Ncert solutions physics mtg class 12

φnet = 0, As the net electric flux with closed surface like cube in uniform electric field is equal to zero, because the number of lines entering the cube is the same as the number of li

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Chapters Page No.

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Electric Charges and Fields 1

1 What is the force between two small charged spheres having charges

of 2 × 10 –7 C and 3 × 10 –7 C placed 30 cm apart in air?

Soln. F = 1

4πε0 1 22

q r

2 The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge –0.8 µC in air is 0.2 N.

(a) What is the distance between the two spheres?

(b) What is the force on the second sphere due to the first?

Soln. (a) Given, F12 = 0.2 N, q1 = 0.4 × 10–6 C, q2 = –0.8 × 10–6 C

(b) By Newton’s third law, F21 = F12 = 0.2 N attractive

3 Check that the ratio Ke

Gm m e p

2

is dimensionless Look up a table of Physical Constants and determine the value of this ratio What does the ratio signify?

Soln. The ratio of electrostatic force to the gravitational force between an

electron and a proton separated by a distance r from each other is F

−

2

2 = 1Thus, the ratio Gm m Ke

e p

2

is dimensionless

Chapter

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2 NCERT Solutions (Class XII) Ke

(b) Why can one ignore quantisation of electric charge when dealing with

macroscopic i.e., large scale charges?

Soln. (a) The net charge possessed by a body is an integral multiple of

charge of an electron i.e q = ± ne, where n = 0, 1, 2, 3, is the number of electrons lost or gained by the body and e = 1.6 × 10–19 C is charge of an electron This is called law of quantization of charge

(b) At macroscopic level, charges are enormously large as compared to

the charge of an electron, e = 1.6 × 10–19C Even a charge of 1 nC contains nearly 1013 electronic charges So, at this large scale, charge can have a

continuous value rather than discrete integral multiple of e, and hence,

the quantization of electric charge can be ignored

5 When a glass rod is rubbed with a silk cloth, charges appears on both

A similar phenomenon is observed with many other pairs of bodies Explain how this observation is consistent with the law of conservation of charge.

Soln. When a glass rod is rubbed with a silk cloth, electrons from the glass rod are transferred to the piece of silk cloth Due to this, the glass rod acquires positive (+) charge whereas the silk cloth acquires negative (–) charge Before rubbing, both the glass rod and silk cloth are neutral and after rubbing the net charge on both of them is also equal

to zero Such similar phenomenon is observed with many other pairs of bodies Thus, in an isolated system of bodies, charge is neither created nor destroyed, it is simply transferred from one body to the other So, it is consistent with the law of conservation of charge

6 Four point charges q A = + 2 µC, q B = –5µC, q C = + 2 µC and q D = –5 µC are

located at the corners of a square ABCD of side 10 cm What is the force on a

charge of 1 µC placed at the centre of the square?

Soln. Forces of repulsion on 1 mC charge at O

due to 2 mC charge, at A and C are equal and

opposite Therefore, they cancel Similarly,

forces of attraction on 1 mC charge at O, due to

–5 mC charges at B and at D are also equal and

opposite Therefore, these also cancel

Hence, the net force on the charge of l mC at

O is zero.

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7 (a) An electrostatic field line is a continuous curve That is, a field line cannot have sudden breaks Why not?

(b) Explain why two field lines never cross each other at any point?

FOD

FOC FOBFOA

+1  C

qB = –5 C

As, FOA= −FOC or FOA+FOC = 0 .(i)

So, the net force on charge of + 1 µC at O is

F = FOA+FOC+FOB+FOD or F = 0

A field line cannot have sudden breaks because the moving test charge never jumps from one position to the other

(b) Two field lines never cross each other at any point, because if they

do so, we will obtain two tangents pointing in two different directions of electric field at a point, which is not possible

8 Two point charges q A = 3 µC and q B = –3 µC are located 20 cm apart in vacuum.

(a) What is the electric field at the midpoint O of the line AB joining the

+1

As, they are directed in same direction, so net electric field at midpoint

O is

E = E OA + E OB = 2E OB = 2 × 27 × 105

or E = 5.4 × 106 N C–1 directed along OB.

(b) When test charge q0 = –1.5 × 10–9 C is placed at point O, it experiences

a force

F = q0E = 1.5 ×10–9 × 5.4 × 106 or F = 8.1 × 10–3N

In direction opposite to that of E i.e along OA

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4 NCERT Solutions (Class XII)

9 A system has two charges q A = 2.5 × 10 –7C and q B = –2.5 × 10 –7 C located

at point A (0, 0, –15 cm) and B (0, 0, +15 cm), respectively What are the total

charge and electric dipole moment of the system?

