chapter 2 electrostatic potential and capacitance

7/28/2019 Chapter 2 Electrostatic Potential and Capacitance

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Question 2.1:

Two charges 5 108 C and 3 108 C are located 16 cm apart. At what point(s) on theline joining the two charges is the electric potential zero? Take the potential at infinity to

be zero.

Answer

There are two charges,

Distance between the two charges, d= 16 cm = 0.16 m

Consider a point P on the line joining the two charges, as shown in the given figure.

r= Distance of point P from charge q1

Let the electric potential (V) at point P be zero.

Potential at point P is the sum of potentials caused by charges q1 and q2 respectively.

Where,

= Permittivity of free space

ForV= 0, equation (i) reduces to


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Therefore, the potential is zero at a distance of 10 cm from the positive charge betweenthe charges.

Suppose point P is outside the system of two charges at a distances from the negativecharge, where potential is zero, as shown in the following figure.

For this arrangement, potential is given by,

ForV= 0, equation (ii) reduces to

Therefore, the potential is zero at a distance of 40 cm from the positive charge outside thesystem of charges.


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Question 2.2:

A regular hexagon of side 10potential at the centre of the

Answer

The given figure shows six ehexagon.

Where,

Charge, q = 5 C = 5 106

Side of the hexagon, l= AB

Distance of each vertex from

Electric potential at point O,

Where,

= Permittivity of free spa

cm has a charge 5 C at each of its vertices. Cexagon.

qual amount of charges, q, at the vertices of a re

C

BC = CD = DE = EF = FA = 10 cm

centre O, d= 10 cm

e

lculate the

gular


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Therefore, the potential at th

Question 2.3:

Two charges 2 C and 2

Identify an equipotential sur

What is the direction of the e

Answer

The situation is represented i

An equipotential surface is tplane is normal to line AB. Tmagnitude of charges is the s

The direction of the electricthe direction of AB.

Question 2.4:

A spherical conductor of radon its surface. What is the el

Inside the sphere

Just outside the sphere

centre of the hexagon is 2.7 106 V.

are placed at points A and B 6 cm apart.

ace of the system.

lectric field at every point on this surface?

the given figure.

e plane on which total potential is zero everywhe plane is located at the mid-point of line ABame.

ield at every point on this surface is normal to t

us 12 cm has a charge of 1.6 107C distributectric field

ere. Thisecause the

e plane in

d uniformly


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At a point 18 cm from the centre of the sphere?

Answer

Radius of the spherical conductor, r= 12 cm = 0.12 m

Charge is uniformly distributed over the conductor, q = 1.6 107 C

Electric field inside a spherical conductor is zero. This is because if there is field insidethe conductor, then charges will move to neutralize it.

Electric fieldEjust outside the conductor is given by the relation,

Where,


Therefore, the electric field just outside the sphere is .

Electric field at a point 18 m from the centre of the sphere =E1

Distance of the point from the centre, d= 18 cm = 0.18 m

Therefore, the electric field at a point 18 cm from the centre of the sphere is


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.

Question 2.5:

A parallel plate capacitor wit1012 F). What will be the cahalf, and the space between t

Answer

Capacitance between the par

Initially, distance between thconstant of air, k= 1

Capacitance, C, is given by t

Where,

A = Area of each plate


If distance between the plate

Dielectric constant of the sub

Hence, capacitance of the ca

h air between the plates has a capacitance of 8 pacitance if the distance between the plates is reem is filled with a substance of dielectric cons

llel plates of the capacitor, C = 8 pF

e parallel plates was dand it was filled with air.

e formula,

e

is reduced to half, then new distance, d =

stance filled in between the plates, = 6

acitor becomes

F (1pF =duced byant 6?

Dielectric


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Taking ratios of equations (i)

Therefore, the capacitance b

Question 2.6:

Three capacitors each of cap

What is the total capacitance

What is the potential differe

120 V supply?

Answer

Capacitance of each of the th

Equivalent capacitance (C)

Therefore, total capacitance

Supply voltage, V= 100 V

Potential difference (V) acro

Therefore, the potential diffe

and (ii), we obtain

tween the plates is 96 pF.

citance 9 pF are connected in series.

of the combination?

ce across each capacitor if the combination is c

ree capacitors, C= 9 pF

f the combination of the capacitors is given by

f the combination is .

ss each capacitor is equal to one-third of the su

ence across each capacitor is 40 V.

nnected to a

he relation,

ply voltage.


