1

EXERCISE-1 (LEVEL-1)

SECTION - 1 (VOCABULARY BUILDER)

Choose one correct response for each question. Q.4 Column I Column II
Q.5
For Q.1-Q.6 : Match the column I with column II. (a) Charged particle at (i) Produces both

Q.1 Column I Column II rest electric and

(a) If a glass rod is rubbed (i) positive charge magnetic field

with a silk cloth, glass (b) Charged particle in (ii) Produces only

rod acquires unaccelerated motion electric field

(b) In case (a) silk cloth (ii) negative charge (c) Charged particle in (iii) Produces

acquires accelerated motion electric,

(c) If an ebonite rod is rubbed magnetic field

with fur, ebonite rod acquires- Code and radiates.

(d) In case (b) fur acquires (A) a-i, b-ii, c-iii (B) a-ii, b-i, c-iii

Code (C) a-iii, b-ii, c-i (D) a-ii, b-iii, c-i

(A) a-i, b-ii, c-i, d-ii (B) a-ii, b-i, c-i, d-ii Column I Column II

(C) a-i, b-ii, c-ii, d-i (D)a-ii, b-ii, c-i,d-ii (a) Force experienced by (i) ElectricIntensity

Q.2 Column I Column II a unit positive point

(a) Positive and negative (i) Coulomb charge

charges (b) Locus of all points (ii) Electric potential

(b) Force between two (ii) Benjamin having same potential

charged bodies Franklin (c) Scalar function whose (iii) Equipotential

(c) Electric field lines (iii) Michael negative gradient surface

Faraday gives intensity.

(d) Electrostatic machine (iv) Van de Graaff Code

which produces large (A) a-i, b-iii, c-ii (B) a-ii, b-i, c-iii

electrostatic potential difference (C) a-i, b-ii, c-iii (D) a-ii, b-iii, c-i

Code Q.6 Column I Column II

(A) a-i, b-ii, c-iii, d-iv (B) a-ii, b-i,c-iii,d-iv (a) Electric Force between (i) inversely

(C) a-iii, b-iv, c-i, d-ii (D)a-iv,b-iii,c-ii,d-i two point charges proportional

Q.3 Column I Column II(Number) to distance

(a) Uncharged body (i) Electrons = Protons (b) Electric field due to a (ii) inversely

(b) Positively charged (ii) Electrons > protons point charge proportional to

body square of

(c) Negatively charged (iii)Electrons < protons distance

body (c) Electric potential due

Code to a point charge

(A) a-i, b-ii, c-iii (B) a-i, b-iii, c-ii (d) Electric potential energy

(C) a-iii, b-ii, c-i (D) a-ii, b-iii, c-i between two point charge

(A) a-i, b-ii, c-i, d-ii (B) a-ii, b-i, c-i, d-ii

(C) a-i, b-ii, c-ii, d-i (D) a-ii, b-ii, c-i, d-i

80 ELECTROSTATICS

12th PHYSICS MODULE - 1

SECTION - 2 (BASIC CONCEPTS BUILDER)

For Q.7 to Q.25 :
Choose one word for the given statement from the list.

True, False, Zero, 8, 9, Finite negative, Inside, Outside, Increases, Remains the same,
Repulsion, Attraction, Expands, Contracts, Metal, Plastic, Normal, Series, Parallel, Same,
Double.

Q.7 A sure test of electrification is _______. Q.18 27 drops of same size and having equal and

similar charge coalesce to form bigger drop,ratio

Q.8 Four charges +q, +q, -q and –q respectively are of potential of bigger drop to the smaller drop is

placed at the corners A, B, C and D of a square _______.

of side a. The electric potential at the centre O Q.19 When a test charge is brought from infinityalong

of the square is _______. the perpendicular bisector of the electric dipole

Q.9 When a soap bubble is charged, it _______. the work done is _______.

Q.10 Lightning rods are made of _______. Q.20 In case of parallel plate capacitor capacity is
Q.11 Q.21 directly proportional to the area of the plate.
Electric lines of forces are _______ to the surface
Q.12 of conductor. (True /False)

A dipole is placed in a uniform electric field with A parallel plate capacitor is connected to a
its axis parallel to the field. It experiences only a battery. If the dielectric slab of thickness equal
torque.(True / False) to half the plate separation is inserted between
the plates, capacitance of the capacitor _______.

Q.13 The work done in moving 500 µC charge Q.22 Capacitors are combined in _______(series/
parallel)when we require a big amount of charge
between two points on equipotential surface is
at a small potential.
_______.

Q.14 Dipole moment is a scalar quantity.(True/False) Q.23 _______(Series/Parallel) combinations are used
Q.15 Q.24 when a high voltage is to be divided on several
A hollow metal ball carrying an electric charge Q.25 capacitors.
Q.16 produces no electric field at points _______ the
Q.17 sphere. If a capacitor is charged by a battery ,then work
done by the battery is _______(same/double)the
At a point on the axis of an electric dipole the energy stored in the capacitor.
electric field is zero.(True /False)
A group of identical capacitors is connected first
The electric field due to uniformlycharged sphere in series and then in parallel.The capacity of
is maximum at the surface.(True /false). parallel combination is 100 times larger than for
the series combination.Number of capacitors in
the group are _______.

81 ELECTROSTATICS

12th PHYSICS MODULE - 1

SECTION - 3 (ENHANCE PROBLEM SOLVING SKILLS)

Choose one correct response for each question. Q.31 If two bodies are rubbed and these pairs are
PART 1 : ELECTRIC CHARGE I. Glass  Silk
Q.32 II. Wool  Ebonite or Plastic
Q.26 When a glass rod is rubbed with silk, the rod Q.33 III. Ebonite  Polythene
acquires one kind of charge and silk acquires Q.34 IV. Dry hair  Comb
the second kind of charge. Now, if the electrified Q.35 Then charges appearing on first member and
rod is brought in contact with silk, with which it second member of list are respectively
was rubbed, they Q.36 (A) positive, positive (B) positive, negative
(A) no longer attract each other. (C) negative, negative (D) negative, positive
(B) they attract other light objects.
(C) they repel other light objects. If a charge on the body is 1 nC, then how many
(D) attract each other very strongly and stick
together. electrons are present on the body?
(A) 1.6 × 1019 (B) 6.25 × 109
(C) 6.25 × 1027 (D) 6.25 × 1028

Q.27 A glass rod rubbed with silk is used to charge a When 1014 electrons are removed from a neutral
gold leaf electroscope and the leaves are
observed to diverge. The electroscope thus metal sphere, the charge on the sphere becomes
charged is exposed to X-rays for a short period.
Then, (A) 16µC (B) – 16µC
(A) the divergenceof leaves will not be affected.
(B) the leaves will diverge further. (C) 32µC (D) – 32µC
(C) the leaves will collapse
(D) the leaves will melt. A conductor has 14.4 1019 coulombs
positive charge. The conductor has
Q.28 The minimum charge on an object is
(Charge on electron  1.6 1019 coulombs )
(A) 1 C (B) 1 stat C (A) 9 electrons in excess
(C) 1.6 × 10–19 C (D) 3.2 × 10–19 C (B) 27 electrons in short
(C) 27 electrons in excess
(D) 9 electrons in short

Q.29 Which of these are properties of charge? The electric charge in uniform motion produces
(A)An electric field only
I. Charges are additive in nature. (B)Amagnetic field only
(C) Both electric and magnetic field
II. Charges are conservative in nature. (D) Neither electric nor magnetic field

III. Charges are quantised in nature.

IV. Charges can be transformed from one type

to another.