Soln. The total charge of the system is,

qnet = q A + q B or qnet = 2.5 × 10–7 – 2.5 × 10–7 or qnet = 0

and the total electric dipole moment of the system is

or p = 7.5 × 10–8 C m directed along BA

10 An electric dipole with dipole moment 4 × 10–9 C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 104 N C –1 Calculate the magnitude of the torque acting on the dipole.

Soln. Torque on dipole is, τ = pE sin 30°

so, 1.875 × 1012 electrons have transferred from wool to polythene, as polythene acquires negative charge

(b) Yes, mass of 1.875 × 1012 electrons

i.e., m = nm e = 1.875 × 1012 × 9.1 × 10–31 kg

= 1.71 × 10–18 kg has transferred from wool to polythene

12 (a) Two insulated charged copper spheres A and B have their centres

separated by a distance of 50 cm What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10 –7C ? The radii of A and B are

negligible compared to the distance of separation.

(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Soln. (a) F = 41

0

1 2 2πε

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14

20

2πε

13 Suppose the spheres A and B in previous question have identical sizes

A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from

both What is the new force of repulsion between A and B?

Soln. Particles 1 and 2 are negatively charged

as they experience forces in direction opposite

to that of electric field E, whereas particle 3 is

positively charged as it experience force in the

direction of electric field E

Particle-3 has the highest charge to mass ratio, as it shows maximum deflection in the electric field

E

F= +q E

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15 Consider a uniform electric field  

E= ×3 10 3iN C −1

(a) What is the flux of this field through a square of 10 cm on a side whose

plane is parallel to the yz plane?

(b) What is the flux through the same square if the normal to its plane

makes a 60° angle with the x-axis?

Soln. (a) Area of square, A = a2 = 102 = 100 cm2 = 100 × 10–4 m2 = 10–2m2

As the plane is along yz plane, so area vector A is directed along x-axis i.e., A = (10−2^i)m2

∴ Electric Flux through the square is

Soln. φnet = 0, As the net electric flux with closed surface like cube in uniform electric field is equal to zero, because the number of lines entering the cube is the same as the number of lines leaving the cube

17 Careful measurement of the electric field at the surface of a black box indicates that the net outward flux though the surface of the box is 8.0 × 10 3 N m 2 C –1

(a) What is the net charge inside the box?

(b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or why not?

18 A point charge +10 µC is at a distance of 5 cm directly

above the centre of a square of side 10 cm, as shown in

figure What is the magnitude of the electric flux though

the square?

Soln. Let us assume that the given square be one face of the cube of edge

10 cm

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As charge of +10 µC is at a distance of 5 cm above

the centre of a square, so it is enclosed by the cube

Hence by Gauss’s theorem, electric flux linked with

×

−

− = 1.13 × 10

(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?

(b) What is the value of the point charge?

Soln. (a) On increasing the radius of the Gaussian surface, charge enclosed

by it remains the same and hence the electric flux linked with Gaussian surface also remains the same

Soln. r = 20 cm = 0.20 m, E = 1.5 × 103 N C–1, R = 10 cm = 0.1 m

E = 41

0 2πε

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22 A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 µC m –2

(a) Find the charge on the sphere.

(b) What is the total electric flux leaving the surface of the sphere?

Soln. R = 2 4 m

2 = 1.2 m, σ = 80.0 µ C m–2 = 80 × 10–6 C m–2(a) As σ = q

×

−

− = 1.64 × 108 N m2 C–1

23 An infinite line charge produces a field of 9 × 10 4 N C –1 at a distance of

2 cm Calculate the linear charge density.

(a) the outer region of the first plate.

(b) the outer region of the second plate, and

(c) between the plates?

Soln. As both the plates have same surface charge

2 0−2 0 or E III = 0

(c) In region-II, E II = E A + E B = σ

ε

σε

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Soln. In equilibrium, force due to electric field on the drop balances the

3 124

qE g

e E g

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27 In a certain region of space, electric field is along the z-direction

throughout The magnitude of electric field is, however, not constant but

increases uniformly along the positive z-direction, at the rate of 105 N C –1

per metre What are the force and torque experienced by a system having a total dipole moment equal to 10 –7 C m in the negative z-direction?

