XII ELECTROSTATICS
1
Mother and daughter are both 1 Concept of Electric charges SKILLS TO MASTER 6
enjoying the effects of electrically
charging their bodies. Each and their conservation. Electric potential, potential
individual hair on their heads difference, electric potential
becomes charged and exerts a 2 due to a point charge,
repulsive force on the other hairs, a dipole and system of
resulting in the “stand-up’’ hairdos Coulomb’s law-force charges, equipotential
that you see here. between two point charges, surfaces.
forces between multiple
A portable defibrillator is being used in charges; superposition 7
an attempt to revive this heart attack principle.
victim. A defibrillator uses the Electrical potential energy
electrical energy stored in a capacitor 3 of a system of two point
to deliver a controlled electric current charges and of electric
that can restore normal heart rhythm. Electric field, electric field dipoles in an electrostatic
due to a point charge and field.
continuous charge
distribution. Electric field 8
lines.
Conductors and insulators,
4 free charges and bound
charges inside a
Electric dipole, electric conductor. Dielectrics and
field due to a dipole; torque electric polarization.
on a dipole in a uniform
electric field. 9
5 Capacitors & capacitance,
combination of capacitors
Electric flux, Gauss’s in series and in parallel,
theorem & its applications capacitance of a parallel
to find field due to infinitely plate capacitor with and
long straight wire, uniformly without dielectric medium
charged infinite plane sheet between the plates, energy
and uniformly charged thin stored in a capacitor,
spherical shell (field inside Van de Graaff generator.
and outside).
3
12th PHYSICS MODULE - 1
* – ve charged body Body has gained
1.1 INTRODUCTION electrons
* + ve charged body Body has lost some
* Electricity is all about charge. (Charge is the electrons
fundamental quantityof electricity.) * + ve & – ve charge named by Benjamin
* The classical study of electricity is generally Franklin.
divided into three general areas. Basic properties of
• Electrostatics (static electricity) : the study electric charge
of the forces acting between fixed arrangements
• of charge in space. (1) Charge is scalar and can be of two types
Electric current: thestudyof the forms of energy
(i.e. + ve or – ve). It adds algebraically.
associated with the flow of charge through
(2) Charge is conserved. During any process
circuits.
(chemical, nuclear, decay etc.) the net electric
• Electromagnetism: the study of the forces
charge of an isolated system remains constant.
acting between charges in motion.
* Electric charge (often just called charge) comes (3) Charge is Quantized (exists as discrete
"Packets") : Robert Millikan discovered that
in two and only two types : positive + and
electric charge always occurs as some integral
negative – .
multiple of fundamental unit of charge (e).
* The term neutral does not refer to a third type of
q = Ne [N is some integer]
charge, but to the presence in a region of positive
and negative charges in equal amount. Charge on a body can never be 13 e, 2 e
* The choice of assignment of positive to one type 3
of charge and negative to the other was etc. as it is due to transfer of electron.
completely arbitrary. Quark particles have charges equal to ± e/3 or
* Electrostatics deals with the studyof forces, fields ± 2e/3 but they are not stable in free state so we
and potentials arising from static charges. take electron only for quantization purpose.
1.2 CONCEPT OF (4) Through large number of experiments it is also
well established that similar charges repel each
CHARGE other while dissimilar attract. (Applicable for
* In nature, atoms are normally found with equal point charges).
numbers of protons and electrons, i.e. atom is
electrically neutral.
* The charge on an electron or a proton is the
smallest amount of free charge that has been Rubber Rubber
discovered.
* Charges of larger magnitude are built up on an F
object by adding or removing electrons.
–– –– – –– –– –
* If in a body there is excess of electrons over its FF
neutral configuration, conventionallythe body is Glass –– –– –– – Rubber
said to be negatively charged and if there is + + ++ ++ + F
deficiency of electron it is said to be positively
charged. (a) (b)
(a) A negatively charged rubber rod suspended by a thread is
attracted to a positively charged glass rod. (b) A negatively
charged rubber rod is repelled by another negatively charged
rubber rod.
4 ELECTROSTATICS
Here it is worth noting that true test of • 12th PHYSICS MODULE - 1
electrification is repulsion and not attraction as Practical units of charge :
amp × hr (= 3600 coulomb),
attraction mayalso take place between a charged Faraday ( = 96500 coulomb)
Charge on 6.25 × 1018 electrons = – 1 C
and uncharged body. •
(5) Charge is always associated with mass i.e.
charge can not exist without mass though mass Methods of Charging
can exist without charge.
(6) Charge is transferable. Process of charge transfer (A) Triboelectricity:
is called conduction.
The rubbing process serves only to separate
(7) Charge is invariant i.e. it is independent on frame
electrons already present in the materials. No
of reference.
electrons or protons are created or destroyed.
(8) Charge at rest produces Electric effect
Materials acquiring charges on rubbing together
Charge in unaccelerated motion produces
+ ve charged – ve charged of
Electric and magnetic fields.
(Vitreous) (Resinous)
Accelerate charge particle Electric &
Glass rod Silk Cloth
magnetic effect + radiate energy (According to
Fur or wool Ebonite or Rubber
electromagnetic theory)
or Amber
(9) Charge resides on the outer surface of a
Wool Plastic
conductor.
Dry hair Plastic comb
(a) (b)
For a conductor, charge For an insulator,
can move freely and find charge cannot
an equilibrium distribution. move easily.
Figure : (a) Charge placed on a conductor spreads Figure : A charged comb attracts bits of paper
out over the entire conductor. (b) Charge placed on because charges in molecules in the paper are
an insulator stays at its original place. realigned.
(10) How to express charge : Conduction :Transfer bycontact with an already
charged object.
• The SI unit for measuring the magnitude of an (B) Induction: When a charged particle is taken near
a neutral metallic object then the electrons move
electric charge is the coulomb (C). to one side and there is excess of electrons on
that side making it negatively charged and
• Current drift of charge per unit time (C) deficiency on the other side making that side
I = q/t q = It positively charged. Hence charges appear on two
1 coulomb 1 ampere × 1 sec sides of the body (although total charge of the
body is still zero). This phenomenon is called
If a charge of 1 coulomb drift per second through induction and the charge produced by it is called
induced charge.
cross-section of conductor, current flowing is
called 1 ampere.
• Charge on electron = – 1.6 × 10–19 C,
• Charge on proton = + 1.6 × 10–19 C
• 1 Coulomb = 3 × 109 esu of charge
(static coulomb or frankline)
= 1 emu of charge
10
5 ELECTROSTATICS
12th PHYSICS MODULE - 1
Ebonite rod Metal Charged
conductor glass rod
Metal sphere Rubber
Insulated stand collar Induced
(a) (b) negative
(a) Electrons are transferred by rubbing charge
the negatively charged rod on the metal sphere.
(b) When the rod is removed, the electrons Induced
distribute themselves over the surface of positive
the sphere. charge
Gold Case
leaf
Stem
Figure : Gold Leaf Electroscope
– –– – –– Ebonite rod ––– ––– ––– –––
––– – –– ++ + +–––––– sphere +++ + + EXAMPLE 1
+ +
++ – Metal ++ –– ++
Grounding + + When a piece of polythene is rubbed with wool,
wire a charge of – 2 × 10–7 C is developed on
polythene. What is the amount of mass, which is
Insulated stand transfered to polythene.
Connection
to ground
(a) (b) (c)
(a) When a charged rod is brought near the metal sphere SOLUTION:
without touching it, some of the positive and negative charges From Q = ne, So, the no. of electrons
in the sphere are separated. (b) Some of the electrons leave the
sphere through the grounding wire, with the result Q 2 107
(c) that the sphere acquires a positive net charge. e 1.6 1019
n = 1.25 × 1012
* Induced charge can be lesser or equal to inducing Now mass of transfered electrons
charge (but never greater) and its maximum value = n × mass of one electron
is given byq' = – q (1 – 1/k), where q' is inducing = 1.25 × 1012 × 9.1 × 10–31
charge and k is the dielectric constant of the = 11.38 × 10–19 kg
material of the uncharged body.
For metals k = q' = – q. EXAMPLE 2
Detecting charge 1012 – particles (Nuclei of helium) per second
falls on a neutral sphere, calculate time in which
* Charge can be detected and measured with the sphere gets charged by 2C.
help of gold-leaf electroscope, voltameter,
ballistic galvanometer. S O LUTION: – particles falls in t second = 1012t
Number of
* Gold leaf electroscope consist of two gold leaves Charge on – particle = +2e ,
attached to a conducting post that has a So charge incident in time t = (1012t).(2e)
conducting disc ball on top. The leaves are Given charge is 2 C
otherwise insulated from the container. 2 × 10–6 = (1012t).(2e)
If a charged body is brought near to it, charge
on the ball of electroscope will be opposite to t 1018 6.25 s
that of body & on leaves similar to that of body 1.6 1019
and leaves will diverge.
6 ELECTROSTATICS
12th PHYSICS MODULE - 1
Checkup 1 Q.6 Three objects are brought close to each other,
two at a time. When objectsAand B are brought
together, they repel. When objects B and C are
Q.1 The planets in the Solar System exert large brought together, they also repel. Which of the
gravitational forces on each other, but only following are true? (a) ObjectsAand C possess
insignificant electric forces. The electrons in an charges of the same sign. (b) Objects A and C
atom exert large electric forces on each other, possess charges of opposite sign. (c)All three of
but onlyinsignificant gravitational forces. Explain the objects possess charges of the same sign.
this difference. (d) One of the objects is neutral. (e) We would
need to perform additional experiments to
Q.2 A stone of mass 1.0 kg rests on the ground.What determine the signs of the charges.
is the net electric force that the ground exerts on Q.7 Three objects are brought close to each other,
the stone?
two at a time. When objectsAand B are brought
Q.3 Suppose that in Fig., the atomic nuclei were together, they attract. When objects B and C
displaced toward each other, instead of away are brought together, they repel. From this, we
from each other.Would the net electric force conclude that (a) objects A and C possess
between the atoms be attractive or repulsive? charges of the same sign. (b) objects A and C
possess charges of opposite sign. (c) all three of
Like charges are the objects possess charges of the same sign.
closer on average. (d) one of the objects is neutral. (e) we need to
perform additional experiments to determine
information about the charges on the objects.
nucleus 1.3 COULOMB’S
LAW
Q.4 Six electrons are added to 1.0coulomb of positive *
charge. The net charge is approximately Force between two point charges (interaction
force) is directly proportional to the product of
(A) 7.0 C (B) –5.0 C magnitude of charges (q1 and q2) and is inversely
proportional to the square of the distance
(C) 1.0 C (D) –6e between them i.e., (1/r2).
This force is conservative in nature.
Q.5 If you rub an inflated balloon against your hair, * This is also called inverse square law.
the two materials attract each other, as shown in The direction of force is always along the line
Figure. Is the amount of charge present in the * joining the point charges.
system of the balloon and your hair after rubbing
(a) less than, (b) the same as, or (c) more than F
the amount of charge present before rubbing?
–
–q
rF q
…as does the
repulsive force
for like charges.
F (a) – q' (b)
+ q' F
For unlike charges, the attractive
force of each lies along the line
joining the charges...
Figure : (a) Two charged particles q and q', one of which is positive and
one of which is negative. (b) Two other particles q and q'; both are negative.
7 ELECTROSTATICS
12th PHYSICS MODULE - 1
* If two point charges q1 and q2 at rest are Y
separated by a distance r in vacuum, the
magnitude of force between them is given by F12
q1
F q1q2 ; F k q1q2
r2 r2 + r1–r2
r1 + q2
where k is a constant. r2 F21
OX
k= 1 = 9 × 109 N-m2 /C2
40
0 = permittivity of free space Z
= 8.85 × 10–12 C2/N-m2
* Dimension of 0 = M–1L–3T4A2 Similarly, electric force on q2 due to charge q1 is
* If the point charges are kept in some other
medium (saykerosene)then Coulomb’s law gives F21 1 r2q1qr21 r2 r1
40
F q1q2 | |3
4 0 r r 2
Here q1 and q2 are to be substituted with sign.
where r is known as relative permitivity of the Position vector of charges q1 and q2 are
medium and is dimensionless. (Note r> 1). r1 x1ˆi y1ˆj z1kˆ and r2 x2ˆi y2ˆj z2kˆ
r is also denoted by K and known as dielectric
respectively. Where (x1, y1, z1) and (x2, y2, z2)
constant of the medium. Also note that the are the co-ordinates of charges q1 and q2.
pSeor,mritoivriKtyionfdtihceatmesedthiuatmthweiflol brceeb0etrw. een two If F1 and F2 are the forces acting on a charge Q
point charges in a medium is decreased K times * due to two other charges, then the resultant force
compared to the forces between them in vacuum. on charge Q is : F2 = F12 + F22 + 2F1F2 cos
where is the angle between F1 and F2 .