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Question 2.7:

Three capacitors of capacita

What is the total capacitance

Determine the charge on eacsupply.

Answer

Capacitances of the given ca

For the parallel combinationalgebraic sum,

Therefore, total capacitance


The voltage through all the t

Charge on a capacitor of cap

q = VC (i)

For C = 2 pF,

For C = 3 pF,

ces 2 pF, 3 pF and 4 pF are connected in parall

of the combination?

capacitor if the combination is connected to a

acitors are

of the capacitors, equivalent capacitor is giv

f the combination is 9 pF.

ree capacitors is same = V= 100 V

citance Cand potential difference Vis given b

l.

100 V

n by the

the relation,


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For C = 4 pF,

Question 2.8:

In a parallel plate capacitorm2 and the distance betweencapacitor. If this capacitor is

plate of the capacitor?

Answer

Area of each plate of the par

Distance between the plates,


Capacitance Cof a parallel p

Where,


= 8.854 1012N1 m2 C2

ith air between the plates, each plate has an arethe plates is 3 mm. Calculate the capacitance ofconnected to a 100 V supply, what is the charg

llel plate capacitor,A = 6 103 m2

d= 3 mm = 3 103 m

late capacitor is given by,

e

a of 6 103

theon each


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Therefore, capacitance of the109 C.

Question 2.9:

Explain what would happensheet (of dielectric constant

While the voltage supply re

After the supply was disconn

Answer

Dielectric constant of the mi

Initial capacitance, C= 1.771


Potential across the plates re

Dielectric constant, k= 6

capacitor is 17.71 pF and charge on each plate

f in the capacitor given in Exercise 2.8, a 3 mm6) were inserted between the plates,

ained connected.

ected.

a sheet, k= 6

1011 F

ains 100 V.

is 1.771

thick mica


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Initial capacitance, C= 1.771

If supply voltage is removed,plates.

Charge = 1.771 109 C

Potential across the plates is

Question 2.10:

A 12 pF capacitor is connectin the capacitor?

Answer

Capacitor of the capacitance,

Potential difference, V= 50

Electrostatic energy stored in

Therefore, the electrostatic e

1011 F

then there will be no effect on the amount of c

given by,

d to a 50V battery. How much electrostatic en

C= 12 pF = 12 1012 F

the capacitor is given by the relation,

ergy stored in the capacitor is

arge in the

rgy is stored


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Question 2.11:

A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supplyand is connected to another uncharged 600 pF capacitor. How much electrostatic energyis lost in the process?

Answer

Capacitance of the capacitor, C= 600 pF

Potential difference, V= 200 V

Electrostatic energy stored in the capacitor is given by,

If supply is disconnected from the capacitor and another capacitor of capacitance C= 600pF is connected to it, then equivalent capacitance (C) of the combination is given by,

New electrostatic energy can be calculated as


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Therefore, the electrostatic e

Question 2.12:

A charge of 8 mC is locatedcharge of2 109 C from a(0, 6 cm, 9 cm).

Answer

Charge located at the origin,

Magnitude of a small charge, 109 C

All the points are represente

Point P is at a distance, d1 =

Point Q is at a distance, d2 =

ergy lost in the process is .

t the origin. Calculate the work done in takingpoint P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), v

q = 8 mC= 8 103 C

which is taken from a point P to point R to poi

in the given figure.

cm, from the origin alongz-axis.

4 cm, from the origin alongy-axis.

a smalla a point R

t Q, q1 = 2


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Potential at point P,

Potential at point Q,

Work done (W) by the electr

Therefore, work done during

Question 2.13:

A cube of side b has a chargfield due to this charge array

Answer

Length of the side of a cube

Charge at each of its vertices

A cube of side b is shown in

static force is independent of the path.

the process is 1.27 J.

q at each of its vertices. Determine the potentiat the centre of the cube.

b

= q

the following figure.

l and electric


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d= Diagonal of one of the si

l= Length of the diagonal of

The electric potential (V) at tat the vertices.

