NEET - PHYSICS ELECTROSTATICS

12th PHYSICS MODULE - 1

(2) Since these field contributions are in strong as that due to either sheet alone.Outside,

perpendicular directions, the resultant magnitude the sheets behave like a single sheet with twice

is E  102 102 N / C  2 10 N / C the charge density.
(9) No. If the motion is at constant velocity, then
= 14 N/C; since the two perpendicular
there is no net force and thus no electric field;
components are equal, the direction of the
but if the straight-line motion is accelerated, then
resultant is at 45° , i.e., preciselyin the northeast
there is an electric field directed along the line of
direction.
motion.
(3) At the center of the square, the contribution to (10) (C). The acceleration of a charged particle in an
the electric field from each point charge points
electric field E is a = F/m = qE/m. So if E is the
away from that point charge. Since the charges
same and the charge q and mass m of the particle
are equal and are equidistant from the center,
both double, then the charge-to-mass ratio q/m
the electricfields due to charges diagonallyacross
from each other exactly cancel, and the net field remains the same, the acceleration is unchanged.

is zero. CHECK UP 4

(4) (A). The fields due to the two equal positive (1) (A, B, C).The field is greatest at pointAbecause
(5) charges that are diagonally across from one this is where the field lines are closest together.
another cancel at the center of the square. (2) The absence of lines near point C indicates that
However, the electric field at the center of the (3) the electric field there is zero.
square dueto the remaining positivecharge points (b). Electric field lines begin and end on charges
away from it, and the electric field due to the and cannot close on themselves to form loops.
negative charge points toward it; these two In Fig. a, there are electric field lines that cross;
contributions are in the same direction, directly since the direction of the electric field is tangent
toward the negative charge. to the field lines, crossing field lines (two
The contribution to the electric field from each directions at one point) are not allowed in field
positive sheet of charge points perpendicularly line diagrams. Fig b shows field lines beginning
out from the corresponding sheet and is in empty space; a field line must begin and end
independent of distance from the sheet. For two on charges (or at infinity). Fig. c shows a field
perpendicular equivalent sheets, these two line looping around and closing on itself; in
vectors areequal in magnitude andat right angles; electrostatics, field lines must have a beginning
they add to produce a field at 45° to the sheets and an end.

(with magnitude 2 times as large as each (4) Thespherehas somepositive charge(in theregion
where fieldlines are emerging) andsome negative
individual sheet field).
charge (where field lines are entering). Since the
(6) Far away from a charge distribution, the field is
net number of field lines is two lines emerging
essentially that of a point charge, in this case E =
(1/40) Q/r2. Very close to the surface, the disk (fouremergingminustwo entering), thenet charge
looks like an infinite plane of charge, with field on the sphere must be one-quarter the value of

(7) E = /20 = (Q/A) / 20 (5) the point charge (which had eight lines emerging).
For an infinite sheet, the electric field is Since the field is pointing downward, the field

independent of distance. So E = E0 at both 2 m lines terminate at the Earth’s surface, and the
and 4 m from the sheet. Earth’s surface charge must be negative.

(8) (B). Since each sheet by itself produces a field (6) (D). Since field lines onlyenter the rod, its charge
is negative. Since there are 12 lines entering the
that points perpendicularly away from the sheet
rod, compared with 8 lines emerging from the
on each side, the total field is zero in between
point charge, the magnitude of the charge on the
the two sheets, where the two equal-magnitude
rod is 12 / 8 = 3 / 2 times that of the point charge.
contributions oppose.Outside the sheets, the two
The charge on the rod is – (3/2)×6.0C = – 9.0C.
contributions are parallel, so the field is twice as

103 ELECTROSTATICS

12th PHYSICS MODULE - 1

CHECK UP 5 CHECK UP 6

(1) If the surface were horizontal, nofield lines would (1) (b). When moving straight from Ato B, E and

cross it, so the flux would be zero (smaller); if ds both point toward the right. Thus, the dot
the surface were vertical, more field lines would product E . ds is positive and V is negative.

cross it, so the flux would be larger. (2) (b, f). The electric potential is inversely

(2) (C). Since all the field lines that enter the surface proportion to the radius. Because the same

also leave it, the net flux through the surface is number of field lines passes through a closed

zero. surface of any shape or size, the electric flux

(3) (D). Since the field is uniform, field lines enter through the surface remains constant.

the sphere on one side and leave it on the other; (3) (a). If the potential is constant (zero in this case),

the net flux through the sphere is zero. its derivative along this direction is zero.