(A) I, II and III (B) I, II and IV

(C) 1, III and IV (D) II, III and IV

Q.30 Two bodies are rubbed and one of them is PART 2 : COULOMB'S LAW

negatively charged. For this body, if mi = initial The figure below shows the forces that three
mass, mf = mass after charging, then charged particles exert on each other. Which of
(A) mi = mf (B) mi < mf the four situations shown can be correct.
(C) mi > mf (D) mi + mf = 2mf

82 ELECTROSTATICS

12th PHYSICS MODULE - 1

(I) (II) Q.40 A force F acts between sodium and chlorine ions
of salt (sodium chloride) when put 1cm apart in
Q.41 air. The permittivity of air and dielectric constant
Q.42 of water are 0 and K respectively. When a piece
Q.43 of salt is put in water, electrical force acting
Q.44 between sodium and chlorine ions 1cm apart is
Q.45
(III) (IV) (A) F FK
K (B) 0
(A) all of the above (B) none of the above
(C) II, III (D) II, III & IV F (D) F0
(C) K0 K

Q.37 Given are four arrangements of three fixed A charge Q is divided into two parts of q and

Q.38 electric charges. In each arrangement, a point Q – q . If the coulomb repulsion between them
Q.39
labeled P is also identified –test charge, +q, is when they are separated is to be maximum, the

placed at point P. All of the charges are of the ratio of Q/q should be

same magnitude, Q, but they can be either (A) 2 (B) 1/2

positive or negative as indicated. The charges (C) 4 (D) 1/4

and point P all lie on a straight line. The distances

between adjacent charges, either between two The force between two charges 0.06m apart is

charges or between a charge and point P, are all 5N. If each charge is moved towards the other

the same. by 0.01m, then the force between them will

I. +++ II. + + P – become

P (A) 7.20 N (B) 11.25 N

III. + + – IV. + – + (C) 22.50 N (D) 45.00 N

P P
Correct order of choices in a decreasing order

of magnitude of force on P is – Two point charges placed at a certain distance r
in air exert a force F on each other. Then the
(A) II > I > III > IV (B) I > II > III > IV distance r' at which these charges will exert the
same force in a medium of dielectric constant K
(C) II > I > IV > III (D) III > IV > I > II is given by
(A) r (B) r/K
Two electrons are a certain distance apart from

one another. What is the order of magnitude of

the ratio of the electric force between them to (C) r / K (D) r K

the gravitational force between them?
(A) 108 : 1 (B) 1028 : 1
(C) 1031 : 1 (D) 1042 : 1 The charges on two sphere are +7C and – 5C
respectively. They experience a force F. If each
Nucleus 92U238 emits -particle (2He4). - of them is given an additional charge of – 2C,
particle has atomic number 2 and mass number the new force of attraction will be
4. At any instant -particle is at distance of
9 × 10–15 m from the centre of nucleus of (A) F (B) F / 2

uranium. What is the force on -particle at this (C) F / 3 (D) 2F

instant ? 92U238 2He4 + 90Th234 When two charges are placed at a distance apart.

(A) 512 N (B) 412 N Findthemagnitudeofthird chargewhich is placed

(C) 325 N (D) 612 N at mid point of the line joining the charge, so that

system is in equilibrium -

(A) –Q/4 (B) –Q/2

(C) –Q/3 (D) – Q1

83 ELECTROSTATICS

PART 3 : ELECTRIC FIELD 12th PHYSICS MODULE - 1

What would be the direction of the resulting
electric field at the center point P

Q.46 The field of an electric field is a cosine function in (A) (B)
Q.47 (C) (D)
Q.48 xy-plane as shown in the diagram, then the

Q.49 representation of electric feld can be

(A) E(x, y)  ˆi  sin (x)ˆj y

(B)   ˆi  cos (x)ˆj (0,0) x Q.50 If the nucleus of a hydrogen atom is considered
E(x, y) Q.51 to be a sphere of radius 10–15 m, then the electric
Q.52
(C)   ˆi  sin (x)ˆj field on its surface will be
E(x, y) Q.53
 Q.54
(D) E (x, y)  ˆi  cos (x)ˆj (A) 14.4 V/m (B) 14.4 × 1011 V/m
(C) 14.4 × 1015 V/m (D) 14.4 × 1020 V/m

Consider a regular + + The intensity of electric field due to a uniformly

cube with positive + +

point charge +Q in all P charged non-conducting sphere at a distance x

corners except for one + + from the centre inside the sphere is proportional

which has a negative – + to –

point charge –Q. Let the distance from anycorner (A) 1/x (B) x
(C) 1/x2 (D) x2
to the center of the cube be r. What is the

magnitude of electric field at point P, the center A charged particle of mass m and charge q is
released in an electric field of magnitude E. Its
of the cube? (B) E = 1keQ/r2 kinetic energy after time t will be.
27kkeeQQ//rr22 (D) E = 6keQ/r
(A) E =
(C) E =

In an ink-jet printer, an ink droplet of mass m is (A) 2E2 t  (B) E2q2t
given a negative charge q by a computer mq 2m
-controlled charging unit, and then enters at speed
v theregion between two deflectingparallel plates (C) Eq2m (D) Eqm
of length Lseparated by distance d (see figure). 2t2 2t

vE qd

L An electron and a proton are in a uniform electric
field, the ratio of their accelerations will be –
All over this region exists a downward electric (A) Zero
field which you can assume to be uniform. (B) Unity
Neglecting the gravitational force on the droplet, (C) The ratio of the masses of proton & electron
the maximum charge that it can be given so that (D) The ratio of the masses of electron and proton
it will not hit a plate is most closelyapproximated
by If an insulated non-conducting sphere of radius
R has charge density . The electric field at a
(A) mv2E (B) mv2d distance r from the centre of sphere (r < R) will
dL2 EL2

(C) 2dmv2 (D) none R r
EL2 (A) 30 (B) 0
–2Q + +Q
– P
Four electrical charges are r
10cm (C) 30 (D) 3 R
arranged on the corners of a 0
+2Q + 10cm – –Q
10cm square as shown.

84 ELECTROSTATICS

Q.55 The electric potential inside a conducting sphere (A) constant 12th PHYSICS MODULE - 1
Q.56 (A) Increases from centre to surface (C) positive
Q.57 (B) Decreases from centre to surface (B) zero
(C) Remains constant from centre to surface (D) negative
(D) Is zero at every point inside
Q.61 The figure shows the electric field lines between
two parallel plates that for all practical purposes
The number of electrons to be put on a spherical extend an infinite distance both to the right and
to the left and into and out of the paper. Four
conductor of radius 0.1m to produce an electric point P, Q, R and S are marked in this figure. At
which point is the electric potential the largest
field of 0.036 N/C just above its surface is
(A) 2.7 × 105 (B) 2.6 × 105
(C) 2.5 × 105 (D) 2.4 × 105 + + + + ++ ++ ++
P• Q• R•
An electron falls a distance of 4 cm in a uniform S•
electric field of magnitude 5 x 104 N/C. The time

taken by electron in falling will be-
(A) 2.99 × 10–7 s (B) 2.99 × 10–8s (A) P (B) Q
(C) 2.99 × 10–9 s (D) 2.99 × 10–10 s (C) R (D) S

Q.58 A sphere of radius 5 cm has electric field Q.62 If a = 30 cm , b = 20 cm, q = + 2.0 nC, and
Q.59 5 × 106 V/m on its surface. What will be the Q.63
force acting on a charge of 5 × 10–8C placed at Q = – 3.0 nC in the figure, what is the potential

distance of 20cm from the centre of sphere- difference VA– VB ? a –
(A) 1.5 × 10–2 N
(B) 40 N ab B Q

(C) 4 N (D) 0 N +

qA

(A) + 60 V (B) + 72 V

In a uniformly charged sphere of total charge Q (C) + 84 V (D) + 96 V
and radius R, the electric field E is plotted as
function of distance from the centre. The graph A uniform electric field points in the positive x
which would correspond to the above will be : direction, as shown.Along the two lines f1, f2 ,
we plot the electric potentials as a function of
EE distance. Choose the correct plot.