Soln. As electric field increases in positive z-direction from A to B, So

E B = E A + dE

dr⋅2a

So, the net force on the electric dipole in electric field is

Fnet = F B – F A = q(E B – E A ) or Fnet = q E dE

directed in direction of FB i.e., from B to A or along –z direction.

As the two forces on charges of electric dipole are collinear and opposite,

so net torque on it is equal to zero

28 (a) A conductor A with a cavity is shown in figure (a) is given a charge

Q Show that the entire charge must appear on the outer surface of the

conductor.

(b) Another conductor B with charge

q is inserted into the cavity keeping B

insulated from A Show that the total

charge on the outside surface of A is

Q + q (figure (b))

(c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment Suggest a possible way.

Soln. (a) Let us consider a closed surface inside the

conductor enclosing the cavity As electric field

inside the conductor is zero, so

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This shows that there cannot be any charge on the inner surface of

conductor at the cavity, so any charge Q given to the conductor will

appear only on its outer surface

(b) Let us consider that the charge –q1 is induced on the inner surface

of conductor A at the cavity, when conductor B of charge q is kept in its cavity Due to this charge of Q + q1 is induced on the outer surface of conductor

Let us construct a closed Gaussian surface S inside

the conductor A enclosing its cavity As, electric

field inside conductor is zero, so

A at the cavity, when conductor B of charge, + q is kept is its cavity Hence, the charge on outer surface of conductor is Q + q1 = Q + q

(c) As we found that electric field inside the cavity of conductor is zero, even on charging the conductor, so the sensitive instrument can

be shielded from the strong electrostatic fields in its environment, by covering it with a metallic cover

29 A hollow charged conductor has a tiny hole cut into its surface Show that the electric field in the hole is σ

n, where n is the unit vector in

the outward normal direction, and σ is the surface charge density near the hole.

Soln. Inside a charged conductor, the electric field is zero

But a uniformly charged flat surface provide an electric field 2σε

0 normal

to its plane

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Now if a hole is made in charged conductor, the field due to small flat

part is absent but the field due to rest of charged conductor is present i.e.,

λπε

λπε4

24

12

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31 It is now believed that protons and neutrons (which constitute nuclei

of ordinary matter) are themselves built out of more elementary units called quarks A proton and a neutron consist of three quarks each Two types of

quarks, the so called ‘up’ quark (denoted by u) of charge +2/3 e, and the

‘down’ quark (denoted by d) of charge (–1/3)e, together with electrons build

up ordinary matter [Quarks of other types have also been found which give rise to different unusual varieties of matter] Suggest a possible quark composition of a proton and neutron.

13

e+ e− e

So, configuration of proton is ‘uud’

Charge of neutron = + 23e−13e−13e

So, configuration of neutron is ‘udd’.

32 (a) Consider an arbitrary electrostatic field configuration A small test

charge is placed at a null point (i.e., where E

= 0) of the configuration Show that the equilibrium of the test charge is necessarily unstable.

(b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed at certain distance apart.

Soln. (a) Let us assume that a radial electric field is present with origin as centre The E is 0 at origin and a small test charge ‘q0’ is placed at origin.Force on test charge at origin is zero

F



= qE = 0

But as the test charge is displaced a little, the

electrostatic force will drift it further away so

charge is in unstable equilibrium

(b) A test charge placed at mid point of equal charges is in stable equilibrium, for lateral displacement

Initially at mid point position of test charge, two

repulsion forces on test charge are equal and

opposite

F A = F B so q0 is in equilibrium

But on displacing q0 close to one of the charge,

one of the forces become stronger and pushes the

charge towards another

F A > F B so q0 moves towards B.

33 A particle of mass m and charge (–q) enters the region between the two charged plates initially moving along x-axis with speed v x The length of JOIN ON TELEGRAM @iitjeeadv

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plate is L and an uniform electric field E

is maintained between the plates Show that the vertical deflection of the particle at the far edge of the plate

is qEL2/(2mv x).

Soln. Let the point at which the charged

particle enters the electric field, be origin

O (0, 0), then after travelling a horizontal

displacement L, it gets deflected by

displacement y in vertical direction as it

comes out of electric field

So, co-ordinates of its initial position are x1 = 0 and y1 = 0

and final position on coming out of electric field are

x2 = L and y2 = y

Components of its acceleration are a x = 0 and a y = F

m =

qE m and of initial velocity are u x = v x and u y = 0

so, by 2nd equation of motion in horizontal direction,

34 Suppose that the particle in question 33 is an electron projected

with velocity v x = 2.0 × 10 6 m s –1 If E between the plates separated by

0.5 cm is 9.1 × 10 2 N C –1 , where will the electron strike the upper plate?