Therefore, the potential at th

The electric field at the centrbecause the charges are distrHence, the electric field is ze

faces of the cube

the cube

e centre of the cube is due to the presence of ei

centre of the cube is .

e of the cube, due to the eight charges, gets canbuted symmetrically with respect to the centrero at the centre.

ght charges

elled. This isf the cube.


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Question 2.14:

Two tiny spheres carrying charges 1.5 C and 2.5 C are located 30 cm apart. Find the

potential and electric field:

at the mid-point of the line joining the two charges, and

at a point 10 cm from this midpoint in a plane normal to the line and passing through themid-point.

Answer

Two charges placed at points A and B are represented in the given figure. O is the mid-point of the line joining the two charges.

Magnitude of charge located at A, q1 = 1.5 C

Magnitude of charge located at B, q2 = 2.5 C

Distance between the two charges, d= 30 cm = 0.3 m

Let V1 andE1 are the electric potential and electric field respectively at O.

V1 = Potential due to charge at A + Potential due to charge at B

Where,

0 = Permittivity of free space


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E1 = Electric field due to q2 Electric field due to q1

Therefore, the potential at mid-point is 2.4 105 V and the electric field at mid-point is4 105 V m1. The field is directed from the larger charge to the smaller charge.

Consider a point Z such that normal distanceOZ = 10 cm = 0.1 m, as shown in thefollowing figure.

V2 andE2 are the electric potential and electric field respectively at Z.

It can be observed from the figure that distance,

V2= Electric potential due to A + Electric Potential due to B

Electric field due to q at Z,


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Electric field due to q2 atZ,

The resultant field intensity a

Where, 2is the angle, AZ

From the figure, we obtain

Therefore, the potential at aand electric field is 6.6 105

Question 2.15:

A spherical conducting shell

A charge q is placed at the cinner and outer surfaces of t

Is the electric field inside a cbut has any irregular shape?

Answer

t Z,

oint 10 cm (perpendicular to the mid-point) ism1.

of inner radius r1 and outer radius r2 has a char

ntre of the shell. What is the surface charge dee shell?

avity (with no charge) zero, even if the shell isxplain.

.0 105 V

ge Q.

sity on the

ot spherical,


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Charge placed at the centre oinduced to the inner surfacethe shell is q.

Surface charge density at the

A charge of +q is induced onplaced on the outer surface oshell is Q + q. Surface charg

Yes

The electric field intensity inany irregular shape. Take a cfield line while the rest is instest charge over a closed looHence, electric field is zero,

Question 2.16:

Show that the normal compoa charged surface to another

Where is a unit vector nor

density at that point. (The di

outside a conductor, the elec

Show that the tangential comcharged surface to another. [done by electrostatic field on

Answer

f a shell is +q. Hence, a charge of magnitude f the shell. Therefore, total charge on the inner

inner surface of the shell is given by the relatio

the outer surface of the shell. A charge of magf the shell. Therefore, total charge on the outer

density at the outer surface of the shell,

side a cavity is zero, even if the shell is not sphlosed loop such that a part of it is inside the cavide the conductor. Net work done by the field i

is zero because the field inside the conductor ihatever is the shape.

nent of electrostatic field has a discontinuity frogiven by

al to the surface at a point and is the surface

ection of is from side 1 to side 2.) Hence sho

ric field is

ponent of electrostatic field is continuous fromint: For (a), use Gausss law. For, (b) use the f

a closed loop is zero.]

will besurface of

,

itude Q isurface of the

rical and hasity along acarrying a

s zero.

m one side of

charge

that just

one side of aact that work


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Electric field on one side ofsame body isE2. If infinite p

due to one surface of the cha

Where,

= Unit vector normal to th

= Surface charge density a

Electric field due to the othe

Electric field at any point du

Since inside a closed conduc

Therefore, the electric field j

When a charged particle isdone by the electrostatic fielfield is continuous from one

Question 2.17:

charged body isE1 and electric field on the otane charged body has a uniform thickness, then

ged body is given by,

surface at a point

that point

surface of the charged body,

to the two surfaces,

or, = 0,

st outside the conductor is .

oved from one point to the other on a closed lois zero. Hence, the tangential component of elside of a charged surface to the other.

er side of theelectric field

p, the workctrostatic


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A long charged cylinder of li

conducting cylinder. What is

Answer

Charge density of the long c

Another cylinder of same lencylinder isR.