(A) (B) y f2

E Rr E Rr

(C) (D) E

Rr Rr i f1
x

1,2potential 2potential

PART 4 :ELECTRIC POTENTIAL (A) (B) 1

AND EQUIPOTENTIAL distance distance
SURFACE
potential potential
Q.60 In a certain region of space, the electric field is (C) 1 2 (D) 1,2
zero. From this we can conclude that the electric
potential in this region is – distance distance

85 ELECTROSTATICS

Q.64 A number of spherical shells of different radii are (A) –80 V 12th PHYSICS MODULE - 1
Q.65 uniformlycharged to same potential. The surface (C) 120 V
Q.66 charge density of each shell is related with its (B) 80 V
radius as (D) –120 V
Q.67
(A )   1 (B)   1 PART 5 : ELECTRIC POTENTIAL
R2 R ENERGY

(C)   R (D)  is same for all Q.70 An electron at a potential of – 10 kV moves to a
point where its potential is – 1 kV. Its potential
Two conducting, concentric, hollow spheres A energy has –
(A) decreased
and B have radii a and b respectively, with A (B) increased
(C) not changed
inside B. They have the same potential V. A is (D) one needs to know the distance b/w the
points to say
now given some charge such that its potential

becomes zero. The potential of B will now be–

(A) 0 (B) V (1–a/b)

(C) Va/b (D) V (b – a) (b + a) Q.71 A charge (–q) and another charge (+Q) are kept
at two points Aand B respectively. Keeping the
A hollow conducting sphere of radius R has a charge (+Q) fixed at B, the charge (–q) at A is
charge (+Q) on its surface. What is the electric moved to another point C such that ABC forms
potential within the sphere at a distance r = R/3 an equilateral triangle of side . The net work
from its centre – done in moving the charge (–q) is –

(A) Zero 1Q 1 Qq 1 Qq
(B) 40 r (A) 40  (B) 40 2

1Q (D) 1 Q 1 Qq
(C) 40 R 40 r2 40
(C) (D) Zero

A spherical conductor of radius 2m is charged

to a potential of 120 V. It is now placed inside Q.72 On moving a charge of 20 coulombs by 2 cm,

another hollow spherical conductor of radius 6m. 2J of work is done, then the potential difference

Calculate the potential to which the bigger sphere between the points is –

would be raised – (A) 0.1 V (B) 8 V

(A) 20 V (B) 60 V (C) 2 V (D) 0.5 V

(C) 80 V (D) 40 V

Q.68 125 identical drops each charged to the same PART 6 : ELECTRIC FIELD LINES
Q.69
potential of 50 volts are combined to form a Q.73 Electric lines of force about negative point charge
are
single drop. The potential of the new drop will (A) Circular, anticlockwise
(B) Circular, clockwise
(A) 50V (B) 250V (C) Radial, inward
(D) Radial, outward
(C) 500V (D) 1250V

Assume that an electric field   30x 2 ˆi exists in
E

space. Then the potential difference VA – VO,
where VO is the potential at the origin and VA
the potential at x = 2 m is –

86 ELECTROSTATICS

12th PHYSICS MODULE - 1

Q.74 A long cylindrical shell carries positive surface Q.77 A uniform electric field   aˆi  bˆj , intersects a
charge  in the upper half and negative surface E
charge – in the lower half. Theelectric field lines
around the cylinder will look like figure given in surface of areaA. What is the flux through this
(figures are schematic and not drawn to scale)
area if the surface lies in the yz plane?

(A) a A (B) 0

(C) b A (D) A a2  b2

(A) +––+–++– (B) +––+–++– Q.78 A uniform electric field E = 2 × 103 NC–1 is

acting along the positive x-axis. The flux of this

field through a square of 10 cm on a side whose

plane is parallel to the yz plane is
(A) 20 NC–1 m2 (B) 30 NC–1 m2
(C) 10 NC–1 m2 (D) 40 NC–1 m2
+––+–++– +––+–++–
(C) (D) PART 8 : GAUSS'S LAW

Q.79 Iftheelectricflux enteringandleavingan enclosed
Q.75 Abhishek, Hritik, John, and Amir are assigned Q.80 surface respectively is 1 and 2, the electric
the tasks of moving equal positive charges slowly
through an electric field, along assigned path charge inside the surface will be –
(shown as dotted line). In each case the charge ((11 ++22))/00 ((22 ––11))/00
is at rest at the beginning. They all have paths of (A) (B)
exactly equal lengths. Who must do the most (C) (D)
positive work?
Which of the following statements is not true
Hritik John Amir about Gauss's law?
(A) Gauss's law is true for any closed surface.
Abhishek (B) The term q on the right side of Gauss's law
Electric field lines
includes the sum of all charges enclosed by
(A) Abhishek (B) Hritik the surface.
(C)Amir (D) John (C) Gauss's lawis not much useful in calculating
electrostatic field when the system has some
symmetry.
(D) Gauss's law is based on the inverse square
dependence on distance contained in the
coulomb's law.

PART 7 : ELECTRIC FLUX Q.81 A 5.0 µC point charge is placed at the center of
a cube. The electric flux in N-m2 /C through one
Electric Flux is a measure of
Q.76 (A) the rate at which moving electric charges side of the cube is approximately :
(B) 7.1 × 104
are crossing an area. (A) 0 (D) 1.4 × 105
(B) the number of electric field lines passing (C) 9.4 × 104

through an area. PART 9 : ELECTRIC DIPOLE
(C) the surface densityof electric charge spread
Q.82 An electric dipole is placed in an electric field
along the area. generated by a point charge
(D) the rate at which electric field lines are (A) The net force on the dipole must be zero.
(B) The net force on the dipole may be zero.
spreading out in space and moves further
and further away from electric charges.

87 ELECTROSTATICS

12th PHYSICS MODULE - 1

(C) The torque on the dipole due to the field Q.88 If Ea be the electric field strength of a short dipole
must be zero. at a point on its axial line and Ee that on the
equatorial line at the same distance, then
(D) The torque on the dipole due to the field
may be zero. (A) Ee = 2Ea (B) Ea = 2Ee
(C) Ea = Ee (D) None of the above
Q.83 An electric dipole of moment p is kept along an
Q.84
Q.85 electric field E. The work done by external agent PART 10 : CONDUCTORS,
Q.86
Q.87 in rotating it from stable equilibrium position by DIELECTRICS &
an angle , is POLARISATION
(A) pE sin  (B) pE cos 
(C) pE (1– sin) (D) pE (1 – cos )

An electric dipole of moment p is placed in the Q.89 Which of the following statements is false for a
perfect conductor?
position of stable equilibrium in uniform electric (A) The surface of the conductor is an

field of intensity E . It is rotated through an angle equipotential surface.
 from the initial position. The potential energy (B) The electric field just outside the surface of
of electric dipole in the final position is
(A) pE cos  (B) pE sin  a conductor is perpendicular to the surface.
(C) pE (1 – cos ) (D – pE cos  (C) The charge carried bya conductor is always

An electric dipole is kept in non-uniform electric uniformlydistributed over the surface of the
field. It experiences – conductor.
(A) A force and a torque (D) None of these
(B) A force but not a torque
(C) A torque but not a force Q.90 I. The molecules of a substance may be polar
(D) Neither a force nor a torque or non-polar.

An electric dipole consisting of two opposite II. In a non-polar molecule, the centres of
charges of 2 × 10–6C each separated by a positive and negative charge coincide.

distance of 3cm is placed in an electric field of III. The non-polar molecule has no permanent
2 × 105 N/C. The maximum torque on the dipole (or intrinsic) dipole moment.

will be (A) I, II are correct, III may be correct.
(A) 12 × 10–1 Nm (B) I and III are correct, II may be correct.
(C) 24 × 10–1 Nm (C) II and III are correct, I is incorrect.
(D) I, II and II are correct.