So, r or K Force in vacuum
Force in the medium
If F1 and F2 are mutually perpendicular then
(between two point charges separated by a
particular distance). F2 = F12 + F22 or F F12 F22
* For vaccum/air, K 1
* For water, K = 80
* For Mica, K = 7 to 10 Superposition Theorem
* For metal, K =
Coulomb’s Law in The interaction between any two charges is
Vector Form
independent of the presence of all other charges.
* Suppose the position vectors of two charges q1 Electrical force is a vector quantitytherefore, the
and q2 are r1 and r2 , then, electric force on
net force on any one charge is the vector sum of
the all the forces exerted on it due to each of the
charge q1 due to charge q2 is, other charges interacting with it independentlyi.e.
1 r1q1qr22 r1 r2 Net force on charge q, F F1 F2 F3 ......
F12 40
| |3
8 ELECTROSTATICS
12th PHYSICS MODULE - 1
(a)
Net force on q is obtained y F2 due to –Q
by parallelogram method. +Q is attractive… …and F1 due to
+Q is repulsive.
F1 + F2 d/2 θ r q
x
x
F2 d/2 θ
Force on q
due to q2. F1 –Q F2 F1
Force on q F1 F2
due to q1.
q
q1 (b)
y The x components
cancel…
pq2Foiinegxut ercrehta:erTlgewectoqr.ipcTofihonertcqnce2hestaFfro1greacsneqdo1nF2q oisn – F2 sin θ F1 sin θ x
θθ
and the vector sum of these forces. F2
the – F2 cosθ F1
– F1 cos θ
Golden Tips …and the y
components add.
* True test of electrification is repulsion and not Figure : (a) The charges +Q and –Q
attraction as attraction may also take place exert forces F1 and F2 on the charge q.
between a charged and an uncharged body and The net force is the vector sum F1 + F2.
also between two similarly charged bodies. (b) The x and y components of F1 and F2.
* In case of irregular conducting body charge SOLUTION:
density is not uniform. It is maximum where the As illustrated in Fig. a, the charge +Q produces
radius of curvature is minimum and vice - versa. a repulsive force on the charge q, and the charge
This is why charge leaks from sharp points. –Q produces an attractive force. Thus, the vector
F1 points away from +Q, and the vector F2
* From coulomb's law in vector form it follows that points toward –Q. From the geometry of Fig. a,
the electrostatic force is central in nature and the the distance from each of the charges +Q and
two charges exert equal & opposite forces on
each other i.e. forms action - reaction pair. –Q to the charge q is r x2 d2 / 4 .
* The force between any two charges is not Hence,the magnitudes ofthe individual Coulomb
affected by presence of another charges. forces exerted by +Q and –Q are
EXAMPLE 3 F1 F2 1 qQ 1 qQ ... (1)
40 r2 40 x2 d2 / 4
Two point charges +Q and –Q are separated by
a distance d, as shown in Fig.. A positive point (magnitudes)
charge q is equidistant from these charges, at a
distancex from theirmid point.Whatis theelectric From Fig. (b) we see that in the vector sum
force F on q?
F = F1 + F2, the horizontal components
(x components) of F1 and F2 cancel, and the
vertical components (y components) add, giving
a net vertical component twice as large as each
individual vertical component. Thus in this case
the net force F has a y component but no
x component.
9 ELECTROSTATICS
12th PHYSICS MODULE - 1
In terms of the angle shown in Fig. (b), the EXAMPLE 5
y component of F1 is ––FF21ccooss.,Sainndcethtehese
y component of F2 is are If three equal charges (Q each) are placed on
the vertices of an equilateral triangle of side a.
equal, the net force is then The resulstant force on any one charge due to
F = Fy = –F1 cos – F2 cos = –2F1 cos the other two is _______.
2 1 x2 qQ cos SOLUTION:
40 d2 /4 The charges are shown in figure.
From Fig. (a), we see that F2 F F1
Q + 30°
cos 1 d and therefore
2 aa
(x2 d2 / 4)1/2
Q+ a +Q
F Fy 1 (x2 qQd The resultant force
40 d2 / 4)3/2
F F12 F22 2F1F2 cos 60
Comments : Note that if the charge q is at a with F1 = F2 = kQ2/a2
large distance from the two charges ±Q, then d2
can be neglected compared with x2, so F 3kQ2
(x2 + d2/4)3/2 (x2)3/2 = x3. a2
The force F is then proportional to 1/x3; that is,
the force decreases in proportion to the inverse From symmetry the direction is shown along
cube of the distance. Thus, although the force
y-axis.
contributed by each point charge ±Q is an
inverse-square force, the net force has a different EXAMPLE 6
behavior, because at large distances the force
contributed by one charge tends to cancel the Four charges Q, q, Q and q are kept at the four
force contributed by the other. corners of a square as shown. The net force on
a charge q is zero. Find q/Q.
EXAMPLE 4
Q
+q
The smallest electric force between two charges SOLUTION: q +
placed at a distance 1 m is _______. Q
SOLUTION: Both the q will have same sign either positive or
Fe 1 . q1q2 .............. (i) negative.Similarlyboth theQwillhavesamesign.
40 r2
Let us make the force on upper right corner q
For Fe to be minimum q1 1q.26sh×o1u0ld–1b9e minimum. equal to zero.
(q1)min = (q2)min = e = C
Lower q will apply a repelling force F1 on upper
Substituting in Eq. (i), we have q (because both the charges have same sign).
(9.0 109 ) (1.6 1019 ) (1.6 1019 ) To balance this force both 'Q' must apply
(1.0)2
(Fe)min =
attractive forces F2 and F3 of equal magnitude
= 2.304 × 10–28 N. (So, Q and q will have opposite signs).
10 ELECTROSTATICS
12th PHYSICS MODULE - 1
Now the resultant of and will be and 4F and 2F are exactly opposite to each other
F2 F3 so its effect will be 2F towards 2Q as shown in
the figure. So resultant force will be (Pythagoras)
F2 2 (Pythagoras theorem) and it will be
of 2F and 2F equal to 2 2 F .
exactly opposite to F1 and same in magnitude. Now, F is basically force between Q and q so
From Coulomb's Law
Q F2 F1
F1 kq2 , F2 kQq + –q that is F 40 Qq 2)2
(d 2)2 d2 d2 (d /
F3
Note that the distance between Q and q is
and F1 2 F2 q– d + Q
Q F1 (d / 2) as the side of the square is d.
+ –q
q2 2Qq So, final answer is 2 2F
(d 2)2 d2 2F 2
2 2F 2 2 Qq 4 2 Qq 2F 2F
40(d / 2)2 40d2
Q q q– + Q
22 Direction is upwards.
But Q and q must have opposite sign so,
q 2 2 Q EXAMPLE 8
EXAMPLE 7 Five point charges, each of value +q are placed
on five vertices of a regular hexagon of side L m.
The charges on the four corners of the square The magnitude of the force on a point charge of
are Q, 2Q, 3Q and 4Q respectively as shown in value –q coulomb placed at the centre of the
the figure then find the force on the charge q kept hexagon is _____.
at the centre of a square.
SOLUTION:
Q + 2Q If there had been a sixth charge +q at the
remaining vertex of hexagon force due to all the
+ six charges –q at O will be zero (as the forces
due toindividual charges will balance each other),
+q d
ED
++ q + +q
4Q 3Q
SOLUTION: F+ O q
If Q applies a force F, then 2Q will apply 2F on q +C
q. Because as per Coulomb’s law the force is –q –
proportional to charge, if rest of the factors are
same. A +q
Similarly 3Q will apply 3F and 4Q will apply 4F FR B
on q. The four forces along with its direction are
shown below. i.e., 0
Now if f is the force due to sixth charge F due
Q 2Q Q 2Q
+ 4F + to remFainfing f0ivei.ceh.aFrges, f
3F + F d ++
or F f 1 qq 1 q 2
2F 2F + 2F 40 L2 40 L
q
++
++
4Q 3Q
4Q 3Q
We can see, 3F and F are exactly opposite to
each other so its net effect will be 2F towards Q
11 ELECTROSTATICS
EXAMPLE 9 (ii) 12th PHYSICS MODULE - 1
(iii)
A thin straight rod of length carryinga uniformly Unstable Equilibrium : If charge is displaced
distributed charge q is located in vacuum. The by a small distance from its equilibrium position
magnitude of the electric force on a point charge and the charge has no tendency to return to the
Q kept as shown in the figure is ______ same equilibrium position then equilibrium is
called unstable equilibrium. Instead it goes away
+Q from the equilibrium position.
Neutral Equilibrium : If charge is displaced by
a a small distance and it is still in equilibrium
codition then it is called neutral equilibrium.
SOLUTION:
As the charge on the rod is not point charge, Equilibrium of point
charge
therefore, first we have to find force on charge
Q, due to charge over a very small part on the
length of the rod. This part, called element of (i) Main charges must be of like nature
(ii)
length dy, can be considered as point charge. x +
dy y P + Q1 q Q2
r
+ Q
a Third charge should be of unlike nature
Charge on element, dq = dy = q dy x Q1 Q1 Q2 r and q ( Q1Q2
Q1 Q2 )2
Electric force on Q due to element
= kdqQ kQqdy Equilibrium of symmetric
y2 y2 geometrical point charge
system
All forces are along the same direction, q q+ a +q
F = dF. This sum can be calculated using a Qa
+
integration,
aa
therefore, F =a kQqdy kqQ 1 aa Q
ya y2 y
+ a q+ q + a + q
q
kQ.q 1 a 1 a kQq Value of Q at centre for which system to be in
a (a ) state of equilibrium
(i) For equilateral triangle Q = q
3
1.4 ELECTROSTATIC (ii) For square Q = q (2 2 1)
EQUILIBRIUM 4
The point where the resultant force on a charged Equilibrium of suspended
particle becomes zero is called equilibrium point charge system
position.
(i) Stable Equilibrium : A charge is initially in For equilibrium position
equilibrium position and is displaced by a small
distance. If the charge tries to return back to the T cos = mg and T sin = Fe = kQ2
same equilibrium positionthen this equilibrium is x2
called position of stable equilibrium.
12 ELECTROSTATICS
tan = Fe kQ2 12th PHYSICS MODULE - 1
mg x 2 mg
If q' is displaced along x-axis towards q then it
will move in the same direction, hence, unstable
equilibrium.
If is small Tcos
Q
sin = x T EXAMPLE 11
2 +
tan Fe Q A simple electroscope for the detection and
+ Tsin
x kQ2 x measurement of electric charge consists of two
2 x 2 mg small foil-covered cork balls of 1.5 ×10–4 kg
so mg
each suspended by threads 10 cm long (see
1/3 Fig.).When equal electric charges are placed on
x3 2kQ2 x Q2 the balls, the electric repulsive force pushes them
mg 20mg
apart, and the angle between the threads indicates
the magnitude of the electric charge. If the
If whole set up is taken into an artificial satellite equilibrium angle between the threads is 60°,
(g ~ 0) what is the magnitude of the charge?
kq2 2 (a)
42 180°
T= Fe = + + Electric force …causing threads
pushes balls to make this
q q apart… equilibrium angle.
EXAMPLE 10 30° 30°
Two point charge +4q and +q are placed at a r
distance Lapart.Athird charge q can be placed
such that all the three charges are in equilibrium SOLUTION:
then find q' and its position. Figure b shows a “free-body” diagram for one
of the balls. The electric force F acts along the
SOLUTION: line joining the two charges and is thus horizontal.
Position of q' will be nearer to charge of smaller
magnitude. Third charge should be placed We resolve the tension
between 4q and q so that force on third charge
to be zero (let at distance x from 4q). Third (b) into components.
charge should be –ve (let –q') for the equilibrium
of other charges. T sin 30°
T 30° T cos 30°
x L–x
+– +
4q –q' q
For equilibrium of third charge
k (4q) (q) k (q) (q) x 2L F
x2 (L x2 ) 3
In equilibrium, vertical
For equilibrium 4q components of forces
sum to zero, as do
k (4q) (q) k (4q) (q) q 4q horizontal components.
(2L / 3)2 L2 9
w
Figure : (a) Two equal charged
balls suspended by threads.
(b) “Free-body” diagram for the right ball.