LetEbe the electric field pr

Electric flux through the Ga

Where, d= Distance of a poi

Let qbe the total charge on t

It can be written as

Where,

q = Charge on the inner sphe

0 = Permittivity of free spa

Therefore, the electric field i

near charged density is surrounded by a hollo

the electric field in the space between the two c

arged cylinder of lengthL and radius ris.

gth surrounds the pervious cylinder. The radius

duced in the space between the two cylinders.

ssian surface is given by Gausss theorem as,

t from the common axis of the cylinders

e cylinder.

re of the outer cylinder

e

the space between the two cylinders is

co-axialylinders?

of this

.


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Question 2.18:

In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 :

Estimate the potential energy of the system in eV, taking the zero of the potential energyat infinite separation of the electron from proton.

What is the minimum work required to free the electron, given that its kinetic energy inthe orbit is half the magnitude of potential energy obtained in (a)?

What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 separation?

Answer

The distance between electron-proton of a hydrogen atom,

Charge on an electron, q1 = 1.6 1019 C

Charge on a proton, q2 = +1.6 1019 C

Potential at infinity is zero.

Potential energy of the system,p-e = Potential energy at infinity Potential energy atdistance, d

Where,

0 is the permittivity of free space


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Therefore, the potential ener

Kinetic energy is half of the

Total energy = 13.6 27.2 =

Therefore, the minimum wor

When zero of potential energ

Potential energy of the syst

Question 2.19:

If one of the two electrons o

. In the ground state of athe electron is roughly 1 frsystem. Specify your choice

Answer

The system of two protons a

y of the system is 27.2 eV.

agnitude of potential energy.

13.6 eV

k required to free the electron is 13.6 eV.

y is taken,

m = Potential energy at d1 Potential energy a

a H2 molecule is removed, we get a hydrogen

, the two protons are separated by roughlyom each proton. Determine the potential energyof the zero of potential energy.

d one electron is represented in the given figur

d

olecular ion

1.5 , andof the

.


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Charge on proton 1, q1 = 1.6

Charge on proton 2, q2 = 1.6

Charge on electron, q3 = 1.

Distance between protons 1

Distance between proton 1 a

Distance between proton 2 a

The potential energy at infini

Potential energy of the syste

Therefore, the potential ener

Question 2.20:

Two charged conducting sphWhat is the ratio of electric fobtained to explain why charhigher than on its flatter port

Answer

1019 C

1019 C

1019 C

nd 2, d1 = 1.5 1010 m

d electron, d2 = 1 1010 m

d electron, d3 = 1 1010 m

ty is zero.

,

y of the system is 19.2 eV.

eres of radii a and b are connected to each otheields at the surfaces of the two spheres? Use thege density on the sharp and pointed ends of a cons.

by a wire.resultnductor is


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Let abe the radius of a sphecapacitance of the sphere. Lesphere, and CBbe the capacita wire, their potential (V) wil

LetEAbe the electric field oftheir ratio,

Putting the value of (2) in (1

Therefore, the ratio of electri

Question 2.21:

Two charges q and +q are l

What is the electrostatic pote

Obtain the dependence of po>> 1.

e A, QAbe the charge on the sphere, and CAbet bbe the radius of a sphere B, QBbe the chargance of the sphere. Since the two spheres are col become equal.

sphere A andEBbe the electric field of sphere

, we obtain

c fields at the surface is .

ocated at points (0, 0, a) and (0, 0, a), respect

ntial at the points?

ential on the distance rof a point from the origi

heon the

nnected with

. Therefore,

ively.

n when r/a


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How much work is done in moving a small test charge from the point (5, 0, 0) to (7, 0,0) along thex-axis? Does the answer change if the path of the test charge between thesame points is not along thex-axis?