(4) No.Although Gauss’Law is still valid, there is (4) (b). If the electric field is zero, there is no change

no surface through which the electric flux may in the electric potential and it must be constant.

be calculated easily. To calculate the electric field This constant value could be zero but does not

for four point charges, you would have to take have to be zero.

the vector sum of the four Coulomb (point (5) The graph would look like the sketch below.

charge) fields. Notice the flat plateaus at each conductor,

(5) (B). The field contribution due to the sheet with representing the constant electric potential inside
charge density is /20; that due to the 2 sheet a conductor.
is /0. Since both charge densities are positive,
V
each field contribution is directed away from its

respective sheet. Thus between the two sheets

these contributions oppose to give a net field of a b– c b+ c x
magnitude /0 – /20 = /20, and outside the
sheets the two contributions are parallel and give
a net field /20 + /0 = 3/20.
(6) (c). The charges q1 and q4 are outside the
surface and contribute zero net flux through S'.
(7) (d). We don’t need the surfaces to realize that

anygivenpoint inspace willexperience an electric

field due to all local source charges. Right edge Left edge Right edge

(8) E  qin of sphere 1 of sphere 2 of sphere 2
0
(6) No charge stays on the inner sphere in

Through S1 : E  2Q  Q   Q equilibrium. If there were any, it would create an
0 0 electric field in the wire to push more charge to
the outer sphere.All of the charge is on the outer
Q  Q
Through S2 : E  0  0 sphere. Therefore, zero charge is on the inner
sphere and 10.0 µC is on the outer sphere.

Through S3 : E  2Q QQ   2Q (7) E   dV = –b = – (– 7.00 V/m)
0 0 dx

Through S4 : E  0 = 7.00 N/C in the + x direction.

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12th PHYSICS MODULE - 1

CHECK UP 7 F  ke (Q / 2) (Q / 2)  4 keQ2
(L  R  R)2 (L  2R)2
(1) The electric field always points in the direction

of the greatest change in electric potential.  8.99 109 Nm2 (60.0 106 C)2
(2) B C, C D, A B, D E. 4C2 (2.01m)2

Moving from B to C decreases the electric

potential by 2 V, so the electric field performs 2 (5) = 2.00 N
J of work on each coulomb of positive charge (a) The charge +q at the center induces charge
that moves. Moving from C to D decreases the –q on the inner surface of the conductor, where
electric potential by 1 V, so 1 J of work is done
by the field. It takes no work to move the charge its surface density is: a  q
from A to B because the electric potential does 4a 2

not change. Moving from D to E increases the (b) The outer surface carries charge Q + q with
electric potential by 1 V, and thus the field does
–1 J of work per unit of positive charge that density b  Qq
moves. 4b2

(3) (f). The electric field points in the direction of (6) The dielectricstrength is ameasureofthe potential

decreasing electric potential. difference per unit length that a dielectric can

(4) (a) EA > EB since E  V withstand without having individual molecules
s ionized,leavingin its wakeaconductingpathfrom
plate to plate. For example, dryair has a dielectric
(b) EB   V   (6  2) V strength ofabout 3 MV/m. Thedielectric constant
s 2cm in effect describes the contribution of the electric

= 200 N/C down dipoles of the polar molecules in the dielectric to

CHECK UP 8 the electric field once aligned.
(7) In water, the oxygen atom and one hydrogen

(1) No. It is possible in an insulator, but not in a atom considered alone have an electric dipole

conductor when in electrostatic equilibrium. moment that points from the hydrogen to the

(2) (D). The point charge Q induces an opposite oxygen. The other O-H pair has its own dipole

charge –Q on the inner surface of the conductor; moment that points again toward theoxygen. Due

this ensures E = 0 inside the conducting shell. to the geometry of the molecule, these dipole

Since the shell is overall uncharged, the charge moments add to have a non-zero component

–Q on the inner surface implies there is a charge along the axis of symmetry and pointing toward

Q on its outer surface. the oxygen. A non-polarized molecule could

(3) If the person is uncharged, the electric field inside either have no intrinsic dipole moments, or have

the sphere is zero. The interior wall of the shell dipole moments that add to zero.

carries no charge. The person is not harmed by An example of the latter case is CO2. The
touching this wall. If the person carries a (small) molecule is structured so that each CO pair has

charge q, the electric field inside the sphere is no a dipole moment, but since both dipole moments

longer zero. Charge –q is induced on the inner have the same magnitude and opposite direction-

wall of the sphere. The person will get a (small) due to the linear geometry of the molecule-the

shock when touching the sphere, as all the charge entire molecule has no dipole moment.

on his body jumps to the metal.

(4) The charge divides equallybetween the identical
spheres, withcharge Q/2 on each. Then theyrepel (1) Nothing happens to the charge if the wires are
disconnected. If the wires are connected to each
like point charges at their centers:
other, charges in the single conductor which now

105 ELECTROSTATICS

12th PHYSICS MODULE - 1

exists move between the wires and the plates (8) (a). When the key is pressed, the plate separation

until the entire conductor is at a single potential is decreased and the capacitance increases.

and the capacitor is discharged. Capacitance depends only on how a capacitor

(2) 336 km. The plate area would need to be is constructed and not on the external circuit.

1 m2. EXERCISE - 1
0
(1) (C).
(3) A capacitor stores energy in the electric field (2) (B).
(3) (B).
between the plates.

(4) (B).