(B) 12 × 10–3 Nm Q.91 Dielectric constant for a metal is
(D) 24 × 10–1 Nm
(A) zero (B) infinite

Two opposite and equal charges 4 × 10–8C (C) 1 (D) 10
when placed 2 × 10–2cm away, form a dipole. If
this dipole is placed in an external electric field Q.92 For linear isotropic dielectric, the polarisation is
4 × 108N/C, the value of maximum torque and 2eEeE –e/EeE
the work done in rotating it through 180° will be (A) P = (B) P =
(A) 64 × 10–4 Nm and 64 × 10–4J (C) P = (D) P =
(B) 32 × 10–4 Nm and 32 × 10–4J
(C) 64 × 10–4 Nm and 32 × 10–4J Q.93 Whichamongthefollowingis anexampleofpolar
(D) 32 × 10–4 Nm and 64 × 10–4J
molecule?

(A) O2 (B) H2
(C) N2 (D) HCl

88 ELECTROSTATICS

12th PHYSICS MODULE - 1

Q.99 A parallel plate capacitor has a capacity C. The

PART 11 : CAPACITORS AND separation between the plates is doubled and a
CAPACITANCE
dielectric medium is introduced between the

plates. If the capacity now becomes 2C, the

dielectric constant of the medium is

Q.94 Two insulated conductors are charged by (A) 2 (B) 1

transferring from one conductor to another. The (C) 4 (D) 8

potential difference of 100V was produced on Q.100 A parallel plate condenser with oil between the
transferring 6.25 × 1015 electrons from one plates (dielectric constant of oil K = 2) has a
capacitance C. If the oil is removed, then
conductor to another. Find the capacitor of the capacitance of the capacitor becomes

system.

(A) 5 µF (B) 10 µF (A) 2C (B) 2C

(C) 15 µF (D) 20 µF

Q.95 The potentials of the two plates of capacitor are (C) C / 2 (D) C/2

+10V and –10 V. The charge on one of the plates Q.101 When a dielectric material is introduced between
the plates of a charged condenser, then electric
is 40 C. The capacitance of the capacitor is field between the plates
(A) Remain constant
(A) 2 F (B) 4 F (B) Decreases
(C) Increases
(C) 0.5 F (D) 0.25 F (D) First increases then decreases

Q.96 One plate of parallel plate capacitor is smaller Q.102 A sheet of aluminium foil of negligible thickness
than other, then charge on smaller plate will be
(A) Less than other is introduced between the plates of a capacitor.
(B) More than other
(C) Equal to other The capacitance of the capacitor –
(D) Will depend upon the medium between them
(A) Remains unchanged (B) Becomes infinite

PART 12 :PARALLEL PLATE (C) Increases (D) Decreases
CAPACITOR
PART 13 :COMBINATION OF
Q.97 The capacity of a parallel plate condenser is C. CAPACITORS

Its capacity when the separation between the Q.103 In four options below, all the four circuits are
arranged in order of equivalent capacitance. Se-
plates is halved will be lect the correct order. Assume all capacitors are
of equal capacitance.
(A) 4C (B) 2C

(C) C/2 (D) C/4

Q.98 Force of attraction between the plates of a parallel
plate capacitor is

q2 q2 (1) (2)
(A) 20AK (B) 0AK

q q2 (3) (4)
(C) 20A (D) 20A2K (A) C1 > C2 > C3 > C4 (B) C1 > C3 > C2 > C4
(C) C1< C2 < C3 < C4 (D) C1 < C3 < C2 < C4

89 ELECTROSTATICS

12th PHYSICS MODULE - 1

Q.104 In the network shown we have three identical Q.108 The capacitor of capacitance 4µFand 6µFare
capacitors. Each of them can withstand a
maximum 100 V p.d. What maximum voltage connected in series. A potential difference of
can be applied across A and B so that no
capacitor gets spoiled? 500V applied to the outer plates of the two

capacitor system. Then the charge on each

capacitor is numerically

C (A) 6000 C (B) 1200 C

(C) 1200 µC (D) 6000 µC

C Q.109 Two capacitances of capacity C1 and C2 are
connected in series and potential difference V is
A B
C

(A) 150 V (B) 120 V applied across it. Then the potential difference
(C) 180 V (D) 200 V
across C1 will be

Q.105 Two capacitors C1 and C2 are connected in (A) V C2 (B) V C1  C2
series, assume that C1< C2. The equivalent C1 C1
capacitance of this arrangement is C, where
(C) V C2 (D) V C1
(A) C < C1/2 (B) C < C2/2 C1  C2 C1  C2
(C) C1 < C < C2 (D) C2 < C < 2C2

Q.106 The equivalent capacitance betweenAand B in Q.110 Four capacitors of each of capacity 3µFare
the figure is 1µF. Then the value of capacitance connected as shown in the adjoining figure. The
C is ratio of equivalent capacitance betweenAand B
and between Aand C will be

AC AB
2.5F
1F

(A) 1.4µF B (A) 4 : 3 C
(C) 3.5µF (C) 2 : 3
(B) 2.5µF (B) 3 : 4
(D) 1.2µF (D) 3 : 2

Q.107 Between the plates of a parallel plate condenser, Q.111 In the circuit shown in the figure, the potential
a plate of thickness t1 and dielectric constant k1 difference across the 4.5F capacitor is
is placed. In the rest of the space, there is another
plate of thickness t1 and dielectric constant k2. 3F
The potential difference across the condenser will 6F

Q  t1 t2  0Q  t1 t2  4.5F
A0  k1 k2  A  k1 k2 
(A)  (B) 

(C) Q  k1  k2  (D) 0Q (k1t1  k 2 t 2 ) 12V
A0  t1 t2  A
(A) 8/3 volts (B) 4 volts
(C) 6 volts (D) 8 volts

90 ELECTROSTATICS

12th PHYSICS MODULE - 1

Q.112 Two condensers C1and C2 in a circuit are joined Q.117 A12pF capacitor is connected to a 50V battery.
as shown in figure. The potential of pointAis V1
and that of B is V2. The potential of point D will How much electrostatic energy is stored in the

A DB capacitor (B) 2.5 × 10–7J
be V1 C1 (A) 1.5 × 10–8J (D) 4.5 × 10–2J
C2 V2 (C) 3.5 × 10–5J

(A) 1 (V1  V2 ) (B) C2V1  C1V2 Q.118 Four condensers each of capacity 4µF are
2 C1  C2 connected as shown in figure. VP – VQ = 15volts.
The energy stored in the system is
(C) C1V1  C2V2 (D) C2V1  C1V2
C1  C2 C1  C2 4F

Q.113 For the circuit shown in the figure, the charge on 4 F 4 F
4F capacitor is : P Q

4F

(A) 2400 ergs (B) 1800 ergs
(C) 3600 ergs (D) 5400 ergs

(A) 30 C +– Q.119 If n capacitor connected in series with a cell of
(C) 24 C emf V volt. The energy of system is –
(B) 4010CV
(D) 54 C (A) 1 n CV2 (B) 1 CV2
2 2 n
Q.114 A parallel plate capacitor is made by stacking n

equally spaced plates connected alternatively. If 1 CV2
2
the capacitance between any two adjacent plates (C) (D) none of above

is 'C' then the resultant capacitance is –

(A) (n – 1) C (B) (n + 1) C Q.120 The work done in placing a charge of 8 × 10–18

(C) C (D) nC coulomb on a condenser of capacity 100 micro-

farad is – (B) 4 × 10–10 J
(A) 3.1 × 10–26 J
PART 14 :ENERGY STORED (C) 32 × 10–32 J (D) 16 × 10–32 J
IN A CAPACITOR

Q.115 The energy stored in a condenser of capacity C PART 15 :VAN DE GRAAFF
which has been raised to a potential V is given GENERATOR

(A) 1 CV (B) 1 CV2 Q.121 Van de Graaff generator is used to –
2 2 (A) store electrical energy
(B) build up high voltages of few million volts.
(C) CV (D) 1 (C) decelerate charged particle like electrons
2VC (D) Both (A) and (B) are correct.