Answer

Zero at both the points

Charge q is located at (0, 0, a) and charge + q is located at (0, 0, a). Hence, they forma dipole. Point (0, 0,z) is on the axis of this dipole and point (x,y, 0) is normal to the axisof the dipole. Hence, electrostatic potential at point (x,y, 0) is zero. Electrostatic potentialat point (0, 0,z) is given by,

Where,


p = Dipole moment of the system of two charges = 2qa

Distance ris much greater than half of the distance between the two charges. Hence, thepotential (V) at a distance ris inversely proportional to square of the distance i.e.,

Zero

The answer does not change if the path of the test is not along thex-axis.

A test charge is moved from point (5, 0, 0) to point (7, 0, 0) along thex-axis.Electrostatic potential (V1) at point (5, 0, 0) is given by,


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Electrostatic potential, V2, at

Hence, no work is done in m0) along thex-axis.

The answer does not changecharge between the two poin

Question 2.22:

Figure 2.34 shows a charge aof the quadrupole, obtain the

results with that due to an el

Answer

Four charges of same magnitshown in the following figur

point ( 7, 0, 0) is given by,

oving a small test charge from point (5, 0, 0) to

because work done by the electrostatic field ins is independent of the path connecting the two

rray known as an electric quadrupole. For a poidependence of potential on rforr/a >> 1, and

ctric dipole, and an electric monopole (i.e., a si

ude are placed at points X, Y, Y, and Z respecti.

point (7, 0,

oving a testpoints.

nt on the axisontrast your

gle charge).

vely, as


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A point is located at P, which is rdistance away from point Y.

The system of charges forms an electric quadrupole.

It can be considered that the system of the electric quadrupole has three charges.

Charge +qplaced at point X

Charge 2qplaced at point Y

Charge +qplaced at point Z

XY = YZ = a

YP = r

PX = r+ a

PZ = r a

Electrostatic potential caused by the system of three charges at point P is given by,

Since ,

is taken as negligible.


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It can be inferred that potenti

However, it is known that fo

And, for a monopole,

Question 2.23:

An electrical technician requdifference of 1 kV. A large ncan withstand a potential difarrangement that requires the

Answer

Total required capacitance,

Potential difference, V= 1 k

Capacitance of each capacito

Each capacitor can withstan

Suppose a number of capacitconnected in parallel (row) t

be 1000 V and potential diffof capacitors in each row is

Hence, there are three capaci

al,

a dipole,

ires a capacitance of 2 F in a circuit across a pumber of 1 F capacitors are available to him eerence of not more than 400 V. Suggest a possiminimum number of capacitors.

= 2 F

= 1000 V

r, C1 = 1F

a potential difference, V1 = 400 V

ors are connected in series and these series circeach other. The potential difference across eac

rence across each capacitor must be 400 V. Heiven as

tors in each row.

tentialch of whichle

its areh row must

ce, number


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Capacitance of each row

Let there are n rows, each haHence, equivalent capacitan

Hence, 6 rows of three capac

capacitors are required for th

Question 2.24:

What is the area of the platesbetween the plates is 0.5 cm?are in the range of F or less.capacitance (0.1 F) because

Answer

Capacitance of a parallel cap

Distance between the two pl

Capacitance of a parallel pla

Where,

ing three capacitors, which are connected in pae of the circuit is given as

itors are present in the circuit. A minimum of 6

e given arrangement.

of a 2 F parallel plate capacitor, given that the[You will realize from your answer why ordinHowever, electrolytic capacitors do have a muf very minute separation between the conducto

acitor, V= 2 F

tes, d= 0.5 cm = 0.5 102 m

e capacitor is given by the relation,

rallel.

3 i.e., 18

eparationry capacitors

ch largers.]


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Hence, the area of the platesin the range of F.

Question 2.25:

Obtain the equivalent capacidetermine the charge and vol

Answer

Capacitance of capacitorC1 i




Supply potential, V= 300 V

Capacitors C2 and C3 are con

e = 8.85 1012 C2N1 m2

is too large. To avoid this situation, the capacita

ance of the network in Fig. 2.35. For a 300 V stage across each capacitor.

s 100 pF.

s 200 pF.

s 200 pF.

s 100 pF.

nected in series. Let their equivalent capacitanc

nce is taken

pply,

e be


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Capacitors C1 and Care in parallel. Let their equivalent capacitance be

are connected in series. Let their equivalent capacitance be C.