C (5) (A).
(6) (D). a-ii, b-ii, c-i, d-i
 C/2 F 1/r2, E 1/r2, V 1/r and U 1/r
(4)
C/3 (7) Repulsion

(8) Zero

Ceq  C 1 1  13  11 C = 1.83 C (9) Expands
2 6 (10) Metal
(11) Normal

 1 1  1 (12) False. A dipole is placed in a uniform electric
5.00 10.0
(5) Cs    3.33F field with its axis parallel to the field, neither

experiences a net force nor a torque.

Cp1  2 (3.33)  2.00  8.66F (13) Zero (14) False.
Cp2  2 (10.0)  20.0 F
(15) Inside (16) False

(17) True

 1 1  1 (18) 9. Q = 27q ; 4 R3  27  4 r 3 
8.66 20.0 3 3
Ceq    6.04 F
k  27q
1 Q2 0A VB  kQ  (27)1/3 r  (27)2/3 kq  9kq  9Vs
2 C d R r r
(6) Use, U  and C
(19) Zero. Potential at all points on bisector is zero.

 1 (20) True (21) Increases
2
If d2 = 2d1, C2 C1 . (22) Parallel (23) Series

Therefore, the stored energy doubles. (24) Double

K10A / 2 K20A / 2 (25) 10. Cp = 100Cs
d d/2
(7) C1  ; C2  nC0 = 100 C0 ; n = 10
n

C3  K30A / 2 rubbed 
d/2 Glass rod
Glass rod – – –
 1  1  1  C2C3  0A  K 2K3  (26) (A). silk cloth silk cloth
 C2 C3  C2  C3 d  K2  K3  rubbed 
Plastic rod–C– a–t+'s+f+ur
Plastic rod
Cat's fur

C  C1   1  1  1  0A  K1  K2K3  It is convention to take charge on glass rod
 C2 C3  d  2 K2  K3  as positive charge and silk cloth as negative
charge.

106 ELECTROSTATICS

12th PHYSICS MODULE - 1

When charged glass rod touches charged q1
silk cloth, afterwards they do not attract

paper pieces because their charges (III) q3 q1 & q2 are unlike,
neutralise each other.
(27) (A). As X-rays are electromagnetic waves and q2
they are neutral photons like light photons,
they do not have charge. So, divergence of q2 & q3 are like, q1 & q3 are unlike
possible

leaves will not be affected because (IV) All three unlike charges
divergence will change onlywhen leaves get

some positive or negative charge.  not possible
(C).
(28) (C). (37)
(29) (A). Charges are additive, conservative and
3 2 1
quantised, in nature. (I) F3 F2 F1
(30) (B). A negatively charged body acquires some + + +
P

electrons, so its mass is more than its neutral 32 F1 1
mass. F3 F2
(31) (B). First members acquires positive charge and (II) + + –

IInd member have negative charge. 3 2 1 F1 F3 F2

1  109 (III) + + –
1.6 1019 1P
(32) (B). q  ne  n  q   6.25 109 3 2
e
3 2 1 F2 F3 F1
(33) (A). Q  ne  1014 1.6 1019
(IV) + – + P

 Q  1.6 105 C  16C (38) (D). Fe  kq1q2  9 1.6 1.6 1029
r2 r2
Electrons are removed, so charge will be

positive. Fg  Gm1m2  6.7  9.1 9.11073
(34) (D). Positive charge shows the deficiency of r2 r2

electrons.

Number of electrons  14.4 1019 9 Fe  1042
1.6 1019 Fg
1 q1q2
(35) (C). A movable charge produces electric field (39) (A). F  40 R2
(36) and magnetic field both.
92U238 has charge 92e. When -particle is
(C). emitted, charge on residual nucleus is
92e - 2e = 90e
q1

(I) q2 q1 & q2 are opposite,  q1 = 90e, q2 = 2e, and R = 9 × 10–15m

q3  F  9  109  (90e) (2e)
(9  1015 )2
 q1 & q3 are like, q2 & q3 are like
not possible
9 109  90  (1.6 1019 )2
 (9 1015 )2  2  512 N

(II) all three like charges (40) (A).When put 1 cm apart in air, the force between

 possible Na and Cl ions = F.
When put in water, the force between Na

and Cl ions = F/K

107 ELECTROSTATICS

12th PHYSICS MODULE - 1

(41) (A). Let separation between two parts be 1 at2, L
2 v
  (Q  q) (48) (C). d= t =
r2
r F k.q 2

For F to be maximum d 1   Eq    Lv  ; q 2dmv2
2 m EL2
Q 2
dF 0  q  1 (49) (B). The given system can be reduced as
dq
–Q –
2
1 F1  r2  . So, resultant is .
(42) (B). F r2  F2   r1 
+Q +

5   0.04  2  F2  11.25 N kQ 9  109  1.6  10–19
F2 0.06
 (50) (D). E = r2 = (10–15 )2

(43) (C). F = F' or Q1Q2  Q1Q2  r '  r = 14.4 × 1020 V/m
(44) 40r2 40r '2 K
K (51) kq Ei  x
(52) (B). Ei = R3 x or
(A). F  1 (7 106 ) (5106 )
40 r2 (53) 1 mv2 = 1 ma2 t2 1  qE  2 t2
2 2 2 m
(B). Ek = = m

  1 35 1012 N = q2E2t2
40 r2 2m

F  1 (5 106 )(7 106 ) (C). a  qE  ae  mp
40 r2 m ap me

  1 35 1012 N (54) (C). For non-conducting sphere
40 r2 r
Qq Q Ein  k.Qr  30
R3
+– + (55) (C). Electric potential inside a conductor is
(45) (A). constant and it is equal to that on the surface
FFoQr, Qsysr/tF2eQm, qeqru0ilibrium of conductor.