Q.116 A capacitor 4µFcharged to 50V is connected to

another capacitor of 2µF charged to 100V with Q.122 Van de Graaff generator is a machine capable of

plates of like charges connected together. The building up potential difference of a few million

total energy before and after connection in volts and fields close to the breakdown field of
multiples of (10–2 J) is
air which is about –
(A) 1.5 and 1.33 (B) 1.33 and 1.5 (A) 3 × 105 V/m (B) 3 × 106 V/m

(C) 3.0 and 2.67 (D) 2.67 and 3.0 (C) 3 × 10–6 V/m (D) 3 × 104 V/m

91 ELECTROSTATICS

12th PHYSICS MODULE - 1

EXERCISE-2 (LEVEL-2)

Choose one correct response for each question. Q.6 A spherical drop of mercury having an electric

Q.1 Number of electrons in 10 g of water is potential of 2.5 V is obtained as a result of
(A) 3.344 × 1024 (B) 5.35 × 105
(C) 3.344 × 1023 (D) 5.35 × 1023 merging 125 identical spherical droplets. The

electric potential of each of the original small

Q.2 Three point charges 3nC, 6nC and 9nC are droplets is

placed at the corners of an equilateral triangle of (A) 0.1 V (B) 0.2 V

side 0.1 m. The potential energy of the system is (C) 0.4 V (D) 0.5 V

(A) 9910 × 10–9 J (B) 8910 × 10–9 J Q.7 Aninfinite conductingplate of thickness 0.0200m
(C) 99100 × 10–9 J (D) 89100 × 10–9 J is surrounded by a uniform field E = 400 V/m
directed left to right. See the figure. Let the
Q.3 A continuous lineof charge of length3d lies along potential V0 = 0 at a distance 0.0200 m to the
right of the plate. What is V3, the potential
the x-axis, extending from x = d to x = + 4d. the 0.0300 m to the left of the plate?
line carries a uniform linear charge density .
EE
y

x V3 V2 V1 V0
d 3d
(A) –28 V (B) –20 V
In terms of d,  and any necessary physical (C) +20 V (D) +28 V
constants, find the magnitude of the electric field

at the origin.
/ /5160d0d / 4/ 80d0d
(A) 3 (B) 3 Q.8 Two point charges + Q and – Q are kept at a
(C) (D) distance d from each other. At the mid point of
the line joining both the charges
Q.4 As shown in the figure, an insulating rod is net (A) Potential is zero but electric field is not zero
(B) Electric field is zero but potential is not zero
into the shape of a semicircle. The left half of the (C) Electric field as well as potential are zero
(D) Electric field as well as potential are non zero
rod has a charge of + Q uniformly distributed

along its length, and the right half of the rod has a

charge of –Q uniformly distributed along its

length. What vector shows the correct direction Q.9 A sphere carrying a A
of the electric field at point P, the centre of the
charge of Q having
semicircle ? +Q D –Q
(A) A weight w falls under B
gravitybetweenapairof
(B) B A •P C 45° C
(C) C verticalplatesatadistance

(D) D B of d from each other. V is appliedd between
When a potential difference

Q.5 The maximum electric field at a point on the axis the plates the acceleration of sphere changes as

of a uniformly charged ring is E0.At how many shown in the figure, to along line BC. The value
points on the axis will the magnitude of electric
of Q is

field be E0/2 (A) w/V (B) w/2V
(A) 1
(B) 2 (C) wd (D) 2wd
V V
(C) 3 (D) 4

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Q.10 A parallel plate capacitor is charged and then 12th PHYSICS MODULE - 1
Q.11 isolated. The effect of increasing the plate
Q.12 separation on charge, potential and capacitance areaA.As suggested by the figure, the plates of
respectively are capacitors A and C are separated by a distance
(A) constant, decreases, increases d while those of B and D are separated by a
(B) constant, decreases, decreases distance 2d. CapacitorsAand B are maintained
(C) constant, increases, decreases in vacuum while capacitors C and D contain
(D) increases, decreases, decreases dielectrics with constant  = 5.

vacuum d(ie=le5c)tric
AB CD

A spherical shell of radius 10 cm is carrying a d 2d d 2d

charge q. if the electric potential at distances 5

cm, 10 cm and 15 cm from the centre of the

spherical shell is V1 , V2 and V3 respectively, Q.15 Which list below places the capacitors in order
(A) V1 = V2 > V3 (B) V1 > V2 > V3 Q.16
(C) V1 = V2 < V3 (D) V1 < V2 < V3 of increasing capacitance?

(A) A, B, C, D (B) B, A, C, D

An electric dipole is placed at the origin O such (C) B, A, D, C (D) A, B, D, C

that its equator is y-axis. At a point P far away Which capacitor has the largest potential
difference between its plates?
from dipole, the electric field direction is along (A) A
y-direction. OP makes an angle  with the x- (B) B
axis such that : (C) D
(A) tan  = 3 (B) tan  = 2 (D)Aand D are the same and larger than B or C

(C) tan  = 1 (D) tan  = 1/ 2

Q.13 A point charge +Q is positioned at the center of Q.17 The plates of a parallel-plate capacitor are
the base of a square pyramid as shown. The flux Q.18 separated by a solid dielectric. This capacitor
through one of the four identical upper faces of and a resistor are connected in series across the
the pyramid is terminals of a battery. Now the plates of the
capacitor are pulled slightly farther apart. When
Q Q equilibrium is restored in the circuit.
(A) 160 (B) 40 (A) the potential difference across the plates has

Q (D) None + increased
(C) 80 (B) the energy stored in the capacitor has
+Q
increased
Q.14 If a rectangular area is rotated in a uniform (C) the capacitance of the capacitor has

electric field from the position where the maximum increased
(D) the charge on the plates of the capacitor
electricflux goes through it toanorientationwhere
has decreased
onlyhalf the maximum flux goes through it, what

has been the angle of rotation? The plates of a parallel plate capacitor are

(A) 30° (B) 60° charged upto 100 volt. A 2 mm thick plate is

(C) 45° (D) 26.6° inserted between the plates, then to maintain the

For Q 15-16 same potential difference, the distance between
The figure below shows four parallel plate
capacitors : A, B, C and D. Each capacitor the capacitor plates is increased by 1.6 mm. The
carries the same charge q and has the same plate
dielectric constant of the plate is

(A) 5 (B) 1.25

(C) 4 (D) 2.5

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Q.19 A capacitor of capacitance of 2 F is charged to 12th PHYSICS MODULE - 1
Q.20 a potential difference of 200 V, after
opposite to the direction of the field.
disconnecting from the battery, it is connected in (B) The tangent drawn to a line of force

parallel with another uncharged capacitor. The represents the direction of electric field.
(C) Field lines never intersect.
final common potential is 20V, the capacitance (D) The electric field lines forms closed loop.

of second capacitor is: (B) 4 F
(A) 2 F
(C) 18 F (D) 16 F Q.25 Two charges +q and – q are arranged as shown

in the figure. The work done in carrying a test

charge q' from X to Y will be

Two spheres carrying charges +6µC and +9µC, a ra

separated by a distance d, experiences a force +q + •X Y• – –q
B
of repulsion F. When a charge of –3µC is given A kqq kqq
r  2a (B) r
to both the sphere and kept at the same distance (A)

as before, the new force of repulsion is 2kqqa 2kqqr
(C) r(r  a) a(a  r)
(A) F/3 (B) F (D)

(C) F/9 (D) 3 F

Q.21 In an isolated charged Q1 Q2 Q3 Q4 Q.26 Two capacitors of 10 µF and 20 µF are
capacitor of capacitance Q.27
Q.22 ‘C’, the four surfaces have Q.28 connected to 200 V and 100 V sources
Q.23 charges Q1, Q2, Q3 and Q.29
Q.24 Q4 as shown. respectively. If they are connected by the wire,
Potential difference between the plates of the
capacitor is what is the common potential of the capacitors?