Hence, the equivalent capacitance of the circuit is

Potential difference across =

Potential difference across C4 = V4

Charge on

Q4= CV


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Hence, potential difference,

Charge on C1 is given by,

C2 and C3 having same capaC2 and C3 are in series, the p

V2 = V3 = 50 V

Therefore, charge on C2 is gi

And charge on C3 is given by

Hence, the equivalent capaci

Question 2.26:

The plates of a parallel plate2.5 mm. The capacitor is cha

How much electrostatic ener

View this energy as stored inenergy per unit volume u. Helectric fieldEbetween the p

1, across C1 is 100 V.

itances have a potential difference of 100 V togtential difference across C2 and C3 is given by,

en by,

,

ance of the given circuit is

capacitor have an area of 90 cm2 each and are srged by connecting it to a 400 V supply.

y is stored by the capacitor?

the electrostatic field between the plates, and once arrive at a relation between u and the magnlates.

ether. Since

parated by

tain theitude of


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Answer

Area of the plates of a parallel plate capacitor,A = 90 cm2 = 90 104 m2

Distance between the plates, d= 2.5 mm = 2.5 103 m

Potential difference across the plates, V= 400 V

Capacitance of the capacitor is given by the relation,

Electrostatic energy stored in the capacitor is given by the relation,

Where,

= Permittivity of free space = 8.85 1012 C2N1 m2

Hence, the electrostatic energy stored by the capacitor is

Volume of the given capacitor,

Energy stored in the capacitor per unit volume is given by,


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Where,

= Electric intensity =E

Question 2.27:

A 4 F capacitor is chargedand is connected to anotherthe first capacitor is lost in th

Answer

Capacitance of a charged cap

Supply voltage, V1 = 200 V

Electrostatic energy stored in

Capacitance of an uncharged

y a 200 V supply. It is then disconnected fromncharged 2 F capacitor. How much electrostate form of heat and electromagnetic radiation?

acitor,

C1 is given by,

capacitor,

the supply,ic energy of


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When C2 is connected to the

According to the conservatiofinal charge on capacitors, C

Electrostatic energy for the c

Hence, amount of electrostat

=E1 E2

= 0.08 0.0533 = 0.0267

= 2.67 102 J

Question 2.28:

Show that the force on each() QE, where Q is the char

between the plates. Explain t

Answer

LetFbe the force applied toof x. Hence, work done by

As a result, the potential ene

circuit, the potential acquired by it is V2.

n of charge, initial charge on capacitorC1 is eqand C2.

ombination of two capacitors is given by,

ic energy lost by capacitorC1

late of a parallel plate capacitor has a magnitude on the capacitor, andEis the magnitude of ele origin of the factor .

separate the plates of a parallel plate capacitorhe force to do so =Fx

gy of the capacitor increases by an amount give

al to the

e equal toectric field

y a distance

n as uAx.


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Where,

u = Energy density


d= Distance between the pla

V= Potential difference acro

The work done will be equal

Electric intensity is given by,

However, capacitance,

Charge on the capacitor is gi

Q = CV

The physical origin of the fa

the conductor, field isEandfield that contributes to the f

Question 2.29:

tes

ss the plates

to the increase in the potential energy i.e.,

en by,

tor, , in the force formula lies in the fact that

inside it is zero. Hence, it is the average value,rce.

just outside

, of the


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A spherical capacitor consists of two concentric spherical conductors, held in position bysuitable insulating supports (Fig. 2.36). Show

that the capacitance of a spherical capacitor is given by

where r1 and r2 are the radii of outer and inner spheres, respectively.

Answer

Radius of the outer shell = r1

Radius of the inner shell = r2

The inner surface of the outer shell has charge +Q.

The outer surface of the inner shell has induced charge Q.

Potential difference between the two shells is given by,

Where,



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Hence, proved.

Question 2.30:

A spherical capacitor has ancm. The outer sphere is earthspace between the concentric

Determine the capacitance o

What is the potential of the i

Compare the capacitance ofExplain why the latter is mu

Answer

Radius of the inner sphere,

Radius of the outer sphere,

Charge on the inner sphere,

Dielectric constant of a liqui

inner sphere of radius 12 cm and an outer sphered and the inner sphere is given a charge of 2.5spheres is filled with a liquid of dielectric cons

the capacitor.

ner sphere?

his capacitor with that of an isolated sphere of rh smaller.