KQ2  KQq  0   Q (56) (C). E 1 . ne  n  Er2 .40
r2 (r / 2)2 4 40 r2 e
; q

(46) (A). y = –cos x  n  0.036  0.1 0.1  360  105
y 9 109 1.6 1019 144
Ex = x = sin x and Ey = y =1
y = 2.5  105 N/C.
When x  (0, ), slope of given graph is
positive thus x & y component of E should (57) (C). y = 1 at2 = 1 eE t2 [ a = Fe = eE ]
positive in this domain. 2 2 m m m

(47) (C). Due to symmetry field due to one pair will t= 2ym = 3 × 10–9 s
be forming the resultant field strength and e.E

equal to E 2 1 Q  2keQ [Putting y = 4 × 10–2 m,
40 r2 r2 m = 9.1 × 10–31 kg,
e = 1.6 × 10–19C, E = 5 × 104 N/C]

108 ELECTROSTATICS

12th PHYSICS MODULE - 1

kq kq (65) (B). Charge must be on outer surface only
(58) (A). E0 = r2 , Es = R2
v  kQ ... (1) Q
b a
qEs R 2
 F = qE0 = r2 = 1.5 × 10–2N kQ  kQ  0
b a
[q = 5 × 10–8 C, Es = 5 × 106 v/m, b
R = 0.05 m, r = 0.20 cm]
Q   Qa ... (2)
b
E
(59) (C). RQ vB  kQ  kQ Q
C b b Q1

Rr vB  v 1  a 
b
(60) (A). E   dV
dr (66) (C). Inside a conducting body, potential is same

E = 0  dV = 0  V = constant everywhere and equals to the potential of
dr it’s surface.
(67) (D). If charge acquired by the smaller sphere is
(61) (A). Potential will be largest at P as we move
Q then it’s potential
towards (–) plate potential will reduce.
 kQ
VA  kq  kQ VB  kq  kQ 120 2 ..... (i)
a ab ab a
(62) (A). ; Also potential of the outer sphere

VA  VB  a kqb  a kQb V  kQ .....(ii)
(a  b) (a  b) 6

 a kb (q  Q) = 60 V From equation (i) and (ii) V = 40 volt
(a  b) (68) (D). V = n2/3 v  V = (125)2/3  50 = 1250 V

(63) (C). Since electricfield lines point inthe direction (69) (A).   30x 2 ˆi
of decreasing electric potential, the potential E

decreases as one travels from initial point dV   E.dx ; vA dV  2 dx
to final point on each line. Since the electric
field is uniform and in the  30x2
x-direction, equipotential lines will be v0 0
perpendicular to the x-axis. Therefore, since
the final position of each line has the same (70) (A). VChAa–ngVeOin=P–.E8.0vUol=t q (V)
x-coordinate, each will have the same final (71) = –1.6 × 10–19 [(–1) – (10)] × 103
potential. Plot (C) best describes this U  – ve i.e. decrease in P.E.
situation.
(D). According to figure, potential at A and C
kQ VR are equal. Hence work done in moving – q
R k
(64) (B). =V ; Q= charge from A to C is zero.

A –q
–
Q VR 1
  4R2  4kR 2   R ll

B+ l C
+Q

109 ELECTROSTATICS

12th PHYSICS MODULE - 1

(72) (A). W = q(V) (85) (A).As the dipole will feel two forces which are
2 = 20(V) V = 0.1 volt.
although opposite but not equal.
 A net force will be there and as these forces
(73) (C). Electric lines force due to negative charge

are radially inward. act at different points of a body, a torque is

also there.

(86) (B). Maximum torque = pE
–– = 2  10–6  3  10–2  2  105
= 12  10–3 N-m.
(87) (D). Dipole moment p = 4  10–8  2  10–4
(74) (D). (A) and (B) is not possible since field lines = 8  10–12 m

should originate from positive and terminate Maximum torque = pE

to negative charge. (C) is not possible since = 8  10–12  4  108 = 32  10–4 Nm

field lines must be smooth. (D) satisfies all Work done in rotating through 180o = 2pE

required condition. = 2  32  10–4 = 64  10–4 J

(75) (D). Most positive work is done when positive 2kp kp
charge is displaced maximum against (88) r3 r3
strongest electric field. (B). Ea  and Ee  ;  Ea  2Ee

(76) (B). Electric Flux is a measure of the number of (89) (D). For a perfect conductor :
electric field lines passing through an area.
* The surface of the conductor is an
(77) (A). Unit vector normal to x-y plane is ˆi , thus
Q  equipotential surface.
E·A = aA * The electric field just outside the surface of