13 ELECTROSTATICS
12th PHYSICS MODULE - 1
In equilibrium, the vector sum of the electric When the ballsaresuspended in aliquid ofdensity
repulsion F, the weight w, and the tension T of and dielectric constant K, the electric force
the thread must be zero. Accordingly, the will become (1/K) times, i.e., F' = (F/K)
horizontal component of the tension must balance mwgh'e=remgis–dFeBns=itmy ogf–liVquigd]
the electric repulsion, and the vertical component while weight
of the tension must balance the weight: [as FB = Vg,
F = T sin 30° i.e., mg mg 1 as V m
mg = T cos 30°
We can eliminate the tension from the problem
by taking the ratio of these equations, yielding So, for equilibrium of ball,
From Fig. a we see that the distance between
the balls is r = 2 sin 30°, so Coulomb’s Law tan F Kmg F ( / )] ........ (4)
mg [1
tells us F 1 q2 According to given information ' = ,
40 (2 sin 30)2 so from eq. (4) and (3), we have,
Equating these two expressions for F, we find K (
)
1 q2
mg tan 30° = 40 (2 sin 30)2 Checkup 2
and q 40mg tan 30 2 sin 30 Q.1 Suppose that the electric force between two
= 3.1 ×10–8 C charges separated by a distance of 1 m is
1 ×10–4 N.What will be the electric force if we
increase the distance to (a) 10 m ; (b) 100 m?
EXAMPLE 12 Q.2 Two balls, separated by some distance, carry
Q.3 equal electric charges and exert a repulsive
Two identical charged spheres are suspended by Q.4 electric force on each other. If we transfer a
strings ofequal length. Each stringmakes an angle fraction of the electric charge of one ball to the
with the vertical. When suspended in a liquid other, will the electric force increase or decrease?
of density (), the angle remains the same. Find
the dielectric constant of the liquid. (Density of Two particles are separated by a distance of
the material of sphere is )
3.0m; each exerts an electric force of 1.0 N on
SOLUTION:
Initially as the forces acting on each ball are the other. If one particle carries 10 times as much
tension T, Weight mg and electric force F. electric charge as the other, what is the magnitude
For its equilibrium along vertical of the smaller charge?
T cos = m g (A) 10 pC (B) 10 µC
........ (1) /////////////////////////// (C) 10 C (D) 10 kC
and along horizontal T Three identical point charges are at the vertices
T sin = F T of an equilateral triangle.Afourth, identical point
........ (2) charge is placed at the midpoint of one side of
the triangle.As a result of the three electric force
Dividing Eqn. F+ +F contributions from the vertex charges, the fourth
charge
(2) by (1), mg mg (A) Is in equilibrium and remains at rest.
(B) Is pushed toward the center of the triangle.
we have tan F ........ (3) (C) Is pushed outside the triangle
mg
14 ELECTROSTATICS
Q.5 Object A has a charge of +2µC, and object B 12th PHYSICS MODULE - 1
has a charge of +6µC. Which statement is true 1.5 ELECTRIC
FIELD
about the electric forces on the objects?
The physical field where a charged particle,
(a) FAB = –3FBA (b) FAB = –FBA * irrespective of the fact whether it is in motion or
(c) 3FAB = –FBA (d) FAB = 3FBA at rest, experiences force is called an electric field.
(e) FAB = FBA (f) 3FAB = FBA Theconcept ofelectric field was given byMichael
Faraday. Characteristics of electric field :
Q.6 Identical point charges are fixed to diagonally * (1) Electric field intensity(shortlywe will call
opposite corners of a square. Where does a third
point charge experience the greater force? electric field).
(a) At the center of the square. (2) Electric potential.
(b) At one of the empty corners. (3) Electric lines of forces.
(c) The question is unanswerable because the
polarities of the charges are not given.
Q.7 The drawing shows three point charges arranged
in three different ways. The charges are +q, –q,
and –q; each has the same magnitude, with one Electric field intensity E
positive and the other two negative. In each of
the arrangements the distance d is the same. Rank *
the arrangements in descending order (largest Definition of the E -field :
first) according to the magnitude of the net One obvious disadvantage of concentrating on
electrostatic force that acts on the positive charge.
force is that its magnitude at everypoint in space
depends not only on the primary charge
distribution, but also on the size of the test charge
+q –q –q * q0.
dd What we really want is a map showing the field
A
–q +q –q of a primary body independent of the detector, a
dd
B map that could be used to compute the force at
–q every point in space when any size charge is
placed there.
d * Regardless of its source, we define the electric
90°
+q field ( E ) at a point in space to be the electric
force experienced by a positive test-charge at
d –q
C
that point divided by that charge
F
Q.8 A particle is attached to one end of a horizontal q0 ........ (1)
spring, and the other end of the spring is attached E
to a wall. When the particle is pushed so that the
spring is compressed more and more, the particle Electric field has the Sl units of newtons per
experiences a greater and greater force from the *
spring. Similarly, a charged particle experiences coulomb (N/C).
a greater and greater force when pushed closer
and closer to another particle that is fixed in The presence of the charge q0 will generally
position and has a charge of the same polarity. change the original distribution of the other
Considering this similarity, will the charged particle
exhibit simpleharmonicmotiononbeingreleased, charges, particularly if the charges are on
as will the particle on the spring?
conductors. However, we may choose q0 to be
small enough so that its effect on the original
charge distribution is negligible.
ql0im0
E F
q0
15 ELECTROSTATICS
12th PHYSICS MODULE - 1
Field direction is radially E (r) = 1 (x2 q × (xˆi yˆj zkˆ)
outward for positive q'. 40 y2 z2 )3/2
Field vector for only one * The three rectangular components of ( r ) are
point is shown. E
Electric field E due to
point charge decreases E as follows :
with inverse square of
distance r. Ex( r ) = 1 (x2 q z2 )3/2 x ,
40 y2
r 1 q
40 y2 z2 )3/2
Ey( r ) = (x2 y
q' and Ez( r ) = 1 (x2 q z2 )3/2 z
40 y2
Figure : A charge q' generates an electric
field E at some distance r. The direction
of this electric field is along the radial line.
* Conversely, known at any location in space Electric field due to Discrete
E distribution of charge
(whatever the source) we can calculate the force Point charges placed at different position, use
vector approach (Better term : superposition rule)
F that would arise on any point charge q placed
at that location; accordingly
F qE ....... (2) n
E E1 E2 Ei
....... i 1
* Notice that F and E point in the same direction
when q is positive.
* A charge will never exert force on itself. with 1 qi
40 ri3
Electric field due to a Ei ri
point charge
EXAMPLE 13
* The electric field produced by a point charge q
can be obtained in general terms from Coulomb's If a charge q is placed at each vertex of a regular
law. polygon, the net electric field at its centre is ___.
* First, note that the magnitude of the force exerted q q
by the charge q on a test charge q0 is +q q+ + +
F = kqq0/r2. Then, divide this value E2 E1
by q0 to q+ q
E2 E1
+
obtained the magnitude of the field. E3 E5
E3 E2 E3 E4
* Sinceq0 is eliminatedalgebraicallyfrom theresult, E4 +
the electric field does not depend on the test E1 (b)
+ (a) + q+ q+ + (c) q
q q q
charge: Point charge q : E kq + q0 SOLUTION:
r2 P The distance of the centre of a regular polygon
* If (x, y, z) are the co-ordinates from ea|||cEEEh222ve||| rte|||xEEEw333 i|||llinb||e(EEas)44a.m|| ien.| Therefore,
| E1 | (b)
of the observation point P, then r | E1 |
r xˆi yˆj zkˆ | E1 | E5 | in (c)
Also, r==(x(x2 2++yy2 2++zz2)23)/12/2qO+ Source Charge
and r3
16 ELECTROSTATICS
12th PHYSICS MODULE - 1
The angle between any two consecutive field EXAMPLE 15
vectors for each of a polygon is also same. Consider two charges ±Q of equal magnitudes
and opposite signs, separated by a distance d.
Hence, E1 E2 E3 .... En 0 Find the electric field at a point equidistant from
the two charges, a distance x from their midpoint
where n = 3, 4, 5, 6........ (see Fig.).What is the dependence on distance
of this field for x >> d?
Note: E1 E2 ... En 0
... E n
E1 E2 En1
Hence, if a charge q is placed at each vertex y
+Q
except one vertex of regular polygon, then the r = x2 + (d/2) 2
θ x
net electric field at its centre, distant r from each
d E1
vertex is 1 q , directed towards or away –Q Px
40 r2
θ
from the empty vertex depending on whether q E E2
is positive or negative. E1 points …and E2 points
toward –Q… awayfrom + Q.
EXAMPLE 14 Resultant has negative y
component,but no x component.
Two positive charges Q1 and Q2 are placed on Sol. The magnitude of each of the two individual
a line as shown in figure. The position of point P,
where the net electric field = 0 is given by electric fields is E1 = E2 = 1 Q , where
40 r2
x = ________ Q1 P Q2
+ +
xR
SOLUTION: r x2 (d / 2)2 .
Let position of P is at a distance x from Q1. Then
the fields at P due to Q1 and Q2 are in opposite At the point P, the electric field due to the +Q
directions. They will add up to give zero, only if charge points away from this charge, and the
their (electric field's) magnitude are equal. That electric field due to the –Q charge points toward
that charge, as shown in Fig. The x components
is kQ1 kQ2 of the electric field at the point P in Fig. cancel,
x2 (R x)2 and the y components are equal, giving
R x Q2 R E Ey 2E1 cos
x Q1 Q2 / Q1
or x = 1 2 1 Q d/2
40 d2 x2 d2 / 4
The distance of point P from charge Q is x2 /4
d = R – x = 1 R 1 (x2 Qd 4)3/2
Q1 / Q2 40 d2 /
If two negative charges are placed on a line For x >> d we can neglect d2 compared with x2
(instead of positive charges), then the position of and obtain
point P where the net electric field is zero, is again E 1 Qd .
40 x3
x = R / 1 Q2 / Q1 ; d = R / 1 Q1 / Q2 .
17 ELECTROSTATICS
12th PHYSICS MODULE - 1
Electric field due to z dE
continuous distribution dEcos
of charge
E-field at point P due to dq:
1 . dq .rˆ P
dE 40 r2
E-field due to charge distribution
P z
r r
++ ++ + ++ + +++
d
++ + + + +++++
Charge distribution R
ds
(2) For each element of length ds, charge
1 dq +++++ +++++
dE 40 r2
E . .rˆ
Volume Volume
(1) In many cases, we can take advantage of the dq . ds
symmetry of the system to simplifythe integral.
(2) To write down the small charge element dq: Linear Circular
1-D, dq = ds ; charge density length element
= linear charge density,
ds = small length element dq =.R d, where is the angle measured on
2-D, dq = dA ; the ring plane.
= surface charge density,
dA = small area element Rd
3-D, dq = dV ; ||
= volume charge density, ds
dV = small volume element
d
R
Let us consider some cases : Net E-field along z-axis due to dq :
Case 1 : Charged ring 1 dq
40 r2
E-field at a height z above a ring of charge of dE . .cos
radius R
NOTE : We are deriving the expression only to Total E-field = dE
understand mathematical approach. For other 2 1
0 40
cases you can remember the expression. We can = . Rd . cos cos z
r2 r
derive some cases with the help of Gauss’s law.
(1) Symmetry considered: For every charge
element dq considered, there exists dq' due to 360° = 2 radian, 180° = radian
Note : Here in this, , R and r are fixed as
which the horizontal E field components cancel. varies but we want to convert, r, to R, z.
Overall E-field lies along z-direction.
18 ELECTROSTATICS
12th PHYSICS MODULE - 1
E 1 . Rz 2 (2) For each element of length dz,
40 r3 charge dq = dz
d
Horizontal E-field at point P due to element dz
0
410.rd2z
E 1 . (2R )z | dE | cos cos
40 (z2 R2 )3/
2 along z-axis dEdz
But : (2R) = total charge on the ring. After integration
E 1 . (z2 Qz E 1 L along x-direction
40 R 2 )3/2 40
2 L 2
2
x x
NOTE Important limiting cases :
* At the centre of the ring z = 0, so E = 0
* For z >> R the ring behaves as a point charge. L
x2
In this case, E 1 1. x >> L : E 1
z2 40
* For z << R the value of E is given by But L = Total charge on rod
System behave like a point charge
E 1 Qz i.e. E z
40 R3 1 L
2. L >> x : E 40
L
* E will be maximum at z R . x. 2
2
An opposite charge kept far off from the centre Ex
20x
on the z-axis (ring axis) will execute oscillatory
and periodic motion but if it is kept very close to
the centre then it will execute SHM. Case 3 : E-field from a disk of
surface charge density
Case 2 : Uniform line of charge
We find the E-field of a disk by integrating
Charge per unit length = concentric rings of charges.
z
dE
+dz P E–dz P
z x z
L
E+dz
–z
r
dr r R
(1) Symmetry considered: The E-field from +z and ELECTROSTATICS
– z directions cancel along z-direction,
Only horizontal E-field components need to be
considered.