(A) 133.3 volt (B) 150 volt

(C) 300 volt (D) 400 volt

(A) Q1  Q2  Q3  Q4 (B) Q2  Q3 A capacitor with air as the dielectric is charged
C C
to a potential of 100 volts. If the space between

| Q2  Q3 | | Q1  Q4 | the plates is now filled with a dielectric of
2C 2C
(C) (D) dielectric constant 10, the potential difference

between the plates will be

(A) 1000 volts (B) 100 volts

In the circuit diagram shown all the capacitors (C) 10 volts (D) Zero
are in F. The equivalent capacitance between
points A & B is (in F) Two conducting spheres of radii 5 cm and 10

637 cm are given a charge of 15µC each. After the
AB
two spheres are joined by a conducting wire,
8 10 4
the charge on the smaller sphere is

(A) 14/5 (B) 7/5 (A) 5µC (B) 10µC
(C) 3/7 (D) None of these
(C) 15µC (D) 20µC

The anglebetweenthe dipolemomentand electric A 10 F capacitor is charged to a potential
difference of1000 V.The terminals of the charged
field at any point on the equatorial plane is
capacitor are disconnected from the power
(A) 180° (B) 0°
supply and connected to the terminals of an
(C) 45° (D) 90° uncharged 6F capacitor. What is the final

Pick out the statement which is incorrect. potential difference across each capacitor
(A) A negative test charge experiences a force
(A) 167 V (B) 100 V

(C) 625 V (D) 250 V

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Q.30 Three charged, metal spheres of different radii (B) Q  1  1 d2 
are connected by a thin metal wire. The potential 2 0 R R2 
and electric field at the surface of each sphere
are V and E. Which of the following is true? (C) zero

12 3 Q  1  1 
4 0  R R2  
(D) d2

wire

(A) V1 = V2 = V3 and E1 = E2 = E3 Q.35 Two insulating plates are both uniformlycharged
(B) V1 = V2 = V3 and E1 > E2 > E3 in such a way that the potential difference
(C) V1 > V2 > V3 and E1 = E2 = E3 between them is V2 – V1 = 20V. (i.e. plate 2 is
(D) V1 = V2 = V3 and E1 < E2 < E3 at a higher potential). The plates are separated
by d = 0.1 m and can be treated as infinitely
Q.31 A voltmeter reads 4V when connected to a large. An electron is released from rest on the
inner surface of plate 1. What is its speed when
parallel plate capacitor with air as a dielectric. it hits plate 2 ?
(e = 1.6 × 10–19 C, me = 9.11 × 10–31 kg) –
When a dielectric slab is introduced between
0.1 m
plates for thesame configuration, voltmeter reads

2V.What is the dielectric constant of the material?

(A) 0.5 (B) 2

(C) 8 (D) 10

Q.32 A spherical conductor of radius 2 cm is uniformly 12 (B) 32 × 10–19 m/s
Q.33 charged with 3 nC. What is the electric field at a (D) 7.02 × 1012 m/s
distance of 3 cm from the centre of the sphere? (A) 1.87 × 106 m/s
Q.34 (A) 3 × 206 V m–1 (B) 3 V m–1 (C) 2.65 × 106 m/s
(C) 3 × 104 V m–1 (D) 3 × 10–4 V m–1
Q.36 Two spherical conductorsAand B of radii 1 mm
Q.37
Four charges equal to – Q are placed at the four and 2 mm are separated by a distance of 5 cm
corners of a square and a charge q is at its centre.
If the system is in equilibrium the value of q is – and are uniformly charged. If the sphere are

connected by a conducting wire then in

Q Q equilibrium condition, the ratio of the magnitude
4 4
(A) – (1 + 2 2) (B) (1 + 2 2) of the electric fields at the surfaces of spheresA

and B is –

(A) 2 : 1 (B) 1 : 4

(C) – Q (1 + 2 2) (D) Q (1 + 2 2) (C) 4 : 1 (D) 1 : 2
2 2
An electric charge 10–3 C is placed at the origin
Two thin wire rings each having a radius R are (0,0) of X - Y co-ordinate system. Two pointsA
placed at a distance d apart with their axes
coinciding. The charges on the two rings are +q and B are situated at ( 2 , 2 ) and (2, 0)
and –q. The potential difference between the
centres of the two rings is respectively.The potential difference between the
(A) QR/40d2
points A and B will be -

(A) 9 volt (B) zero

(C) 2 volt (D) 4.5 volt

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Q.38 Charges are placed on the vertices of a square Q.42 Capacitance (in F) of a spherical conductor with
Q.39  Q.43
as shown. Let E be the electric field and V the radius 1 m is – (B) 10–6
Q.44 (A) 1.1 × 10–10 (D) 10–3
potential at the centre. If the charges onAand B (C) 9 × 10–9

are interchanged with those on D and C
A
respectively, then + B A fully charged capacitor has a capacitance 'C'.
 q + It is discharged through a small coil of resistance
(A) E remains unchanged, wire embedded in a thermally insulated block of
q specific heat capacity 's' and mass 'm'. If the
temperature of the block is raised by 'T', the
(B) VBE ocathnhadnEVgaenrsedmVaicnhuanngcheangedD-q– C potential difference 'V' across the capacitance is
(C) –
-q

(D)  changes, V remains unchanged
E

A thin spherical shell of radius R has charge Q (A) 2m CT (B) m CT
s s
spread uniformly over its surface. Which of the

following graphs most closely represents the msT 2msT
C C
electric field E (r) produced by the shell in the (C) (D)
range 0  r < , where r is the distance from the
centre of the shell?
E(r) A parallel plate condenser with a dielectric of
E(r) dielectric constant K between the plates has a
capacity C and is charged to a potential V volts.
(A) R (B) r The dielectric slab is slowly removed from
between the plates and then reinserted. The net
O r work done by the system in this process is –

OR

E(r) E(r) (A) ½ (K – 1) CV2 (B) CV2 (K – 1) /K
(C) (K –1) CV2 (D) zero

(C) (D) Q.45 A parallel plate capacitor with air between the
Q.46
OR r OR r plates has a capacitance of 9 pF. The separation

Q.40 Two points P and Q are maintained at the between its plates is ‘d’. The space between the
Q.41
potentials of 10 V and –4V, respectively. The plates is now filled with two dielectrics. One of

work done in moving 100 electrons from P to Q the dielectrics has dielectric constant k1 = 3 and
thickness d/3 while the other one has dielectric
is –
(A) –9.60 × 10–17 J (B) 9.60 × 10–17 J constant k2 = 6 and thickness 2d/3. Capacitance
(C) –2.24 × 10–16 J (D) 2.24 × 10–16 J of the capacitor is now –

(A) 45 pF (B) 40.5 pF

A charge Q is placed at each of the opposite (C) 20.25 pF (D) 1.8 pF
corners of a square.Acharge q is placed at each
of the other two corners. If the net electrical force Two capacitors C1 and C2 are charged to 120V
on Q is zero, then Q/q equals – and 200V respectively. It is found that by

connecting them together the potential on each

(A) – 2 2 (B) – 1 one can be made zero. Then –

(C) 1 (D) – 1 (A) 5C1 = 3C2 (B) 3C1 = 5C2
2 (C) 3C1 + 5C2 = 0 (D) 9C1 = 4C2

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Q.47 What is the nature of Gaussian surface involved ///////////
Q.48 2 L2
Q.49 in Gauss law of electrostatic? ///////////
Q.50 L1 1
Q.51 (A) Scalar (B) Electrical

Q.52 (C) Magnetic (D) Vector M1 + + M2
+Q1 +Q2
What is the electric potential at a distance of 9
(A) M1  M2, but Q1 = Q2 (B) M1 = M2
cm from 3nC? (C) Q1 = Q2 (D) L1 = L2