(78) (A).  = E s cos  = 2 × 103 × 10–2 cos 0° a conductor is perpendicular to the surface.
= 20 NC–1 m2 * The charge carried bya conductor is always

uniformlydistributed over the surface of the

(79) (D). By Gauss Theorem 2  1  q conductor.
0 (90) (D).
(91) (B). Permittivity of metals is very high
(80) (C). Gauss's law is often useful towards a much
comparable to permittivity of free space.
easier calculation of electrostatic field when
So dielectric constant for metal is infinite.
the system has some symmetry. This is (92) (A). wFohrelrien,eaeriissoatcroonpsictadniteclehcatrraicctPer=isticeEof the
facilitated by the choice of a suitable dielectric and is known as the electric
Gaussian surface.

(81) (C). By symmetry,  = q susceptibility of the dielectric medium and
60 it is possible to relate e to the molecular
 5  106 properties of the substance.
6 × 4 × 109 × 9 = 9.4 × 104 (93) (D). In polar molecule the centres of positive and
negative charges are separated even when

(82) (D). Net force on dipole can never be zero in a there is no external field. Such molecule

nonuniform field. But torque can be zero if have a permanent dipole moment.

forces on charges are along axis of dipole. Ionic molecule like HCl is an example of
p . 
(83) (D). U() = – E polar molecule.

W = U =coUs ()––(–Up0E) (94) (B). Q = CV  C  Q  ne
= –pE V V
= pE (1 – cos ) Given V = 100 volt; n = 6.25 × 1015

(84) (D). Potential energy of dipole in electric field  C 6.25 1015 1.6 1019  10F
U = –pE cos where  is the angle between 100
electric field and dipole.

110 ELECTROSTATICS

12th PHYSICS MODULE - 1

(95) (A). The potential difference across the parallel (105) (B). C1 < C2
plate capacitor is 10V – (–10V) = 20V. C1 C2
 C1  C2  1 and C1  C2  1
Q 40 2 2
Capacitance  V  20  2F.
C1C2 C2 C1
(96) (C). Because the charges are produced due to C C1  C2  C1  C1  C2  2

induction and moreover the net charge of Similarly, C < C2
the condenser should be zero. 2

(97) (B). C  0A .C'  0A  C' = 2C (106) (A). From the given figure, total capacitance is
d d/2
(98) (A). Force on one plate due to another is 1  1  1  1 1  1
 1 C  2.5) C 3.5
F = qE = q  20 (1

K  C  3.5  1.4 F
2.5
 q  q  q2 (107) (A). Potential difference across the condenser
 2AK0  2AK0 V  V1  V2  E1t1  E2t2

(where  is the electric field produced   t1   t2
20K K10 K20
by one plate at the location of other).
   t1  t2   Q  t1  t2 
(99) (C). C1  0 A and C2  K0 A  V 0  K1 K2  A0  K1 K2 
d1 d2
 C1C2  2.4F.
 C1  1  d2  C  1  2d (108) (C). Ceq C1  C2
 C2 K d1 2C K d
K=4 Charge flown = 2.4  500  10–6 C
=1200 C.
Cmedium C
(100) (D). Cair  K  2 (109) (C). Charge flowing  C1C2 V .
C1  C2
(101) (B). In general electric field between the plates

of a charged parallel plate capacitor is given So potential difference across

by E   C1  C1C2V  1  C2V
0K C1  C2 C1 C1  C2

(102) (A). Remains unchanged (110) (A). CAB  3  3  4F
(103) (B). C1 = 3C, C2 = 2C/3,C3=3C/2, C4 = C/3 3
(104) (A). When applied p.d. is V across A & B
CAC  3  3  3F  CAB : CAC  4 :3
Assuming VAC = V1 & VCB = V2 2 2

C

C (111) (D). The given circuit can be redrawn as follows
AB potential difference across 4.5 F capacitor

C 9 4.5 F   9 F  9 F
 9 2
We have, V1 = V2/2
hence V1 = V/3 & V2 = 2V/3 V   9  12  8V
As V1 & V2 both must not exceed 100 V, 2
the maximum value of applied p.d. across
A & B would be 150 V. 12 V

111 ELECTROSTATICS

12th PHYSICS MODULE - 1

(112) (C). Charge on C1 = charge on C2 EXERCISE - 2
 C1(VA  VD )  C2 (VD  VB )
 C1(V1  VD )  C2 (VD  V2 ) (1) (A). Molar mass of water (H2O)
= 2 × 1 + 16 = 18g.
  C1V1  C2V2 (2)
VD C1  C2 (3)  Number of mole of water in 10 g =1108 mole
(4) 1 Mole contain NA = 6.023 × 1023 number
(113) (C). Potential on 4F capacitor = 6 ×10 = 6V (5) of molecules.
10
Hence charge q = (4F) × (6V) = 24 C  Number of molecules in 10 g
(114) (A). This is (n – 1) capacitors in parallel.
10
So Ceq = (n – 1) C = 18 × 6.023 × 1023 molecules