19
Total charge of ring 12th PHYSICS MODULE - 1
dE the electric field at P due to the small element.
dq .(2rdr) dE kdq . Here, dq Q dx
(x2 r2) L
Area of the ring
E-field from ring : dEcos
dE 1 (z2 dq z dE
40 r2 )3/2 P dEsin
After integration, E z2 z R2
20 1
r
Very important limiting case ++++++++++++++++++++++
If R >> z, that is if we have an infinite sheet of dx x
charge with charge density :
After integration,
E = 20
kQ (cos 2 cos 1)
++ + + In x-direction, Ex Lr
+
+ + + Ey kQ (sin 1 sin 2)
++ + Lr
++ In y-direction,
Figure : E-field due to an infinite Case 5 : Electric field due to a
sheet of charge, charge density = . charged circular arc at centre
E 1 z R2 1 z Figure shows a circular arc of radius R which
20 z2 20 R subtend an angle at its centre.
E Rd
20
X + ++
E-field is normal to the charged surface. ++ + + d
Case 4 : Electric field due to ++++
a uniformly charged rod
Consider P as anygeneral point in the surrounding +
of rod, to find electric field strength at P, consider
an small element on rod of length dx at a distance R
x from point O.
C
P dEsin dEcos
dE Y
1 2
r
++++++++++++++++++++++ To find electric field strength at C, consider a
O small segment on arc of angular width d at an
angle from the angle bisector XY as shown.
L
20 ELECTROSTATICS
12th PHYSICS MODULE - 1
The length of elemental segment is R d, the E
charge on this element dq is dq = Q .d . Graphically : kQ/R2 E 1/r2
Due to this dq, electric field at centre of arc C is r=R r
given as dE kdq
R2
After integration, EC 2kQ sin
R 2 2
R
* For semi-circular ring = .
* E 2kQ
R 2
Figure : The electric field inside a
Case 6 : Spherical uniformly charged spherical shell is zero.
distribution of charge
(b) Volume distribution of charge :
(i) Conducting sphere (Hollow, solid)
(Solid non-conductor)
(ii) Non-conducting sphere (Hollow, solid)
Inside r < R : Uniform volume distribution :
Case (a) Charge on surface.
Charge inside volume Q +++++++R+++++r++++++++
Case (b) Volume distribution of charge.
4 Q 4
(a) Charge on surface : 3 r3 4 R3 3 r3 = Q'
3
Hollow/solid conductor or hollow non-
coductor : Imagine a sphere passing through
desired point (point where E is to be calculated), Qr3 kQ kQ
or Q' = R3 ; E = r 2 = R 3 r
calculate charge inside it and assume it to be
concentrated at centre and use point charge
+ ++
formula. Q+ + +
Inside r < R : E = 0
++ Inside sphere : E 30 rˆ
(No charge inside imagined + rR +
+
sphere) + + = Volume charge density
Q
+ + ++ +++++++++R+++++r+++
kQ +Q+ + + Outside sphere : E kQ : r R
r2 r2
Outside r > R : E + r +
++
+++
+ + R Surface E= kQ
+ R2
++
kQ
Surface r = R : E R2
E r E
E 1/r 2
Graphically :
r=R r
21 ELECTROSTATICS
12th PHYSICS MODULE - 1
1.6 TABLE : ELECTRIC FIELD INTENSITIES DUE TO VARIOUS
CHARGE DISTRIBUTIONS
Name/Type Formula Particular Graph
Point charge E
kq .rˆ kq r qr is source charge. source charge
+ | r |2 r3 is vector drawn from r
E
Infinitely long to the test point.
line charge r
Electric field is nonuniform, radially
E
outwards due to + charges & inwards 20
due to – charges. r
rˆ 2k rˆ is a linear charge density
2 0 r r (assumed uniform)
r is perpendicular distance of point
from line charge.
rˆ is radial unit vector drawn from the
charge to test point.
Semi-infinite 2k , Ex k , E y k At a point above the end of wire at an
r r r angle 45°
Finite line of Ex k [sin sin ] Where is the linear charge density
charge r
k
+++++++++ x Ey r [cos cos ]
P E If =
E
E|| 0, E
20r
Infinite non- nˆ is surface charge density
conducting 20 (assumed uniform)
thin sheet
nˆ is unit normal vector.
Electric field intensity is
independent of distance.
Uniformly charged Q is total charge of the ring.
ring x = distance of point on the axis
from centre of the ring.
E kQx Electric field is always along the axis.
(R 2 x2 )3/2
Maximum at x R / 2
Ecentre 0
22 ELECTROSTATICS
12th PHYSICS MODULE - 1
Infinitely large nˆ is the surface charge.
charged conducting 0
sheet nˆ is unit normal vector
perpendicular to the surface.
Electric field intensity is
independent of distance.
Uniformly charged hollow (i) for r R R is radius of the sphere.
conducting/ r is a vector drawn from centre
nonconducting/ kQ rˆ of sphere to the point.
solid conducting E | r |2
sphere
(ii) for r < R Sphere acts like a point charge,
++ + + ++
++ placed at centre for points
++ E
++ + + + + 0 outside the sphere.
E is always along radial direction.
Q is total charge ( 4 R 2 )
( = surface charge density)
Uniformly charged (i) for r R r is a vector drawn from centre
solid nonconducting of sphere to the point.
sphere (insulating kQ
material) E | r |2 rˆ Sphere acts like a point charge, placed
+++++++++++++++++++++++++++++ at centre for points outside the sphere.
(ii) for r R E is always along radial direction.
kQr r Q is total charge (. 4 4 R 3 )
E R3 3 0 3
( = volume charge density)
Inside the sphere E r
Outside the sphere E 1/r2
Uniformly charged cylinder with for r < R, Ein r
a charge density (R = radius 20
of cylinder)
for r > R, E R 2
R 20r
E
for r < R, Ein = 0, E 1/r
for r > R, E r
0r
Uniformly charged cylindrical
shell with surface charge den-
sity is
23 ELECTROSTATICS
12th PHYSICS MODULE - 1
1.7 MOTION OF CHARGED 1.8 ENERGY STORED IN
PARTICLE IN UNIFORM AN ELECTRIC FIELD
ELECTRIC FIELD
* If the intensity of an electric field is E, then the
= Effective lenght of electic field energystored in that electric field per unit volume
L = Distance of screen from centre of electric
i.e., the energy density U 1 0E2
field 2
Ions enter horizontally. Uniform electric field * The energy associated with a spherical charge
causes projectile
motion between plates.
e v0 E v distribution of radius R is, U q2 .
θ 80R
After exiting, ions Golden Tips
move with constant
velocity at angle θ.
Figure : Motion of a charged particle; between * Electric field inside a solid conductor is always
the parallel plates is a region of uniform electric zero.
field.
* Electric field inside a hollow conductor may or
Particle at time t may not be zero (E 0 if non zero charge is
For x direction x = vt inside the sphere).
For y direction y = uyt + 1 ayt2 * The electric field due to a circular loop of charge
2
and a point charge are identical provided the
uy = 0 ; t = x ; ay = qE distance ofthe observation point from the circular
v m loop is quite large as compared to its radius i.e.
x >>> R.
y = 0 + 1 qE x 2 qE * Charged particle in an electric field always
2 m v 2mv2 experiences a force either it is at rest or in motion.
or y= x2 * In presence of a dielectric, electric field decreases
This is equation of parabola.
Special Results
1. Time taken by the particle to cover length of and becomes 1 times of its value in free space.
electric field, in other words, time for which r
particle moves under the influence ofelectric field
EXAMPLE 16
T= v Can a sphere of radius 1 cm hold a charge of
2. Total deviation in the trajectoryunderelectric field 1 coulomb in air ?
1 qE 2 SOLUTION:
2 m v Electric field at the surface of the sphere.
y= kQ 9 109 1 = 9 × 1013 V
R2 (1102 )2 m
3. For proton , duetron , & –Particle E
(When K. E. is same) y q
For Proton and deutron track same This field is much greater than the dielectric
4. For electron and proton if K.E. same then strength of air (3 × 106 V/m), the air near the
sphere will get ionised and charge will leak out.
y-same. Thus a sphere of radius 1 cm cannot hold a charge
Curvature, sharpness of curve same of 1 coulomb in air.
24 ELECTROSTATICS
12th PHYSICS MODULE - 1
EXAMPLE 17 (iii) r R2 is k (Q1 Q 2 ) ˆr
r2
Three large conducting parallel sheets are placed
at a finite distance from each other as shown in SOLUTION:
figure. Find the electric field intensityat pointsA (i) For r < R1,
and B. Q –2Q 3Q therefore, point lies inside both the spheres
y (ii) Enet = Einrn<erR+2 Eouter = 0 + 0 inner
x For R1 . Point lies outside
AB sphere but inside outer sphere:
Enet = Einner + Eouter
kQ1 rˆ 0 kQ1 rˆ
r2 r2
(iii) For r R2,Point lies outside inner as well
SOLUTION: as outer sphere.
For pointA :
Therefore, Enet = Einner + Eouter
Q –2Q 3Q
kQ1 rˆ kQ2 rˆ k (Q1 Q2 ) rˆ
E–2Q r2 r2 r2
EQ E3Q A
Q.1 Checkup 3
E net EQ E3Q E2Q
A test charge of +3µC is at a point P where an
Q ˆi 3Q ˆi 2Q ˆi Q ˆi external electric field is directed to the right and
2A0 2A0 2A0 A0 has a magnitude of 4 × 106 N/C. If the test charge
For point B : is replaced with another test charge of –3µC,
Q –2Q 3Q the external electric field at P (a) is unaffected
(b) reverses direction (c) changes in a way that
E3Q E–2Q Q.2 cannot be determined.
B EQ
Given that an electric field of 10 N/C has a
northward direction and another electric field of
E net E3Q E 2Q EQ 10 N/C has an eastward direction, what are the
magnitude and direction of the superposition of
3Q ˆi 2Q ˆi Q ˆi 0 these two electric fields?
2A0 2A0 2A0
Four equal positive charges are located at the
EXAMPLE 18 Q.3 corners of a square.What is the magnitude of the
Q.4 electric field at the center of the square?
Twoconcentricuniformlycharged spherical shells
Three equal positive charges are located at three
of radius R1 and R2 (R2 > R1) have Q2 corners of a square, and a negative charge is
total charges Q1 & Q2 Q1 located at the fourth corner.What is the direction
respectively. Electric field at (if any) of the electric field at the center of the
R1 R2 square?
(i) r < R1 is 0 kQ1
r2
(ii) R1 r < R2 is rˆ
25 ELECTROSTATICS
12th PHYSICS MODULE - 1
(A) Toward the negative charge. * 1.9 ELECTRIC LINES
(B) No direction, since field is zero. OF FORCES
(C)Away from the negative charge.
(D) Perpendicular to the diagonal through the The concept of electric lines of forces was
introduced by Michael Faraday.
negative charge. The magnitude of electric field strength at any
point is measured by the number of electric line
Q.5 Two infinitelylarge flat sheets with equal uniform * of force passing per unit small area around that
positive charge distributions intersect at right point normally and the direction of field at any
angles (see Fig.).What is the direction of the point is given by the tangent to the line of force at
electric field in each quadrant? the point.
An electric line of force is that imaginarysmooth
* curve drawn in an electric field along which a
free isolated unit positive (initially at rest) charge
moves.
Q.6 A small disk (approximate diameter 1 cm) has a at any point on a line of force.
charge Q uniformly distributed over its area E
A.What is theelectric field far from this disk (e.g.,
meters away)? What is the field very close to its Properties
surface (e.g., 10–3 cm) and far from the edges? (1)
The lines of force diverge out radially from a
Q.7 The electric field at a distance of 1 m from a + ve charge and converge at a – ve charge.
uniformly charged infinite sheet has the value
E = E0. What is the value of E at a distance of Electric field vectors
2m? At 4 m? are shown for a few
representative points.
Q.8 Suppose that two large parallel sheets have equal
surface charge densities with the same sign,
say, positive.What is the magnitude of the electric
field in the space between the sheets? Outside q
the sheets? ((DB))0,/2/0,0/20 Field vectors point away
//00,, from positive charge.
(A) 0
(C) /0 Magnitude of field vector
decreases with inverse
Q.9 A charged particle moves along a straight line. square of distance.
Does this mean that no electric field is present?
Figure : Electric field vectors surrounding
Q.10 In a region of uniform electric field E, a charged a positive point charge. The field vectors are
directed radially outward. The positions to
particle experiences an acceleration a. If a second which the field vectors belong are marked
in black.
particle with twice the charge and twice the mass
of the first particle enters that same region, it will
experience an acceleration
(A) a / 4 (B) a / 2
(C) a (D) 2a
26 ELECTROSTATICS
12th PHYSICS MODULE - 1
Equal numbers of
field lines…
–q
q
Field vectors point q
toward negative charge.
Figure : Electric field vectors
surrounding a negative point charge. The
field vectors are directed radially inward.
(2) The tangent drawn at any point on line of force
gives the direction of force acting on a positive …emerge from equal
charge placed at that point. positive charges.