= 12 cm = 0.12 m

= 13 cm = 0.13 m

,

of radius 13C. Thetant 32.

adius 12 cm.


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Where,


Hence, the capacitance of th

Potential of the inner sphere

Hence, the potential of the in

Radius of an isolated sphere,

Capacitance of the sphere is

The capacitance of the isolatThis is because the outer sphdifference is less and the cap

Question 2.31:

Answer carefully:

Two large conducting sphere

Is the magnitude of electrostwhere ris the distance betwe

e =

capacitor is approximately .

is given by,

ner sphere is .

r= 12 102 m

iven by the relation,

d sphere is less in comparison to the concentriere of the concentric spheres is earthed. Hence,acitance is more than the isolated sphere.

s carrying charges Q1 and Q2 are brought close

tic force between them exactly given by Q1 Q2en their centres?

spheres.the potential

o each other.

4 r2,


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If Coulombs law involved 1/r3 dependence (instead of 1/r2), would Gausss law be stilltrue?

A small test charge is released at rest at a point in an electrostatic field configuration. Willit travel along the field line passing through that point?

What is the work done by the field of a nucleus in a complete circular orbit of theelectron? What if the orbit is elliptical?

We know that electric field is discontinuous across the surface of a charged conductor. Iselectric potential also discontinuous there?

What meaning would you give to the capacitance of a single conductor?

Guess a possible reason why water has a much greater dielectric constant (= 80) than say,mica (= 6).

Answer

The force between two conducting spheres is not exactly given by the expression, Q1

Q2/4 r2, because there is a non-uniform charge distribution on the spheres.

Gausss law will not be true, if Coulombs law involved 1/r3 dependence, instead of1/r2,

on r.

Yes,

If a small test charge is released at rest at a point in an electrostatic field configuration,then it will travel along the field lines passing through the point, only if the field lines arestraight. This is because the field lines give the direction of acceleration and not ofvelocity.

Whenever the electron completes an orbit, either circular or elliptical, the work done bythe field of a nucleus is zero.

No

Electric field is discontinuous across the surface of a charged conductor. However,electric potential is continuous.

The capacitance of a single conductor is considered as a parallel plate capacitor with oneof its two plates at infinity.


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Water has an unsymmetricalmoment, it has a greater diel

Question 2.32:

A cylindrical capacitor has t1.4 cm. The outer cylinder isDetermine the capacitance oend effects (i.e., bending of f

Answer

Length of a co-axial cylinder

Radius of outer cylinder, r1

Radius of inner cylinder, r2

Charge on the inner cylinder,

Where,


Potential difference of the in

space as compared to mica. Since it has a permctric constant than mica.

o co-axial cylinders of length 15 cm and radiiearthed and the inner cylinder is given a chargethe system and the potential of the inner cylind

ield lines at the ends).

, l= 15 cm = 0.15 m

1.5 cm = 0.015 m

1.4 cm = 0.014 m

q = 3.5 C = 3.5 106 C

e =

er cylinder is given by,

nent dipole

.5 cm andof 3.5 C.er. Neglect


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Question 2.33:

A parallel plate capacitor is tdielectric constant 3 and dielmaximum electric field a maconduct electricity through pexceed, say 10% of the dielehave a capacitance of 50 pF?

Answer

Potential rating of a parallel

Dielectric constant of a mate

Dielectric strength = 107

V/

For safety, the field intensity

Hence, electric field intensit

Capacitance of the parallel p

Distance between the plates i

Where,

o be designed with a voltage rating 1 kV, usingectric strength about 107 Vm1. (Dielectric strenerial can tolerate without breakdown, i.e., with

artial ionisation.) For safety, we should like thetric strength. What minimum area of the plates

late capacitor, V= 1 kV = 1000 V

ial, =3

never exceeds 10% of the dielectric strength.