(115) V  1 CV2
(B). U  0 CV dV 2 Each molecule contains 10 electrons

(116) (A). The total energy before connection Number of electrons

 1  4  106  (50)2  1  2 106  (100)2 = 10 × 10 × 6.023 × 1023
2 2 18
= 3.344 × 1024 electrons.
 1.5 102 J
+ 6nc

When connected in parallel 1 Q1Q2
40 d
4  50  2 100  6  V  V  200 (B). PE  + +
3
Total energy after connection 3nc 9nc

 1  6 106   200  2  1.33 102 J  9  109 [18  54  27] 1018
2 3 0.1
= 90 × 99 × 10–9 = 8910 × 10–9 J
(117) (A). U  1 CV 210–812J  12  1012  (50)2
= 2 4d k dx 4d
1.5 ×  (C). E  d x2  k d x 2 dx

(118) (B). Total capacitance of given system

Ceq  8 F    1 4d  3
5 40  x d 16 0d

U  1 Ceq V 2  1  8  106  225 (C). Direction of field will be along C. Vertical
2 2 5
components of field will be cancelled out
 180 106 J = 1800 erg
net field will be towards C.

USys  1 CSys VS2ys  1  C  V2  E
2 2 n E
(119) (B). (D). -x graph: x

(120) (C). W  Q2 E0 E0/2
2C
|  | -x graph ... (1)
E

(121) (B).Van de Graaff generator is used to build up Hence 4 times.

high voltages of few million volts.

(122) (B). The breakdown field of air which is about (6) (A). n 4 r3  4 R3  r3  1
3 × 106 V/m 3 3 R3 n

112 ELECTROSTATICS

12th PHYSICS MODULE - 1

vo  kq ...(2) ; v  knq ...(3)  m=ax6/02°= EAcoscos = 1/2
r R  Angle of rotation is 60°

v= nvor  nvo 1 (15) (C). CA = 0 A ; CB = 0 A ; CC = 5 0 A
R n1/3 d 2d d

 vo  v  2.5 = 0.1 V CD = 5 0 A ; CB < CA < CD < CC
n2/3 (125)2/3 2d
(B). q = CV = constant ; V  1/C
(7) (C). E goes from higher potential to lower (16) V  maximum ; C  minimum
(17)
potential E  V  V3  V0 (D). Due to increase in distance between the
x 0.05 plates, capacitance decreases.
 V3 = 400 × 0.05 = 20V As battery remains connected
(8) (A). Q V = constant ; Q = CV ; C Q 
E0;V=0 –Q (18) (A). Potential difference between plates remains

same. Decrease in potential difference is

(9) (C). QE = Mg QE counteracted by potential difference due to
the extra distance.
QV wd 45°
d  w  Q  V t  E  E   Ed  t 1 1   d  k  t t d
mg k k 

(10) (C). After separation charge = constant E is original electric field, k dielectric

Capacity [C] = E0A/d constant of plate, t thickness of plate & d
Capacitydecreases with increase in distance
extra distance.
V = Q/C
CV 2  200
(11) Potential increases (19) (C). V  C  C ; 20  2  C
(A). Inside the spherical shell potential remains

same C  18F
V1 = V2 ; V 1/d V1 = V2 > V3
(A). F Q1Q2 ; F  69  F  F
(12) (B). Given E=R9a0l°on–gy-axis ER (20) F1 36 3
thus  
Ev El (21) (C). From charged is tribution Q1 = Q4 net
elctric field between plates is E × d
Also tan = 1 tan
2 Potential Difference

or tan  = 2 Q p = Q2  Q3 d  Q2  Q3
2A0 2C

(13) (C). Flux going in pyramid = 20
which is devided equally among all 4 faces
(22) (A). 63 7

 Flux through one face = Q 8 10 4
80
balanced wheatstone bridge
(14) (B).  = Flux = EAcos, 
E and A. 63
where  is the angle between

For maximum flux,  = 0 cos 0 = 1 8 4 Ceq = 14/5
 max = EA

113 ELECTROSTATICS

12th PHYSICS MODULE - 1

(23) (A). Dipole moment and electric field are KQ2
r2
opposite to each other on equatorial line. KQ2
–Q
(24) (D). Electric lines never form closed loops. – –Q  2r 2
–
(25) (D). W = q´ (V2 – V1) KQ2
r2

(33) (B). KQq
r 2
 q q q q  q
 a a  2

= q´ k a r    a r – –Q

–Q –

= 2Kqq´  a 1 r  1  For system equilibrium
 a Fnet = 0

or |W |= 2Kqqr   KQ2   KQ2  2KQq  0
a(a  r) 2  r2  2r2 r2

(26) (A). V  C1V1  C2V2   q  Q 1 2 2
C1  C2 4

 10  200  20  100  133.3 V (34) (B). Potential difference = V1 – V2
10  20
V1  kq  kq ; V2  kq  kq
(27) (C). New potential difference R R2  d2 R R2  d2