(3) Two lines of force never intersect. If they are Figure : Field lines generated by two
assumed to intersect, there will be two directions positive charges of equal magnitudes.
of electric field at the point of intersection : which
is impossible. (6) The number of lines originating or terminating on
(4) These lines have a tendencyto contract in tension a charge is proportional to the magnitude of
like a stretched elastic strong. This actually charge. There is no rule as to how many lines are
explains attraction between opposite charges. to be shown. However, it is customary to draw
number of lines proportional to the charge. Thus
Equal-magnitude positive if N number of lines are drawn from or into a
and negative charges… charge Q, 2N number of lines would be drawn
for charge 2Q.
q
Twice as many field lines
emerge from +2q charge…
–q
2q q
…have equal numbers …as terminate
of field lines. on – q charge.
Figure : Field lines generated by positive Figure : Field lines generated by positive and negative
and negative charges of equal magnitudes. charges of unequal magnitudes. The positive charge
has 2 times the magnitude of the negative charge.
(5) These lines have a tendency to separate from
each other in the direction perpendicular to their
length. This explains repulsion between like
charges.
27 ELECTROSTATICS
12th PHYSICS MODULE - 1
(7) Total lines of force may be fractional as lines of Q –Q' Q' –Q Q –Q' Q' –Q
+– +– +– + –
force are imaginary. +– +– +– +–
+– Îr +–
(8) Lines of force ends or strarts normally on the + – + – Q'=Q 1– 1r
+ – E=0 + –
surface of a conductor. + – E<E 0 + –
+– + –
(9) If there is no electric field there will be no lines of +– +–
force. Metal Dielectric
(10) Lines of force per unit area normal to the area at The lines of force never intersect each other due
a point represents magnitude of intensity, * to superposition principle.
The property that electric lines of force contract
crowded lines represent strong field while distant * longitudinallyleads to explain attraction between
lines represent weak field. opposite charges.
The property that electric lines of force exert
(11) Electric lines of force differ from magnetic lines lateral pressure on each other leads to explain
repulsion between like charges.
of force. *
(a) Electric linesof forcenever form closed loop
while magnetic lines are always closed or
extended to infinity.
Electric line of force Magnetic line of force EXAMPLE 19
(A) (B)
A metallic solid sphere is placed in a uniform
(b) Electric lines of force always emerge or electric field. The lines of force follow which
terminate normallyonthe surfaceof charged path(s).
conductor, while magnetic lines emerge or
terminate on the surface of a magnetic 1A
material at any angle. 2B
(c) Electric lines of force do not exist inside a 3C
conductor but magnetic lines of force may 4D
exist inside magnetic material.
SOLUTION:
Golden Tips Electricfieldlines neverenter ametallicconductor
(E = 0, inside a conductor) and they fall normally
on the surface of a metallic conductor. Hence,
line of force follows path shown in figure 4.
EXAMPLE 20
* Lines of force starts from + ve charge and ends Three positive charges of equal value q are
on – ve charge. placed at the vertices of an equilateral triangle.
Sketch the possible resulting lines of force.
* Lines of force starts and ends normally on the SOLUTION:
surface of a conductor.
+ + – – +
+ – ++
+ – E=0
E=0 +
––– – – – + –
+ –
Fixed point charge near Edge effect
infinite metal plate
* Lines of force do not exist inside a conductor
(as field inside a conductor is zero) the field
between the plates is as shown.
28 ELECTROSTATICS
12th PHYSICS MODULE - 1
EXAMPLE 21 Q.3 Figure shows diagrams of hypothetical (but
incorrect) field lines corresponding to some static
A few electric field lines for a system of two
charge distributions, which are beyond the edge
charges Q1 and Q2 fixed at two different points
on the x-axis are shown in the figure. Show that of the diagram. What is wrong with each of these
these lines suggest that |Q1| > |Q2| and at a finite diagrams of field lines?
distance to the right of Q2 the electric field is
(a) (b)
zero.
Q1
+ + (c)
Q2
SOLUTION: Q.4 A field line diagram shows a point charge with
From the diagram, it can be observed that Q1 is Q.5 eight field lines emerging from it. Nearby in the
positive, Q2 is negative. same diagram is a metal sphere with four lines
Number of lines on Q1 is greater and number of emerging and two lines entering the sphere.What
lines is directly proportional to magnitude of can we say about the arrangement of charge on
charge. So, | Q1 | > | Q2 | the sphere? The net charge on the sphere?
Electric field will be zero to the right of Q2 as it
has small magnitude & opposite sign to that of During a day of fair weather, the Earth has an
Q1. atmospheric electric field that points vertically
Checkup 4 down. This electric field is due to charges
Q.1 Rank the magnitudes of the electric field at points distributed over the surface of the Earth.What
must be the sign of these charges?
A, B, and C shown in Figure (greatest magnitude
first). Q.6 A field line diagram shows a +6.0 C point charge
B with eight field lines emerging from it. Nearbyin
the diagram is a short, charged rod with 12 field
lines entering it. The charge on the rod is
A (A) + 4.0 C (B) – 4.0 C
+
C+ (C) + 9.0 C (D) –9.0 C
1.10 ELECTRIC
DIPOLE
Q.2 Which of the following statements about electric * In some molecules, the centre of + ve and
field lines associated with electric charges is
false? (a) Electric field lines can be either straight – ve charge do not coincide. This results in the
or curved. (b) Electric field lines can form closed formation of electric dipole. Atom is non-polar
loops. (c) Electric field lines begin on positive
charges and end on negative charges. (d) Electric because in it the centre of + ve and – ve
field lines can never intersect with one another. charges coincide. Polarity can be induced in an
atom by the application of electric field. Hence it
can be called as induced dipole.
29 ELECTROSTATICS
12th PHYSICS MODULE - 1
Electric dipole moment When r >> d , 4 1 2p rˆ
Ea 0 r3
* An electric dipole is a system formed by two The axial field is parallel to dipole moment
equal and opposite charges placed at a short Electric field at an
equatorial point
distance apart. Product of one of the two charges Electric field at P due to negative charge
and the distance between them is called "electric
dipole moment" p . p
q 2
• It is a vector quantity, p E1 = 1 q
40 (r2 d2 / 4)
directed from – ve to – +
–q +q
+ ve charge. 2 E2 E2sin
• Practical unit is Debye p of two equal and E2cos
E1cos
opposite point charges each having charge P
10–10 frankline and separation of 1 Å. i.e.
1D = 10–10 × 10–10 = 10–20 Fr × mt. r2 d2 / 4 E1 r2 d2 / 4
r E1sin
1020
= 3 x109 Cb × mt 3.3×10–30 Cb × mt. A – q+ B
-q p dO
Electric field at an Electric field at P due to positive charge
axial point
4
Electric field at P due to negative charge E2 1 r2 q /
40 d2
1 q d
E1 40 (rˆ) pO Fields E1 and E2 are equal in magnitude
r d 2 r Resolving E1 and E2 into two components one
2 along O and other perpendicular to OP
– + P
WToetaflinfdieEld1sEin===2EEE122ccosoisns+ E2cos = 2E1cos
–q +q
Electric field at P due to positive charge
1 q (rˆ)
E2 40
r d 2 q1 d/2
2
E = 2 4 0 (r2 d2 / 4) r2 d2 / 4
Total electric field at axial point P is = q.d = p
40 (r2 d2 / 4)3/2 40 (r2 d2 / 4)3/2
Ea E1 E2
= q 1 2)2 1 2)2 rˆ If r >> d , 1 p rˆ i.e. field at
Ea 4 0 d/ d/ E 4 0 r3
(r (r
q 2.r.d equatorial point is antiparallel to dipole moment.
Ea 4 0 (r2 d2 / 4)2
or = rˆ Electric field at an
arbitrary point
= 2pr / 4)2 rˆ
4 0(r2 d2 We resolve dipole moment p in two components
one along r and another perpendicular to r.
30 ELECTROSTATICS
The radial component of electric field * 12th PHYSICS MODULE - 1
*
Er = 1 2p cos * The electric field at axial point is parallel to dipole
4 0 r3 moment vector.
* The electric field at equatorial point is antiparallel
1 p sin * to dipole moment vector.
Magnitude of resultant field is E = 4 0 r3 Theratio offield at axial point to field at equatorial
point is Ea : E = 2 : 1.
Er
Dipole in E-field
P E Consider the force exerted on the dipole in an
E
y external E-field.
x
A + +q r Assumption: E-field from dipole doesn't affect
psin pcos
the external E-field.
p
p Dipole moment : qd
B – -q
Force due to the E-field on + ve and – ve
The magnitude of resultant field is
charge are equal and opposite in direction. Total
external force on dipole = 0.
E= Er2 E2 = 1p 1 3cos2 In uniform external
4 0 r3 field, forces on dipole
sum to zero…
The direction of resultant field is
tan = E 1 tan d + F+
Er 2
Case I : at axial point =0° p
F– –
Ea = 1 p 1 3cos2 0 = 1 2p …but tend to rotate
4 0 r3 40 r3 dipole, so there is a
net torque.
Case II : at equatorial point = /2
Figure : Electric dipole placed in a
E = 1 p 1 3cos2 / 2 = 1 p uniform electric field. The dipole
4 0 r3 4 0 r3 moment vector p is directed from the
den Tips negative charge to the positive charge.
Gol There is an external torque on the center of the
* Intensity due to a dipole varies as (1/r3) and can deritpsoalet.oFrqourceeFexr ertFs at point P. The force ex-
never be zero unless r or p 0. on point P with respect
to point O. Direction of the torque vector is
* E will be maximum for end on, axial or tan A
determined from the right-hand rule. F
position Emax 1 2p : Into
40 r3
the P
page r
* E will be minimum for broad on, equatorial or Net torque ; direction : clockwise torque
tan B position Emin 1 p magnitude : ve ve
40 r3
d d
= F. 2 sin F. 2 sin qE.d sin = pE sin
31 ELECTROSTATICS
p 12th PHYSICS MODULE - 1
E
SOLUTION:
Angular SHM : When a dipole is suspended In special orientations the force between two
dipoles can be zero or a force of repulsion. In
in uniform field, it will align itself parallel to the general each dipole will exert a torque on the
other, tending to align its axis with the field
field. Now if it is given a small angular created by the first dipole.After this alignment,
displacement about its angular position, the each dipole exerts a force of attraction on the
restoring couple will be other.
= – pE sin . if is small sin .
= – pE –(Angular SHM).
for balanced condition :
deflecting = restoring EXAMPLE 23
I = – pE = – pE = – 2 An electric dipole of dipole moment 1C-m is
I placed making an angle of 30° with electric field
++q of 2 N/C. Calculate the torque acting on dipole.
pE S O L UTION: p
I Torque acting E
= – E on the dipole = pE sin
– = 1 × 2 × sin 30° Nm = 2 × ½ N-m =1 Nm
T 2 2 I –q
pE
[I moment of inertia]
EXAMPLE 24
Golden Tips Two point dipoles pkˆ p kˆ
2
and are located at
* If dipole is placed in a non-uniform electric field, (0, 0, 0) and (1m, 0, 2m) respectively. Find the
resultant electric field due to the two dipoles at
it performs rotatory as well as translatorymotion, the point (1m, 0, 0)
because now a net force also acts on the dipole SOLUTION: p
2
along with the torque. The given point is at axis of dipole and at
In Uniform electric field, Total force = 0,
Torque may or may not be zero. ( ( 0 if 0) equatorial line of p dipole so that field at given
In Non-uniform electric field, +q point. x
Total force 0, –q F'
Torque may or may +F (1,0,2)
not be zero. Q p/2
For situation shown
(1,0,0)
in figure,
Torque = 0 (Force along same axis). z
pkp 2k(p / 2) 7p
(1)3 (2)3 320
EXAMPLE 22
E
Consider two electric dipoles in empty space.
Each dipole has zero net charge. Does an electric
force exist between the dipoles-that is, can two
objects with zero net charge exert electric forces
on each other? If so, is the force one of attraction
or of repulsion?
32 ELECTROSTATICS
12th PHYSICS MODULE - 1
= Surface integral over surface S
1.11 ELECTRIC FLUX S
= Integration of integral over all area elements
Latin word flux means “to flow”. on surface S.
Graphically: Electric flux E represents the
Golden Tips
number of E-field lines crossing a surface. Electric flux is a scalar quantity
Mathematically: Units (V – m) or (N – m2/Cb)
* Dimensions : [M1 L3 T–3 A–1]
* The value of is zero in the following
(a) Electric flux E is the product (b) E is the product of the
of E , the component of E magnitude of the electric field… *
normal to the surface,…
E E circumstances :
E (a) If a dipole is enclosed by a closed surface.
Acos
(b) Magnitude of + ve and – ve charges are
AA equal inside a closed surface.