(A) 270 V (B) 3 V

(C) 300 V (D) 30 V Q.53 There is a uniform electric field of intensity E which
Q.54 is as shown. How many labelled points have the
The polarised dielectric is equivalent to – Q.55 same electric potential as the fully shaded point?
(A) two charged surface with induced surface Q.56
(B) only single charged surface with induced E

surface (A) 2 (B) 3
(C) either (A) or (B) (C) 8 (D) 11
(D) neither (A) nor (B)

When air is replaced by a dielectric medium of All capacitors used in the diagram are identical
constant K, the maximum force of attraction and each is of capacitance C. Then the effective
between two charges separated by same capacitance between the point A and B is –
distance –
(A) increases K times (B) remains unchanged AB
(C) decreases K times (D) increases K–1 times

I. In an external electric field, the positive and (A) 1.5C (B) 6C
negative charges of a non-polar molecule (C) C (D) 3C
are displaced in opposite directions.
Two identical conducting balls A and B have
II. In non-polar molecules displacement stops
when the external force on the constituent positive charges q1 and q2 respectively. But
charges of the molecule is balanced by the q1  q2. The balls are brought together so that
restoring force.
they touch each other and then kept in their
III. The non-polar molecule develops an
induced dipole moment. original positions. The force between them is–

(A) I, II are III are correct. (A) less than that before the balls touched
(B) I, II and III are incorrect.
(C) I and II are correct, III is incorrect. (B) greater than that before the balls touched
(D) I and III are correct, II is incorrect.
(C) same as that before the balls touched

(D) zero

Two small spheres of masses M1 and M2 are Two identical charged spheres of material density
suspended by weightless insulating threads of , suspended from the same point byinextensible
strings of equal length make an angle  between
the strings. When suspended in a liquid of density
lengths L1 and L2. The spheres carry charges of  the angle  remains the same. The dielectric
Q1 and Q2 respectively. The spheres are constant K of the liquid is –
suspended such that they are in level with one 
 (B)   
another and the threads are inclined to the vertical (A) 
at angles of 1 and 2
following conditions is as shown. Which of the  
essential, if 1=2 (C)  (D)   

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Q.57 Two equal and opposite charges of masses m1 Q.62 In the uniform electric field of E = 1 × 104 N/C,
and m2 are accelerated in an uniform electric field
through the same distance. What is the ratio of an electron is accelerated from rest. The velocity
their accelerations if their ratio of masses is
of the electron when it has travelled a distance of
2 × 10–2 m is nearly ___ms–1
(e/m of electron = 1.8 × 1011 C kg–1)
m1 (A) 0.85 × 106 (B) 0.425 × 106
m2 = 0.5 ? (C) 8.5 × 106 (D) 1.6 × 106

(A) a1  0.5 (B) a1 1 Q.63 In this diagram, the P.D betweenAand B is 60V,
a2 a2 the P.D across 6 µF capacitor is –

(C) a1 2 (D) a1  3 3µF
a2 a2
3µF
Q.58 In the given network, the value of C, so that an A•
•B
equivalent capacitance betweenAand B is 3µF, 6µF 3µF

C 2µF

3µF

6µF 4µF (A) 5V (B) 20V
2µF (C) 4V (D) 10V

12µF Q.64 A small oil drop of mass 10–6 kg is hanging in at
Q.65
1µF 2µF rest between two plates separated by 1 mm

having apotential difference of 500V. The charge

8µF on the drop is
(A) 36µF (B) 48µF (g = 10ms–2)
(A) 2 × 10–9 C (B) 2 × 10–11 C
(C) (31/5) µF (D) (1/5) µF (C) 2 × 10–6 C (D) 2 × 10–8 C

Q.59 Acceleration of a charged particle of charge ‘q’
Q.60
and mass ‘m’ moving in a uniform electric field A uniform electric field in the plane of the paper
as shown. Here A, B, C, D are the points on the
of strength ‘E’ is – circle. V1, V2, V3, V4 are the potentials at those
points respectively. Then
(A) m/qE (B) mqE
AE
(C) q/mE (D) qE/m C

Two fixed charges A and B of 5µC each are D
separated by a distance of 6m. C is the mid point
of the line joiningAand B.Acharge ‘Q’of –5µC B
is shot perpendicular to the line joiningAand B
through C with a kinetic energy of 0.06 J. The (A) VA = VC, VB = VD (B) VA = VC, VB > VD
charge ‘Q’ comes to rest at a point D. The (C) VA > VC, VB > VD (D) VA = VB, VC > VD
distance CD is –

(A) 3m (B) 3 3m Q.66 Two metal spheres of radii 0.01 m and 0.02 m
(C) 4m (D) 3 m
are given a charge of 15 mC and 45 mC

respectively. They are then connected by a wire.

Q.61 A capacitor of capacitance 10µF is charged to The final charge on the first sphere is
(A) 40 × 10–3 C (B) 30 × 10–3 C
10 V. The energy stored in it is – (C) 20 × 10–3 C (D) 10 × 10–3 C

(A) 500 µ J (B) 1000 µ J

(C) 1 µ J (D) 100 µ J

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EXERCISE-3 (LEVEL-3)

Choose one correct response for each quest–ion. (A) 3K0A (B) 4K0A
Q.1 Between two infinitely 4d 3d

long wires having linear AB K  1 0A (D) K2KK31d0A
charge densities  & – aaa
there are two pointsAand 2d

B as shown in the figure. (C)

The amount of work done by the electric field in

moving a point charge q0 fromAto B is equal to Q.5 Three capacitors each of capacity 4µF are to be
connected in such away that the effective
(A) q0 ln 2 (B)  2q0 ln 2 capacitance becomes 6µF. This can be done by
20 0 connecting –
(A) all of them in series.
(C) 2q0 ln 2 (D) q0 ln 2 (B) all of them in parallel.
0 0 (C) two in series that the third parallel to the

Q.2 A neutral spherical conductor (radius r2) has a combination.
concentric spherical cavity (radius r1). A point (D) two in parallel and the third in series with
charge Q is placed at a distance ‘r’(less than r1)
from the centre. The potential at the centre is : the combination.

kQ kQ kQ 2kQ  kQ Q.6 Four identical particles each of mass m and
r2 r1 r r2 r charge q are kept at the four corners of a square
(A)   (B) of length L. The final velocity of these particles
after setting them free will be.
kQ
(C) r (D) None of these  Kq2 (5.4)1/2  Kq2 (1.35)1/2
 mL mL
Q.3 Which of the following represents the (A) (B)
equipotential lines of a dipole ?
 1/ 2
(A) (B) (C) Kq2 (2.7) (D) Zero
mL

(C) (D) Q.7 A capacitor of capacitance C is charged to a
potential difference V from a cell and then
disconnected from it.Acharge +Q is now given
to its positive plate. The potential difference

Q.4 A parallel plate capacitor of area ‘A’ plate across the capacitor is now –

separation ‘d’ is filled with two dielectrics as (A) V (B) V  Q
shown. What is the capacitance of the C
arrangement ?