 V  100  10V (35) (C). e (V2 – V1) = 1 mv2
K 10 2

(28) (B). Charge on smaller sphere (36) (A). For equipotential spheres

= Total charge  r1 r1   30 5 510  10C E 1
  r2  Radius

(29) (C).After charging, total charge on the capacitor
 Q = CV
= 10  10–6 F  1000 V = 10–2 C. (37) (B). V at 2, 2 = V at (2, 0)
Common potential
 Potential difference = 0

qq –q –q
102 + + E
V  C1V1  16 106 = 625V.
C1  C2 (D). E qq
(38) –q –
(30) (B). When spheres are connected by a wire then – –q
charge will flow untill they have same
(39) (D). Inside shell E = 0
potential V1 = V2 = V3
(40) (D). Wext = q (Vf – Vi )
KQ = (–100 × 1.6 10–19)
and for sphere V= R = ER × (–4 – 10)
= –1.6 × 10–17 × (–14)
So for same potential E 1/R. = 2.24 × 10–16 J
So, E1 > E2 > E3
Q q
(31) (B). r  Va  4  2
Vm 2 + –

(41) (A). a

 109  3 109 – Q+
(3  102 )2
q

(32) (C). E  9  109  q  9 Q2
r2 40 (2a2
 qQ 2 0 ; Q  2 2
= 3 × 104 V m–1 ) 4 0a 2 q

114 ELECTROSTATICS

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(42) (A ). C = 40R (lines of force and equipotential surface
are perpendicular)
(43) (D). 1 CV2  msT (54) (A). Capacitors are in parallel, Cp = 3C
2

(44) (D). Once slab is removed and then reinserted
So no change in energy  W = 0
Another such circuit in series
(45) (B). CAir ppc = 9F = 0A  CAB = Cp/2 = 1.5 C
d (55) (B). When two charged spheres are brought in

C'  0A = 4.5 CAir = 40.5 F contact they attain equal charge = q1  q2 .
2
t1  t2
 r1  r2 For a given total charge q1 + q2, the force
between the charges is maximum when
(46) (B). For potential to + C1–
be made zero, charges are equal.

after connection 120V (56) (D). r  Fa  mg
Common voltage + C2– Fm (m  m1) g
200V
= C1V1  C2V2  ( V V    
C1  C2  ) 

120 C1 = 200 C2  3C1 = 5C2 (57) (C). Force is same in magnitude for both
(47) (D). Area vector
a1 F / m1 m2
(48) (C). v  9  109  q  9 109  3 109  300V a2  F / m2  m1  2
r 9 102

(49) (A). The polarised dielectric is equivalent to two 3 16 C  C  48F A C 16/5
charged surfaces with induced surface (58) 5 B
charge densities, say p and –p. (B). 16
5 
C

E (59) (D). a  F  qE
–+ –+ –+ m m

–p – + –+ –+ + p (60) (C). Total energy at C = Potential energy at D
–+ –+ –+
–+ –+ –+ 0.06  9  109  5 106  2  5 106
3
P

A uniformly polarised dielectric amount to  9 109  5 106  2 ; x = 4m
induced surface charge density but not 9 x2
volume charge density.

(50) (C). As Fm  F0 (61) (A). U  1 CV2  1 10 106  100  500J
(51) K (62) 2 2
 The maximum force decreases by K times.
(A). (C). v2 = u2 + 2as  v = 2as

(52) (B). tan   F ; F is same on both the  2   Eme s  2 104 1.8 1011  2 102
Mg  2 1.8 1013
= 2 × 4.25 × 106 = 8.5 × 106 m/s
charges;  is same only if M is equal
(53) (B). Take vertical plane through the shaded

circle, which is equilpotential.

115 ELECTROSTATICS

12th PHYSICS MODULE - 1

A B A/2 A/2

(63) (D). V1 V2 V1 V2

2V1 + 2V2 = 60 and V2 = 2V1 c1 c3
( V Kd
1/C) d c2 K
 6V1 = 60 V1 = 10V 2

(64) (B). Eq = mg ; Vq  mg  q  mgr  ceq.  c1  c2  c3  32dKK KA10
r V c1  c2

(65) (D). Line perpendicular to E-field: Equipotential (c1 and c2 are in series and resultant of

In direction of E-field potential decreases. these two in parallel with c3)

(66) (C). R1  0.01  1 ; Q R ; 44
R2 0.02 2
(5) (C).
Q1   13 Q  20mC (6)
4
EXERCISE - 3
(C). Potential energy of the system

(1) (D). Electric field at P =4× K  q2  q2  q2 2  = 5.4 Kq2
2  L L L  L
    
E  20 1  1 = Final kinetic energy of the system
x 3a  x
xP E 1
2
2a 3a-x =4× mv2 = 2 mv2

W   q0E dx  Kq2 1/ 2
a mL (2.7)

q0 2aa dx 2a dx x  q0  v=
20 x a 3a  0
    ln 2 (7) (C). After redistribution half of total charge
remains on outer surface and then apply
Q conservation of charge on each plate.
–Q
Or V0  kQ  kQ  kQ CV CV +Q/2 +Q/2 +Q/2
r r1 r2 +–
(2) (A). r1 Q
–Q/2
r2