(c) If no charge is enclosed bya closed surface.
…and the surface area A. …and the area A cos facing
perpendicular to the field. (d) In coming flux ( – ve) = out going flux
Figure : (a) Flat rectangular surface * ( + ve). ds E
immersed in a uniform electric field. The ||cioinur|ct|u=l=arR=R=22EEctoutravle=d 0 ds
perpendicular to the surface makes an angle R E
with the field lines. The normal component ds
of the electric field is E = E cos . The electric E
flux is = E A. In terms of field lines, the
electric flux is proportional to the number of
lines intercepted by the area. (b) The area A cos
is the projection of the area A onto a plane
perpendicular to the electric field. RE
* out = in = R2E E
Vector of the area A is perpendicular to the *
area A. For non-uniform E-field & surface,
direction of the area vector A is not uniform.
y ds E
dA =Area vector for small area element dA. itnota=l out = Ea2 a
= 0. ds
Electric flux dE E.dA za ds
x
Electric flux of through surface S :
E a
E E
E.dA R
S * T = 0
EE
E
dA dA dA dA
E 1.12 GAUSS’ LAW
Surface S * The total flux linked with a closed surface is
(1/0) times the charge enclosed by the closed
surface (Gaussian surface).
33 ELECTROSTATICS
12th PHYSICS MODULE - 1
E q Total area of surface S1
E.dA 0
S for any closed surface S L rˆ
E (2rL) 0 E 20r ; E 20r
and q is the net electric charge enclosed in closed (B) Infinite sheet of charge
surface S.
* Gauss' Law is valid for all charge distributions Uniform surface charge density :
and all closed surfaces. (Gaussian surfaces) Planar symmetry : E-field directs perpendicular
* Coulomb's Law can be derived from Gauss' Law. to the sheet of charge.
* For system with high order of symmetry, E-field Construct Gaussian surface S in the shape of a
can beeasilydetermined if weconstruct Gaussian cylinder (pill box) of cross-sectional areaA.
surfaces with the same symmetry and apply
Gauss' Law.
E-field Calculation + +
with Gauss' Law + +
+ + +
+ + + + +
+ + + + + +
+++S3+++ + + + + +
(A) Infinite line of charge + + + dA1 + + + Gaussian
+ + + + + surface
E + + + + +
+ + + +
Linear charge density : dA++3 + + S2+ + E
Cylindrical symmetry : E-field directs radially + +
outward from the rod. Construct a Gaussian + + +
surface S in the shape of a cylinder, making up + + + + + S+1 + dA2
of a curved surface S1, and the top and bottom + + + +
circles S2, S3. + + +
+ + + ++
+ + + +
+ + +
+
+
Gauss’ law : A ;
E.dA 0 E.dA
dA + 2 r 0
+ r Gaussian
S2 + surface S S1
L S1 + dA
+ E
S3 + E dA over whole surface S1
+
+
+
+ E.dA E.dA 0 2EA
+
+ S2 S3
+
+
+
+ (E || dA2 , E || dA3)
+
dA + Note : For S2, both point up
E & dA2
Gauss' Law: Total charge L For S3, both E & dA3 point down.
E.dA 0 0
S
A
2EA 0 E 20
E.dA E.dA E.dA E.dA
S S1 S2 0E S3dA (C) Uniformly charged sphere
E||dA
E dA L Spherical symmetry : Total charge = Q
S1 0
(a) For r > R: Consider a spherical Gaussian
surface S of radius r : E || dA || rˆ
34 ELECTROSTATICS
12th PHYSICS MODULE - 1
Gauss’ law : Q ; E.dA Q 1 . Q r rˆ ; for r R
E.dA 0 0 E 40 R3
S
S
E dA Q
S 0
E
Surface area of S = 4r2
E Q rˆ ; for r > R
40r2
Rr
Enclosed Gaussian (D) Spherical shell
charge is Q S surface
E First we know that all charges move to the surface
dA
of conductors.
R
(i) For r < R : Consider Gaussian surface S2
r charge inside)
E.dA (no
0
S2 + + +
E = 0 everywhere
(b) For r < R : (ii) For r R + +
Consider a spherical Gaussian surface S' of +R +
radius r < R, then total charge included q is
proportional to the volume included by S' Consider Gaussian + S2 +
+ +
surface S1 : S1 + +
+
Q + +
E.dA 0
q Volume enclosed by S S1 Total charge = Q
Q Total volume of sphere
q 4 / 3r3 q r3 Q E Q Fora conductor
Q 4 / 3R3 R3 dA 0 ( E || dA || rˆ)
S1 E Spherical symmetric
Gauss’ law : 4r2
q ; E r3 1 .Q Q 1/r2
E.dA 0 dA R3 0 40r2 r
S E
S
den Tips O R
Gol
Enclosed Gaussian
charge is q S' surface
E
R * R +q
r
E kq 2R 2 q R2 q
E R2 4 0 2 0
Surface area of S' = 4r2
35 ELECTROSTATICS
12th PHYSICS MODULE - 1
Note : Here electric field is radial.
* + Total = q q
0
q +
* q Total = q
0
+ hemisphere = q +
2 0
q q
= q
4 0
* +q Total = q
0
R * If a charge q is placed at a distance a/2 above a
square plate of side a then flux passing through
+q cylinder = q the epnlactleoissed6qth0e. charge assume cube to be a
* R 2 0 To
+ qq Gaussian surface flux through the cube will be
q q and will be equally divided in six faces.
0 0
Total =
+q cube = q EXAMPLE 25
2 0
A flat sheet of paper measuring 22 cm × 28 cm
is placed in a uniform electric field of
100N/C.What is the flux through the paper if the
paper makes an angle of 90° with the electric
field? If the paper makes an angle of 30°? (See
Fig.)
* +q Total = q Electric field, perpendicular
0 to sheet of paper, is parallel
to surface normal.
(a)
A
+q = q 90°
8 0
36 ELECTROSTATICS
12th PHYSICS MODULE - 1
Electric field, at 30° to sheet EXAMPLE 27
of paper, makes an angle of
60° with surface normal. An infinite, insulating slab of thickness d has a
(b) uniform volume charge density , as shown in
Fig. Find the electric field inside and outside the
A slab.
60°
30° E is perpendicular Electric field is perpendicular
to circular ends of to slab, pointing outward on
Gaussian surface. both sides of center.
Figure : A sheet of paper placed in a d
uniform electric field (a) at an angle
of 90° with the electric field and E A
(b) at an angle of 30 °. The black E E
arrow A indicates the perpendicular
to the surface. A
For protruding surface,
SOLUTION:
For a uniform field and a flat area, the flux is E charge occupies volume
merely the product E = EA cos , and since an d × A.
angle of 90° between the sheet of paper and the
electric field means = 0. See Fig. a. Within slab, charge x
E = E A cos inside Gaussian 0
= (100 N/C) × (0.22 m × 0.28m) × 1 surface occupies
= 6.2 N.m2/C volume 2x × A.
An angle of 30° between the paper and the
electric field means = 60°. (see Fig. b), so Figure : An infinite slab of thickness d with a uniform volume
E = E A cos distribution of positive charge. A small, imaginary Gaussian
= (100 N/C) × (0.22 m × 0.28m) × cos 60° cylinder with end area A is used to find the field inside the
= 3.1 N.m2/C slab, a distance x from the center; a longer Gaussian cylinder
with end area A is used to find the field outside the slab.
EXAMPLE 26
A point charge q is located at the center of a SOLUTION:
uniform ring having linear charge density and With x = 0 at the center of the slab, “inside” the
radius a, as shown in Figure. Determine the total slab means and “outside” means | x | < d/2 and
electric flux through a sphere centeredat the point “outside” means | x | > d/2.
charge and having radius R, where R < a. Consider the Gaussian surfaces shown in Fig.
To obtain E inside the slab, we use the smaller
cylinder to evaluate Qinside = Vinside.
Since its volume is
R Vinside = [length] × [base area] = 2x A,
qa Gauss’ Law becomes
2EA Qinside 2xA
0 0
and solving for E we obtain
SOLUTION: E x (| x | d / 2)
Only the charge inside radius R contributes to 0
the total flux.
Thus the E field is zero at the center of the slab
E q
0 and increaseslinearlywith distanceinsidetheslab.
37 ELECTROSTATICS
12th PHYSICS MODULE - 1
Outside the slab, we use the larger Gaussian Q.3 A sphere of radius R is in a region of uniform
cylinder. The amount of volume inside it which electric field E. The net flux through the surface
contains charge is V inside = d × A, so Gauss’ of the sphere is (B) E × 2R2
Law implies (A) E × R2 (D) Zero
(C) E × 4R2
2EA Qinside dA
0 0
Q.4 A charge distribution consists of four equal
and E d (| x | d / 2) charges at the corners of a square. Does this
20 arrangement have enough symmetry so you can
use Gauss’ Law to calculate the electric field at
Note that this constant electric field is the same some distance from the square?
dsheenestitoyfch=arged,(Eas=itm/ u2st0)bwe.ith
as for a surface Two parallel infinite sheets carrypositive uniform
charge surface charge densities and 2.What is the
Q.5
Checkup 5 magnitudeoftheelectric field between the sheets?
Q.1 Consider Fig. a. For the surface with the Outside both sheets? //20;0;03/20
orientation shown in that figure, theflux has some 3//02;30; //020
(A) (B)
(C) (D)
particular value.Would the flux be larger or
smaller if the surface were horizontal and Q.6 Consider the charge distribution shown in Figure.
The charges contributing to the total electric flux
Vertical?
through surface S' are (a) q1 only (b) q4 only (c)
Electric flux E is the product q2 and q3 (d) all four charges (e) none of the
of E , the component of E charges. S
normal to the surface,…
q4
q2
E q1
q3 S'
A
…and the surface area A. S"
Q.7 Again consider the charge distribution shown in
Q.2 Consider the point charge and surface of Fig. previous Figure. The charges contributing to the
The net flux through this surface is total electric field at a chosen point on the surface
With a point charge S' are (a) q1 only (b) q4 only (c) q2 and q3 (d) all
outside a cube… four charges (e) none of the charges.
Q.8 Four closed surfaces, S1 through S4, together
with the charges –2Q , Q , and –Q are sketched
in Figure.(The coloured lines arethe intersections
q of the surfaces with the page.) Find the electric
flux through each surface.
S1
(A) Positive …what is the net flux S4 –2Q S3
through the cube? +Q
–Q
(B) Negative (C) Zero
S2
38 ELECTROSTATICS
12th PHYSICS MODULE - 1
For particle moving from 1 to 2 :
1.13 CONCEPT OF 2 dU U2 U1 2 . ds ,
ELECTRIC POTENTIAL F
11
Potential Energy and where U1, U2 are potential energy at position
Conservative Forces 1, 2.
Electric force is a conservative force. Work done Suppose charge q2 moves from point 1 to 2.
F r2
by the electric force as a charge moves an + r1 dr +
infinitesimal distance ds along PathA= dW
q1 +
q2 F q2
* 2 From definition :
* ds
* Path A 2 dr r2 dr )
* F U2 – U1 = – F F
ds . =– F . dr ( ||
ds
1 r1
F
r2 dr 1 C
1 F = 1 q1q2 dr r2 r
40 r2
dW . ds
F r1
Note: ds is in the tangent direction of the curve 1 q1q2 r2 1 q1q 1 1
of Path A. 40 r r1 40 r2 r1
2
F
Total work done W by force in moving the
particle from Point 1 to Point 2 1 1 1
2 40 r2 r1
W . ds –W = U = q1q 2
F
1 Note :
Path A
* I(IffqqW21,m=q2oWvaeroesrkoawfdsoaanymefreboysmiegqlne1,c,tthtrhiecennrerp2Uu><lsri10v,,ewfeWohrca>ev)0e
2 • IfWq1<, q02 are of different sign, then U > 0,
(W = Work done by attractive force)
Path Integral = Integration over Path A •
1 *
Path A •
•
from Point 1 to Point 2. IIIfffWqqq112,,>mqq22o0avarreeesootoffwsdaiafmrfdeesrseqing1tn,st,ihtghenne,ntrh2e<Unr>1,U0w,<e0hW,av<e0
*
A force is conservative if the work done on a
particle by the force is independent of the path
taken.
The work done by a conservative force on a It is the difference in potential energy that is
particle when it moves around a closed path important.
returning to its initial position is zero. Reference point :
F U(r = ) = 0 U – U1 =
Since the work done by a conservative force 1 q1q2 1 1
40 r2 r1
is path-independent, we can define a quantity,
potential energy, that depends only on the U (r) 1 q1q2
40 r
position of the particle.