,E= 10% of 107 = 106 V/m

ate capacitor, C= 50 pF = 50 1012 F

s given by,

a material ofgth is theut starting tofield never tois required to


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Hence, the area of each plate

Question 2.34:

Describe schematically the e

a constant electric field in th

a field that uniformly increas

a single positive charge at th

a uniform grid consisting of l

Answer

Equidistant planes parallel to

Planes parallel to thex-yplathe planes get closer, the fiel

Concentric spheres centered

A periodically varying shapegradually reaches the shape

Question 2.35:

e =

is about 19 cm2.

uipotential surfaces corresponding to

z-direction,

es in magnitude but remains in a constant (say,

origin, and

ong equally spaced parallel charged wires in a

thex-yplane are the equipotential surfaces.

e are the equipotential surfaces with the exceptincreases.

at the origin are equipotential surfaces.

near the given grid is the equipotential surface.f planes parallel to the grid at a larger distance.

) direction,

lane.

on that when

This shape


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In a Van de Graaff type geneThe dielectric strength of theminimum radius of the spherone cannot build an electrostcharge to acquire a high pote

Answer

Potential difference, V= 15

Dielectric strength of the sur

Electric field intensity,E=

Minimum radius of the sphe

Hence, the minimum radius

Question 2.36:

A small sphere of radius r1 acharge q2. Show that ifq1 isshell (when the two are conn

Answer

According to Gausss law, ththe charge q1 on a small spheand the shell is independentalways positive.

rator a spherical metal shell is to be a 15 106 gas surrounding the electrode is 5 107 Vm1.

ical shell required? (You will learn from this exatic generator using a very small shell which rential.)

106 V

ounding gas = 5 107 V/m

ielectric strength = 5 107 V/m

ical shell required for the purpose is given by,

f the spherical shell required is 30 cm.

d charge q1 is enclosed by a spherical shell ofositive, charge will necessarily flow from the s

ected by a wire) no matter what the charge q2 o

e electric field between a sphere and a shell is dre. Hence, the potential difference, V, betweenf charge q2. For positive charge q1, potential di

electrode.What is theercise whyuires a small

adius r2 andhere to thethe shell is.

etermined byhe sphereference Vis


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Question 2.37:

Answer the following:

The top of the atmosphere iscorresponding to an electricearth, the field is about 100out of our house into the opeinside!)

A man fixes outside his housits top a large aluminium shemetal sheet next morning?

The discharging current in thbe 1800 A on an average ove

itself completely in due courkeeps the atmosphere charge

What are the forms of energdissipated during a lightningits surface in the downwardC m2. Due to the slight conit is good conductor), aboutThe earth, however, does notcontinually all over the glob

Answer

We do not get an electric shoequipotential surfaces of ope

potential.

Yes, the man will get an elecsteady discharging current inits voltage rises gradually. Tcapacitor formed by the alu

The occurrence of thunderstHence, even with the presendischarged completely. The tremains electrically neutral.

at about 400 kV with respect to the surface of tield that decreases with altitude. Near the surfam1. Why then do we not get an electric shock? (Assume the house to be a steel cage so there

e one evening a two metre high insulating slabet of area 1m2. Will he get an electric shock if h

e atmosphere due to the small conductivity of ar the globe. Why then does the atmosphere not

e and become electrically neutral? In other word?

into which the electrical energy of the atmosp(Hint: The earth has an electric field of about

irection, corresponding to a surface charge denuctivity of the atmosphere up to about 50 km (b1800 C is pumped every second into the earthget discharged since thunderstorms and lightnipump an equal amount of negative charge on t

ck as we step out of our house because the origiair changes, keeping our body and the ground

tric shock if he touches the metal slab next morthe atmosphere charges up the aluminium sheee raise in the voltage depends on the capacitaninium slab and the ground.

rms and lightning charges the atmosphere contie of discharging current of 1800 A, the atmospwo opposing currents are in equilibrium and th

e earth,e of theas we stepis no field

arrying one touches the

r is known todischarge

ds, what

ere is00 Vm1 atity = 109

eyond whichas a whole.g occurringe earth.)

nalat the same

ing. The. As a result,e of the

nuously.ere is notatmosphere


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During lightning and thundedissipated in the atmosphere.

storm, light energy, heat energy, and sound energy are

chapter 2 electrostatic potential and capacitance

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