(3) (D). – x y 2a – x Charge on outer plates
(0,0)
–q + Q  CV  CV Q
2  2
+q

=

Vp  kq  kq  0 Charge of innerface of first plate = Q + CV
y2  x2 y2  (2a  x)2
– Q = Q +CV
(4) (D). c1  A / 20  A0 , 2 2
d/2 d
A0 A0
c2  K d , c3  K 2d

116 ELECTROSTATICS

12th PHYSICS MODULE - 1

Charge on innerface of second plate K(2q)q  K(8q)q
x2  10–2 –
 Q CV Q  CV  (9 x)2
2 2 C
=   ; V   V Q or (9 × 10–2 – x)2 = 4x2
2C or 9 × 10–2 – x = 2x

1 +CV –CV  x = 3 × 10–2 m = 3 cm
2 (11) (D). When the charge q is displaced through
(8) (B). U1  CV2
small displacement x, then the resultant force

–2CV +2CV acting on it is

q qq
+x
Ur  1 C (2V)2  2CV2 3CV + O +Q
2
P

Wbattery = 3CV × 2CV = 6CV2 F= kq2  kq2  4kq2 x = ma
Heat produced = Wbattery – (Uf – Ui) (  x)2 (  x)2 3

= 6CV2 – 3 CV2 = 9 CV2 or a =  4kq2  x = 2x or a  x
2 2  m3 

Heat produced  9  2.25  The motion of the particle will be S.H.M.
Uf 4
(12) (D). Let ‘n’ such capacitors are in series and

such ‘m’ such branch are in parallel.
 250 × n = 1000  n = 4
(9) (A). AO = 2 a sin 60º = 2a 3  a …(i)
3 3 3
2 8 m  16 ; m  16  n =8 …(ii)
n 8
Resultant force on charge q  No. of capacitor = 8 × 4 = 32
placed atA (outwards) +

F= 2Kq2 cos 30º – 3Kq2 3µF A 5µF
a2 a2 B

–

= Kq2 ( 3 – 3) + (13) (C). 1µF
a2 4µF
+

Kq2 24V
a2
= – 1.27 Potential difference acrossAand B

Therefore every charge situated at corners VAB  3  24  8V
9
will experience a force towards O. So all

the three charges will move in the inward Energy stored in 1µF capacited

direction. = 1 × 1µF (8V)2 = 32µJ
(10) (A). Forminimum potentialenergyfirstlycharges 2

of larger magnitude should be kept at largest
distance. Therefore charges 2q and 8q (14) (D). Length of the diagonal of a cube having each

should be at the ends of the line of length 9 side b is 3 b. So distance of centre of
cm. If charge q is placed at a distance x
3 b
from the charge 2q then for minimum cube from each vertex is 2 .
potential energy the resultant force on it
must be zero.

117 ELECTROSTATICS

12th PHYSICS MODULE - 1

Hence potential energyof the given system E  Qr12
of charge is 40 R 4

U  8   1 . (q) (q)   4q2 (20) (A). The circuit contains one balanced Wheat-
40 3b/2 30b (21) stone bridge and capacitor in series.

(15) (D). V = 2x + 3y – Z (A). Initial charge on first capacitor is

   V  V ˆi  V ˆj  V kˆ  CV = Q1.
E  x y z  Initial charge on second capacitor is

  [2ˆi  3ˆj  kˆ ] which is constant. 2CV = Q2.
E Final capacitance of first capacitor is KC

(16) (C). Let K be the dielectric constant of solid If V' is the common potential then

capacitor and if C0 is capacitance of air V  Q1  Q2
capacitor then solid capacitor will have C1  C2
capacitance KC0. After charge sharing the
common potential becomes V.  V  CV  2CV  3V
KC  2C 2K
 V  C1V1  C2 V2
C1  C2

 V CV0  KC(0)  K  V0  V (22) (D). Uc  3 KQ q ; Us  KQ q
C  KC V 2 R R

 U  KQ q  1 . 1  4R 3 q  R 2q
2R 40 2R 3 60
(17) (D). P 1 Q
2
3 R
4 
5
6
7
8

12

P 3 Q (23) (B). For practical purposes, the earth is used as
4

87 65 a reference at zero potential in electrical

As encircled capacitor is short circuited circuits.
The electrical potential of a sphere of radius
Hence, CPQ  0A  0A  30A R with charge Q uniformly distributed on
d 2d 2d
Q
(18) (C). First capacity increases and then decreases. the surface is given by 40R .
So for constant supply voltage charge will
flow first fromAto B then from B toA. Hence, S-1 is True, S-2 is True; S-2 is NOT
a correct explanation for S-1
(19) (C). Charge enclosed in a radius

r1= r1 Q r.4r2dr  Qr14
0 R 4 R4

Using Gauss’s law, E.4r12   Qr14  1
 R4  0

118 ELECTROSTATICS