We define potential energy U such that
dU W F . ds If q1, q2 same sign, then U(r) > 0 for all r
If q1, q2 opposite sign, then U(r) < 0 for all r
39 ELECTROSTATICS
12th PHYSICS MODULE - 1
For discrete distribution of charges Usys = k qiq j k
2 ij a 2a
U =k qiq j 1 (ii) = [q1q2 + q1q3 + q2q1
2 i j rij 2
( is used as each term in q1
+ q2q3 + q3q1 + q3q2] +
summation will appear twice) Number of pairs Np aa
For continuous distribution of charges
= N (N 1)
dU = Vdq U = Vdq 2 + a +
N = Number of charges q2 q3
Geometrical point For three charge system
charge system Usys k q1q 2 q2q3 q3q1
a a a
kq2 a
(i) Wsys = Usys = a + – 3 pairs
–q
q
(iii) Number of pairs = 43 = 6 q a (+) q
2
+ +
(–) a
Usystem kq2 1 – 1 1 – 1 –1 –21 2kq2 (–) (–)
a a 2a (–)
–
–
–q
(+) –q
(iv) Variety distance How many times NP = 87 = 28
2
side 1a 12 q + +q
Face diagonal 2a 12
Main diagonal 3a 4 q + +q a
so Usystem = 12kq2 12kq2 4kq2 +q +q
a 2a 3a q+ a a
+q
(v) Variety distance How many times NP = 98 = 36
2
Side 1a 12 q + +q
Face diagonal 2a 12
Main diagonal 3a 4 q+ +q a
q q+
Centre to corner 3a 8
2 q+ +q
a
a +q
So Usystem = 12kq2 12kq2 4kq2 16kq2
a 2a 3a 3a
40 ELECTROSTATICS
12th PHYSICS MODULE - 1
Electric potential V 1 N qi
40 i1 ri
Consider a charge q at center, we consider its NOTE
effect on test charge q0. E, F
Definition : We define electric potential V so * For , we have a sum of vectors.
U W For V, U, we have a sum of scalars.
q0 q0
that V Potential difference
(V is the P.E. per unit charge) (i) Definition : Vab = – b E .dr
* We take V (r = ) = 0. (ii)
* Electric Potential is a scalar.
* Unit: Volt (V) = Joules/Coulomb Its value does not adepend on the frame of
* For a single point charge:
reference, hence it is an absolute quantity.
1 q
V(r) 40 r (iii) As the electric field is
conservative, work done
* Energy Unit: U = q V and hence potential
electron – Volt (eV ) = 1.61019 J difference between to
charge of electron points is path independent
and depends only on the
Potential for a system of charges position of points.
For a total of N point charges, the potential V at (iv) W1 = W2 = W3.
any point P can be derived from the principle of Potential is theoretically zero at infinite &
superposition. practically zero at earth's surface.
Relation between Electric
+ q1 Field E and Electric
Potential V
+ q2
+ q4 + q3 (A) To get V from :
E
r1 r2
+ qN r4 r3 V V2 V1 2 E .d s
1
Note: The integral on the right hand side of the
above can be calculated along anypath from point
P 1 to 2. (Path-Independent)
Recall that potential due to q1 at point P: Convention: V 0 VP P s
E.d
V1 1 q1
40 r1
(B)
Total potential at point P due to N charges: To get E from V :
Again, use the definition of V :
V = V1 + V2 + ............ + VN U q0V W
(principle of superposition)
Work done
= 1 q1 q2 ...... qN
40 r1 r2 rN
41 ELECTROSTATICS
12th PHYSICS MODULE - 1
However, W .s EXAMPLE 28
q0E
If small droplets each of radius r and charge q
Electric force coalesce to form a single drop then find the
potential of the bigger drop
= q0 Es s
where Es is the E-field component along SOLUTION:
the path s
V= kQ , where Q = nq and 4 R 3 n 4 r3
q0V = –q0Es s R 3 3
Es V
s
EXAMPLE 29
E Comment on the electric field and potential at
the mid point of a line joining an electron and a
Es proton.
s SOLUTION:
+ E= kq2 kq2 2kq2 towards electron
x2 x2 x2
q0
V V+ V V= kq kq =0
surface surface xx
(i.e. Potential = V on the surface) At mid point, E 0, V = 0
For infinitesimal s, Es dV EXAMPLE 30 +Q –3Q
ds –
Four point charges are placed +
C
NOTE at the corners of a square of
+
(1) The E-field component along anydirection is the side . Find the potential at
+4Q
negative derivativeof the potential alongthe same the centre of square. –
direction. i.e. in direction of E-field potential SOLUTION: –2Q
V = 0 at C, [Use V = kq/r]
decreases.
ds
(2) If E , then V = 0 EXAMPLE 31
E
(3) V is biggest/smallest if ds || Over a certain region of space, the electric
potential is V = 5x – 3x2y + 2yz2. Find the
Generally, for a potential V (x; y; z), the relation expressions for the x, y, and z components of
the electric field over this region. What is the
magnitude of the field at the point P that has
between E (x, y, z) and V is coordinates (1, 0, –2) m?
Ex V , Ey V , Ez V
x y z
S O L UTION:
V = 5x – 3x2y + 2yz2
x , y , z are partial derivatives
Evaluate E at (1, 0, –2)
For V (x, y, z), everything y, z are treated like Ex V 5 6xy = –5 + 6 (1) (0) = –5
x x
a constant and we only take derivative with Ey V 3x 2 2z2
y
respect to x.
= 3 (1)2 – 2 (–2)2 = –5
42 ELECTROSTATICS
Ez V = –4yz = – 4 (0) (–2) = 0 12th PHYSICS MODULE - 1
z
Total charge on the ring = . (2R)
After integration,
E E 2 E 2 E2z (5)2 (5)2 02 V Q
x y 40 R2 z2
= 7.07 N/C Limiting case : z >> R
Electric potential of V Q z2 Q z |
continuous charge 40 40 |
distribution
For anycharge distribution, we writethe electrical Case 2 : Uniformly charged disk
potential dV due to infinitesimal charge dq:
Using the principle of superposition, we will find
dV 1 . dq dq the potential of a disk of uniform charge density
40 r by integrating the potential of concentric rings.
V charge410 . dq r
r P z
distribution P
Similar to the previous cases on E-field, for the rz
case of uniform charge distribution: dx x
1-D long rod dq = dx
2-D charge sheet dq = dA Total charge = Q
3-D uniformly charged body dq = dV Charge density =
Let us consider some cases : dV 1 dq
40 r
Case 1 : Uniformly-charged ring disk
Length of the infinitesimal ring element = ds = Rd After integration, V ( z2 R2 | z |)
charge dq = ds = R d 20
dV 1 . dq = 1 . Rd
40 r 40 R2 z2
Case 3 : Electric potential due to a shell
z
A shell of radius R has a charge Q uniformly
P
distributed over its surface.
rz (a) At points outsidea uniform spherical distribution,
the electric field is +++
++
1 Q rˆ ,
dq d E 40 r2 ++
ds R
+ + ++
since is radial, . dr Edr +
E E
Charge density since V () = 0, we have
V() V(r) E.dr
43 ELECTROSTATICS
12th PHYSICS MODULE - 1
0V Q dr V 1 Q But Q
40r2 40 r
4 R 3
3
r
We see that the potential due to a uniformly 1 Q
40 2R3
charged shell is the same as that due to a point V [3R 2 r2 ]
charge Q at the centre of the shell.
(b) At an internal Point : At points inside the shell, Alternatively expression can be derived using
E =0. So, theworkdone in bringingaunit positive
charge from a point on the surface to any point dV E dr and E r
inside the shell is zero. 30
Thus, the potential has a fixed value at all points
within the spherical shell and is equal to the Golden Tips
potential at the surface. V
V 1 Q 1Q * Potential at earth is assumed to be zero as it is a
40 R 4 0 R
large conductor and its potential is approximately
O r=R r constant with respect to given charge to it .
Variation of electric potential with the distance * Potential is a relative parameter, while potential
*
from the centre (r). All the above results hold difference is absolute i.e. potential depends on
for a conducting sphere also whose charge lies reference chosen. That's why potential of any
entirely on the outer surface. object [+ve, –ve, neutral] may have (+ ve) or
(–ve) or zero potential.
Case 4 : Electric potential due to a Two concentric spherical shells of radius R1 and
non-conducting charged sphere R2 (R2>R1) are having uniformly distributed
charges Q1 and Q2 respectively. Potential
A charge Q is uniformlydistributed throughout a B C
R1
non-conducting spherical volume of radius R. R2A Q2
Let us find expressions for the potential at an Q1
(a) external point (r > R); Q +++++++R+++++++++++++ (i) at R1, V kQ1 kQ2
(b) internal point (r < R); R1 R2
where r is the distance of the
point from the centre of the (ii) at R1 r R2 ; V kQ1 kQ2
sphere. r R2
(a) At an external point :
V 1 Q (iii) For r R2 ; V kQ1 kQ2
40 r r r
(b) Potential at an internal point :
V (3R2 r 2 ) 1 3QV
60 4 0R
1Q
4 0R
R r
44 ELECTROSTATICS
12th PHYSICS MODULE - 1
1.14 TABLE : ELECTRIC POTENTIAL DUE TO VARIOUS
CHARGE DISTRIBUTIONS
Name/type Formula Note Graph
Point charge V
kq * q is source charge.
r * r is the distance of the point r
from the point charge.
V
Ring (uniform/nonuniform at centre : kQ * Q is source charge.
charge distribution) R * x is the distance of the point
kQ
at the axis : on the axis from centre of ring. r
R2 x2
V
Uniformly charged hollow For r R, V= kQ * R is radius of sphere kQ/R
conducting/nonconducting r * r is the distance from centre
/solid conducting sphere kQ
for r R, V R of sphere to the point
* Q is total charge 4R2
Rr
Uniformly charged solid For r R, V= kQ * R is radius of sphere
nonconducting sphere. r
for r R, * r is distance from centre 3kQ/2R V
kQ/R
kQ (3R2 r2) to the point
2R3 Rr
V * Vcentre 3 Vsurface
2
4
60 (3R 2 r2) * Q is total charge 3 R 3
* Inside the sphere potential
varies parabolically
* Outside the sphere potential
varies hyperbolically.
Infinite line charge * Absolute potential * Potential difference between two
is not defined. points is given by formula :
VB VA 2k ln (rB /rA )
* Potential difference between two
* Absolute potential points is given by formula :
Infinite nonconducting thin is not defined. VB VA 0 (rB rA )
sheet
Infinite charged conducting * Absolute potential * Potential difference between two
thin sheet is not defined.
points is given by formula :
VB VA 0 (rB rA )
45 ELECTROSTATICS
12th PHYSICS MODULE - 1
Checkup 6 Q++++++++++++++++++++++++++++++++ Q = 0–––––––––+ ++
Q.1 In Figure, two pointsAand B are located within +++
+
a region in which there is an electric field.
The potential difference V = VB – VA is ++
(a) positive (b) negative (c) zero.
E
B
A
Q.6 Two concentric spherical conducting shells of
radii a = 0.400 m and b = 0.500 m are connected
bya thin wire as shown in Figure. If a total charge
Q.2 A spherical balloon contains a positivelycharged Q = 10.0 µC is placed on the system, how much
object at its center.As the balloon is inflated to a charge settles on each sphere?
greater volume while the charged object remains
at the center, does the electric potential at the b
surface of the balloon (a) increase, (b) decrease, a
or (c) remain the same? Does the electric flux
through the surface of the balloon (d) increase, q1 Wir e
(e) decrease, or (f) remain the same? q2
Q.3 In a certain region of space, the electric potential
is zero everywhere along the x axis. From this Q.7 The potential in a region between x = 0 and
we can conclude that the x component of the x = 6.00 m is V = a + bx, where a = 10.0 V and
electric field in this region is (a) zero (b) in the +x b = –7.00 V/m. Determine the magnitude and
direction (c) in the –x direction. direction of the electric field at x = 0, 3.00m,
Q.4 In a certain region of space, the electric field is and 6.00 m.
zero. From this we can conclude that the electric 1.15 ELECTRIC POTENTIAL
DUE TO A DIPOLE
potential in this region is (a) zero (b) constant (c)
positive (d) negative.
Q.5 Consider starting at the center of the left-hand V() kq kq , [rl r d cos ]
sphere (sphere 1, of radius a) in Figure and r1 r d cos
moving to the far right of the diagram, passing V() kq kq , [r2 r d cos ]
through the center of the right-hand sphere r2 r d cos
(sphere 2, of radius c) along the way. The centers
of the spheres are a distance b apart. Draw a
graph of the electric potential as a function of
position relative to the center of the left-hand r2 r r1
sphere.