A/2 A/2 (C) V Q (D) V  Q , if V < CV
2C C

d Kd
2
K

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Q.8 A capacitor of capacitance C is initially charged required to form a composite 16 mF, 1000V is :

to a potential difference of V volt. Now it is (A) 2 (B) 4

connected to a battery of 2V volt with opposite (C) 16 (D) 32

polarity. The ratio of heat generated to the final

energy stored in the capacitor will be – Q.13 In the circuit shown, the energy stored in 1µF

(A) 1.75 (B) 2.25 capacitor is – 3µF 5µF
(C) 2.5 (D) 1/2

Q.9 Three charges each of charge + q are placed at (A) 40 µJ 1µF
the corners of an equilateral triangle. A fourth (C) 32 µJ
charge Q is placed at the centre of gravity of the 4µF
triangle. If Q = – q, then –
(A) The charges placed at A, B & C will move 24V
towards centre.
(B) The charges placed at A, B & C will move (B) 64 µJ
away from the centre. (D) None
(C) The charge at A will move away and the
charges at B and C will move towards the Q.14 If identical charges (–q) are placed at each corner
centre. Q.15 of a cube of side b, then electric potential energy
(D) All will remaininequilibrium. Q.16 of charge (+q) which is placed at centre of the
cube will be
Q.10 Three charges q, 2q and 8q are to be placed on
a 9 cm long straight line. Where the charges (A) 8 2q2 8 2q2
should be placed so that the potential energy of 40b (B) 0b
this system is minimum ?
(A) q charge between 2q and 8q charges and (C) 4 2q2 4q2
3 cm from charge 2q. 0b (D) 30b
(B) q charge between 2q and 8q charges and
5 cm from the charge 2q. In a certain region the electric potential at a point
(C) 2q charge between q and 8q charges and (x,y,z) is given by the potential function
7 cm from the charge q. V = 2x + 3y – z. Then the electric field in this
(D) 2q charge between q and 8q charges and region
9 cm from the charge q. (A) Increases with increase in x and y
(B) Increases with increase in y and z
Q.11 A particle of mass m and charge q is lying at the (C) Increases with increase in z and x
mid point of two stationary particles distant 2 (D) Remains constant
and each carrying a charge q. If the free charged
particle is displaced from its equilibrium position An uncharged capacitor with a solid dielectric is
through distance x(x <<), then the particle will connected to a similar air capacitor charged to a
(A) move in the direction of displacement. potential of V0. If the common potential after
(B) stop at its equilibrium position. sharing of charges becomes V, then the dielectric
(C) oscillate about its equilibrium position. constant of the dielectric must be –
(D) execute S.H.M. about its equilibrium
(A) V0 V
position. V (B) V0

Q.12 From a supply of identical capacitors rated 8 mF, (C) V0  V (D) V0  V
250V, the minimum number of capacitors V V0

100 ELECTROSTATICS

12th PHYSICS MODULE - 1

Q.17 Five identical metal plates each of area A are Q.21 Two parallel plate capacitors of capacitances C
Q.18 held parallel to each other with successive and 2C are connected in parallel and charged to
Q.19 separation d as shown in figure. The effective a potential difference V. The battery is then
Q.20 capacitance of the system between points P and disconnected and the region between the plates
Q is – of the capacitor C is completely filled with a
material fo dielectric constant K. The potential
Q difference across the capacitors now becomes
P

2200AA//d3d 3300AA//d2d (A) 3V (B) KV
K2 (D) 3/KV
(A) (B)
(C) (D) (C) V/K

An insulator plate is passed For Q.22-Q.23
(A) Statement-1 is True, Statement-2 is True;
between the plates of a AB Statement-2 is a correct
explanation for Statement-1
capacitor. Then current : (B) Statement-1 is True, Statement-2 is True;
Statement-2 is NOT a correct explanation
(A) Always flows from Ato B for Statement-1
(C) Statement-1 is True, Statement-2 is False
(B)Always flows from B toA (D) Statement-1 is False, Statement-2 is True

(C) First flows fromAto B and then from B toA Q.22 An insulating solid sphere of radius R has a
uniformly positive charge density.As a result of
(D) First flows from B toAand then fromAto B this uniform charge distribution there is a finite
value of electric potential at the centre of the
Let P(r) = Q r be the charge density sphere, at the surface of the sphere and also at a
R 4 point out side the sphere. The electric potential
at infinite is zero.
distribution for a solid sphere of radius R and Statement-1 : When a charge ‘q’ is take from
the centre of the surface of the sphere its potential
total charge Q. For point 'p' inside the sphere at energy changes by q/30.
Statement-2 : The electric field at a distance r
distance r1 from the centre of the sphere, the (r < R) from the centre of the sphere is r/30.
magnitude of electric field is –
Q.23 Statement-1 : For practical purposes, the earth
(A) 0 (B) Q r12 is used as a reference at zero potential in electrical
4 0 circuits.
Statement-2 : The electrical potential of a sphere
(C) Qr12 (D) Qr12 of radius R with charge Q uniformly distributed
4 0 R 4 3 0 R4

Find equivalent capacitance between X andYif
each capacitor is 4 F.

C

CCCC Q
bd on the surface is given by 40R .
xa c y

C

(A) 4 µF (B) 8 µF
(C) 12 µF (D) 1 µF

101 ELECTROSTATICS

12th PHYSICS MODULE - 1

SOLUTIONS

CHECK UP 1 from 1 m to 10 m will decrease the force by a
factor of 100 to 1 ×10–6 N, and the increase

(1) Planets have large masses, and so exert large from 1 m to 100 m will result in a force of
1 ×10–8 N.
gravitational forces. However, planets have

negligible net charge (if any); overall, they are (2) Consider any two equal charges q, with an
electric force F = kq2/r 2. If we transfer a charge
electricallyneutral. In an atom, on the other hand,  from one to the other, the charges become
(q + ) and (q – ), so the force becomes
the electrons have (relatively) large charges and F' = k (q + ) (q – ) / r2.
(relatively) small masses. But (q + ) (q – ) = q2 – 2, which is less than
(2) The ground exerts the normal “contact” force on

the stone, which is electric in origin; for a stone q2, so the force decreases.

at rest, the magnitude of the contact force is the

same as the gravitational force it cancels: (3) (B). For the given force and distance, and with
F = mg = (1.0 kg) (9.8 m/s2) = 9.8 N.
q' = 10q, Coulomb’s Law gives
1.0 N = F = kqq' / r2 = (9.0 ×109 N.m2/C2)
(3) The signs of the charges that are, on average, × 10q2 / (3.0 m)2

closest to each other determine the direction of = 1.0 × 1010 N/C2 × q2.

the net force. Thus with two nuclei (like charges)

closest to each other, the net electric force is Thus q = 1.0 1010 C2
repulsive. The overall effect is somewhat subtle, = 1.0 ×10–5 C = 10 µC
as it depends on summing four forces.
(4) (C). The addition of a few elementary charges (4) (C). Is pushed outside the triangle. Like charges
repel. The forces from the two charges at vertices
will cause only negligible change (here, in the
adjacent to the fourth charge are equal but
eighteenth decimal place) in the net charge of any
opposite and cancel. The net force is thus the
macroscopic quantity of charge, such as one
same as the contribution from the charge at the
(5) coulomb. (5) opposite vertex;that repulsion will pushthe fourth
(6) charge away, out of the triangle.
(b). The amount of charge present in the isolated (6) (b). From Newton’s third law, the electric force
(7) system after rubbing is the same as that before (7) exerted by object B on object A is equal in
because charge is conserved; it is just distributed (8) magnitude to the force exerted by object A on
differently. object B and in the opposite direction.
(a, c, e). The experiment shows that A and B (b)
have charges of the same sign, as do objects B C, A, B
and C. Thus, all three objects have charges of No, because the force of the spring changes
the same sign. We cannot determine from this direction when the spring is stretched compared
information, however, whether the charges are to when it is compressed, while the electrostatic
positive or negative. force does not have this characteristic.
(e). In the first experiment, objectsAand B may
have charges with opposite signs, or one of the

objects may be neutral. The second experiment (1) CHECK UP 3
shows that B and C have charges with the same
signs, so that B must be charged. But we still do (a). There is no effect on the electric field if we
not know ifAis charged or neutral. assume that the source charge producing
the field is not disturbed by our actions.
CHECK UP 2
Remember that the electric field is created

(1) Since the electric force varies in proportion to by source charge(s) (unseen in this case),
the inverse square of distance, 1 / r2, the increase not the test charge(s).

102 ELECTROSTATICS