– +
-q +q
46 ELECTROSTATICS
12th PHYSICS MODULE - 1
V V1 V2 kq r 1 1
d cos d cos
r θ is angle between dipole
moment vector and electric field.
kq r d cos r d cos U U is largest when dipole is
r2 d2 cos2 antiparallel to field…
pE
V kp cos kp cos
r2
r 2 1 d 2 cos2 θ
r2 360°
0 0° 90° 180° 270°
We can use E dV to calculate electric field –pE
dr
…and smallest when dipole
from potential function. is aligned with field.
Dipole in electric field Figure : Potential energy of an
electric dipole as a function of angle.
Energy consideration : When the dipole p
rotates d, the E-field does work.
+ F = qE
(d/2)sin + F = qE F = –qE p E
O E
p –
d
–
F = –qE = 90° ; Torque | | pE ; U = 0 (define)
(based on definition)
Work done by external E-field on the dipole: F = –qE – p + F = qE
dW = – d E
Negative sign here because torque byE-field acts = 0° ; Torque | | 0 ; U = – pE
to decrease . (minimum energyconfiguration)
BUT: Because E-field is a conservative force
field, we can define a potential energy (U) for
the system, so that dU = – dW
For the dipole in external E-field:
dU = –dW = pE sin d
U () = dU pE sin d pE cos U0
Set U ( = 90°) = 0, 0 = – pE cos 90° + U0
PUo0te=nt0ial energy :
U pE cos p.E * Work done in rotating dipole from the position
1 to 2 is given by W = pE (cos 1 – cos 2).
Case (i) : If the dipole be rotated through 90°
from the direction of the field, then the work done
W = pE (1 – cos 90°) = pE (1 – 0) = pE
47 ELECTROSTATICS
Case (ii) : The work done in rotating through * 12th PHYSICS MODULE - 1
180° from the direction of the field
W = pE (1 – cos 180°) = pE [1 – (–1)] = 2pE Electric field lines are perpendicular to
equipotential surfaces (or curves) and point in
* Microwave oven : It is used to cook the food * the direction from higher potential to lower
in a short time. When the oven is operated, the potential.
microwaves are generated, which in turn produce In the region where the electric field is strong,
a non-uniform oscillatingelectric field. The water * the equipotentials are closely packed as the
molecules in the food which are the electric gradient is large.
dipoles are excited by an oscillating torque. For a point charge the equipotential surface is
Hence few bonds in the water molecules are spherical. For a line charge equipotential surface
broken, and heat energy is produced. is cylindrical and for uniform field the equipotential
surface is planar.
1.16 EQUIPOTENTIAL
SURFACE Properties of
* Equipotential surfaces are defined as surfaces (i) Equipotential surface
over wVh(irc)h the potential is constant (ii) Potential difference between two points in an
= constant equipotential surface is zero.
If a test charge q0 is moved from one point to
* At each point on the surface, the electric field is the other on such a surface, the electric potential
energy q0V remains constant.
perpendicular to the surface sincetheelectricfield, (iii) No work is done by the electric force when the
being the gradient of potential, does not have test charge is moved along this surface.
Two equipotential surfaces can never intersect
component along a surface of constant potential. (iv) each other because otherwise the point of
intersection will have two potentials which is of
(v) course not acceptable.
(vi) Field lines and equipotential surfaces are always
mutually perpendicular.
Charged conductors is always equipotential.
Connected
conducting spheres
Two conductors connected can be seen as a
single conductor. R1
Potential everywhere is identical.
Potential of radius R1 sphere q1
V1 = q1
40R1
Potential of radius R2 sphere
(c) V2 = q2 R2
40 R 2 q2
Figure : Equipotential surfaces and electric field
lines for (a) a uniform electric field produced by an
infinite sheet of charge, (b) a point charge, and (c)
an electric dipole.
48 ELECTROSTATICS
V1 V2 q1 q2 q1 R1 12th PHYSICS MODULE - 1
R1 R2 q2 R2
10V 20V 30V 40V
Surface charge density E
30° 30° 30° 30°
x(cm)
1 q1
4R12
Surface area of radius R1 sphere As we know, that, |E| dV
dx
1 2
2 q1 . R 2 R2 V
q2 R1 x
R12 or for uniform electric field E = ;
If R1 < R2 then 1 > 2 V = potential difference
And the surface electric field E1 > E2 x = distance between equipotential surfaces
For arbitrary shape conductor:
20 10 10
At every point on the conductor, we fit a circle. E = 10sin 30 5 = 2 V/m
The radius of this circle is the radius of curvature.
r2 P2 EXAMPLE 33
P1 r1 r3 A, B and C are three points in a uniform electric
field then find the relation among VA,VB and VC
P3
B A
E3 < E2 < E1 E
C
Note: Charge distribution on a conductor does
not have to be uniform. SOLUTION: Line perpendicular to electric field will
be equipotential and in the direction of electric
EXAMPLE 32 field potential decreases. VB = VC ; VA < VB.
Some equipotential surfaces are shown in figure.
Find the electric field.
y(cm) 10V 20V 30V 40V
O 30° 30° 30° 30° Checkup 7
10 20 30 40 x(cm)
Q.1 Explain why equipotential surfaces are always
perpendicular to electric field lines.
Q.2 The labeled points in Figure are on a series of
equipotential surfaces associated with an electric
S O L UT I ON: the direction of electric field is field. Rank (from greatest to least) the work done
(i) We know by the electric field on a positively charged
particle that moves from A to B; from B to C;
to the equipotential surfaces. from C to D; from D to E.
(ii) In the direction of electric field potential
decrease.
Keeping these 2 things in mind we get B
direction of electric field is 120° from 9V A
8V
x-axis i.e. as shown in figure. E
C D
7V
6V
49 ELECTROSTATICS
12th PHYSICS MODULE - 1
Q.3 For the equipotential surfaces in Figure, what is – +
the approximate direction of the electric field? – +
(a) Out of the page (b) Into the page (c) Toward Eint
the right edge of the page (d) Toward the left –+
edge of the page (e) Toward the top of the page –+
(f) Toward the bottom of the page. –+
–+
Q.4 Figure shows several equipotential lines each –+
labeled by its potential in volts. The distance * –+
between the lines of the square grid represents * –+
1.00 cm. (a) Is the magnitude of the field larger –+
at A or at B? Why? (b) What is E at B?
A* Eext
Charge density inside a conductor is zero.
Free charges exist only on the surface of a
conductor: Since there is no net charge inside,
free charges, if any, have to be on the surface.
At the surface of a conductor, the electric
field is normal to the surface:
8 E=0
6
4
2
0
B
1.17 CONDUCTORS If this were not so, the charges on the surface
would move along the surface because of the
* A conductor (typically, a metal like Cu,Ag etc. tangential component of the field, disturbing
equilibrium.
or ionic conductors like HCl or NaCl dissolved * Electric field just outside the conductor :
in water) allows free movement of charges.
* They have low resistivity 10–8m as compared
to typical insulators like quartz, glass etc. which
have resistivity of the order of 1017m.
* However, the property that really distinguishes a
metal from insulators or semi-conductors is the
fact that their temperature coefficient of resistivity
is positive while that of semi-conductors is
negative. Figure : A gaussian surface in the shape of a
* The electric field inside a conductor is zero. small cylinder is used to calculate the electric
* In an equilibrium situation, there cannot be an field just outside a charged conductor. The flux
through the gaussian surface is EA.
electric field inside a conductor as this would Remember that E is zero inside the conductor.
cause charges (electrons or ions) to move around.
* In the presence of an external field, there is charge Applying Gauss’s law , we obtain
separation inside a conductors with opposite
charges accumulating on the surface. This creates E E dA EA qin A
an internal electric field which cancels the effect used 0 0
of the external field in such a way that the net where we have the fact that qin = A.
electric field inside the conductor volume is zero.
E
0
50 ELECTROSTATICS
12th PHYSICS MODULE - 1
Note : E and at sharp points is more while SOLUTION: 2 2
potential is same. 20 20
Pressure = and force = R 2
1.18 ELECTROSTATIC
PRESSURE
Force on small element ds of charged conductor 1.19 EQUILIBRIUM
dF = (Charge on ds ) × Electric field OF SOAP BUBBLE
= ds) +++ + + ++ Pressures (forces) act on a charged soap bubble,
2 0 ++
+++ + + ++ ds due to ++ + + +R+++++
Surface tension PT (inward) + + + + ++
or dF 2 ds (i) Air outside the bubble Po ++
2 0 (ii) (inward)
Inside E1 = E2 = 0 E1 = E2 (iii) Electrostatic pressure Pe (outward)
(iv) Air inside the bubble Pi (outward)
0 2 0 In state of equilibrium
Just outside E = E1 + E2 = E2
Inward pressure = Outward pressure
The electric force acting per unit area of charged or PT + Po = Pi +Pe
surface is defined as electrostatic pressure Excessive pressure of air in side the bubble
2 E2 E1 E2 (Pex) = Pi – Po = PT – Pe
2 0
Pele dF E1 + 4T 2 Pex = 4T 2
dS r 2 0 r 2 0
but PT = and Pe =
* Electrostatic pressure directs normally outwards If Pi = Po then 4T 2
to the surface. r 2 0
* Energyassociated per unit volume ofelectric field
of intensity is defined as energy density
u dW 0 E2 2 J / m3 1.20 MIXING OF IDENTICAL
dV 2 2 0 CHARGED TINY DROPS
U = u.dV = 0 E2 dV ; Let , number of tiny drops = N
2
V is the volume of electric field. R
N tiny drops Big drops
EXAMPLE 34 For each tiny drop For Big drop
A uniformlycharged thin spherical shell of radius (r, q, , E,V) (R, Q, B, EB, VB)
R carries uniform surface charge densityof per (i) Charge conservation Q = Nq
unit area. It is made of two hemispherical shells,
held together by pressing them with force F (see (ii) Volume conservation 4 N.r3 = 4 R3.
figure). Find force F. 3 3
Hence R = N1/3 r , Q = Nq
FF B = N1/3 , EB = N1/3 E , VB = N2/3 V
* When a soap bubble is charged (either positive
or negative) then the size (radius) some what
increases.
51 ELECTROSTATICS
12th PHYSICS MODULE - 1
EXAMPLE 35 * Polarisation : The alignment of dipole moments
* of permanent or induced dipoles in the direction
Eight charged water droplets, each with a radius applied electric field is called polarisation.
of 1mm and a charge of 10–10 C, coalesce to * The dipole moment per unit volume is called
form a single drop. Calculate the potential of the polarisation and is denoted by P. For linear
bigger drop.
isotropic dielectrics, P eE
SOLUTION:
Let R and r be the radii of the bigger drop and where e is a constant characteristic of the
one droplet respectively. dielectric and is known as the electric
Volume of 8 droplets = Volume of bigger drop
susceptibility of the dielectric medium.
8 4 r3 4 R 3 or R = (8)1/3r = 2r The chargeappearing on the surfaceof a dielectric
3 3
when placed in an electric field is called induced
or R = 2 × 1 × 10–3m = 2 × 10–3m charge.As the induced charge appears due to a
Charge on bigger drop, Q = 8 × 10–10C shift in the electrons bound to the nuclei, this
Potential of the bigger drop charge is also called bound charge.
Because of the induced charges, an extra electric
8 1010
1 Q 109 2 103 field is produced inside the material. If E0 be
= 4 0 R 9 volt 3600 volt
EP
the applied field due to external sources and
be the field due to polarization.
1.21 DIELECTRICS AND –+ –+ – +–
POLARISATION –+ –+ –+ + –+ –+
– + –+ –+ –+ –
(a)–++ + –+ –+ –+ –
* In dielectric materials, effectivelythere are no free + –
electrons. –+ –+ –+ –+
* Polar dielectrics : –+ + –+ –+ –+ –
(i) In absence of external field the centre of positive
+–
and negative charge do not coincide-due to
asymmetric shape of molecules. E0
(ii) Each molecule has permanent dipole moment. (b)
(iii) The dipole are randomly oriented so average
dipole moment per unit volume of polar dielectric +– +–
in absence of external field is nearly zero.
(iv) In presence of external field dipoles tends to align +–
in direction of field.
(v) Ex. water, alcohol, CO2, HCl, NH3 + E0 –
+ –
+ Eind –
+ –
–ind
ind
* Non polar dielectrics : Figure : (a) Polar molecules are randomly
(i) In absence of external field the centre of positive oriented in the absence of an external
electric field. (b) When an external
and negative charge coincides in these atoms or electric field is applied, the molecules
molecules because they are symmetric. partially align with the field. (c) The
(ii) The dipole moment is zero in normal state. charged edges of the dielectric can be
(iii) In presence of external field theyacquire induced modeled as an additional pair of parallel
dipole moment. plates establishing an electric field Eind
(iv) Ex. nitrogen, oxygen, benzene, methane in the direction opposite to that of E0.
52 ELECTROSTATICS
The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.
NEET - PHYSICS ELECTROSTATICS
Discover the best professional documents and content resources in AnyFlip Document Base.
Search