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Electrostatic Potential and Capacitance 15

1 Two charges 5 × 10 –8 C and –3 × 10 –8 C are located 16 cm apart At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Soln. (i) Let C be the point on the line joining the two charges, where

electric potential is zero, then

V C = 0

or V CA + V CB = 0 or V CA = – V CB

or 41

0πε

q

r CA A = −4πε10r q CB B or 5 1010

8 2

Cm]

So, electric potential is zero at distance of 10 cm from charge of 5 × 10–8 C

on line joining the two charges between them

If point C is not between the two charges, then

×+ ×

8 2

C m]

or 5x = 48 + 3x or 2x = 48 or x = 24 cm

So, electric potential is also equal to zero at a distance of 24 cm from charge of –3 × 10–8 C and at a distance of (24 + 16) = 40 cm from charge of

5 × 10–8 C, on the side of charge of –3 × 10–8 C

2 A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices Calculate the potential at the centre of the hexagon.

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Soln. V = 6 × 1

4πε0

q r

or V = 6 × 9 × 109 × 5 10

10 10

6 2

3 Two charges 2 µC and –2 µC are placed at points A and B 6 cm apart.

(a) Identify an equipotential surface of the system.

(b) What is the direction of the electric field at every point on this surface?

Soln. (a) Since it is an electric dipole, so a plane normal to AB and passing

through its mid-point has zero potential everywhere

(b) Normal to the plane in the direction AB.

4 A spherical conductor of radius 12 cm has a charge of 1.6 × 10 –7 C distributed uniformly on its surface What is the electric field

(a) inside the sphere

(b) just outside the sphere

(c) at a point 18 cm from the centre of the sphere?

Soln. R = 12 cm, q = 1.6 × 10–7C

(a) Ein = 0

(b) Eout = 41

0 2πε

.( ) or Eon = 4.44 × 104 N C–1

5 A parallel plate capacitor with air between the plates has a capacitance

of 8 pF What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

= pF or C = 3 pF

+q

+q +q +q +q +q r r

r r r r

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(b) Net charge stored in combination of capacitors is

(a) What is the total capacitance of the combination.

(b) Determine the charge on each capacitor if the combination is connected

to a 100 V supply.

Soln. (a) C = C1 + C2 + C3 = 2 + 3 + 4 or C = 9 pF

(b) Since the capacitors are in parallel, so potential difference across each

of them is same i.e.

8 In a parallel plate capacitor with air between the plates, each plate has

an area of 6 × 10 –3 m 2 and the distance between the plates is 3 mm Calculate the capacitance of the capacitor If this, capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

9 Explain what would happen if in the capacitor given in above question,

a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates.

(a) While the voltage supply remained connected.

(b) After the supply was disconnected.

Soln. C = KC0 = 6 × 18 = 108 pF

(a) If the voltage supply remained connected, then the potential

difference across the capacitor will remain the same i.e V = 100 V and

hence charge on the capacitor becomes

Q = CV = 108 × 10–12 × 100 or Q = 1.08 × 10–8 C

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(b) If the voltage supply was disconnected, then charge on the capacitor remains the same

On connecting charged capacitor to uncharged capacitor, the common

potential V across the capacitors is

Soln. As electric field is conservative

field, so work done in moving a

charge in electric field is independent

of path chosen to move the charge in

electric field and depends only on the

electric potential difference between

the two end points So

W PQ = q0[V Q – V P ] = q0 41 41

q r

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or W PQ = qq r r

0 04

13 A cube of side b has a charge q at each of its vertices Determine the

potential and electric field due to this charge array at the centre of the cube.

Soln. The length of diagonal of the cube of each side b is

q

r = ×8 41 3

20πε

q

b or V = 34

0

q b

πε

E = 0, as electric field at centre due to a charge at any corner of cube is

just equal and opposite to that of another charge at diagonally opposite corner of cube

14 Two tiny spheres carrying charges 1.5 µC and 2.5 µC are located 30 cm apart Find the potential and electric field:

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

(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.

Soln. (a) At mid point C of line joining two charges, electric potential is

V C = V CA + V CB

or V C = 1

4πε0

q r

1010

6 2

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or E C = 1

q r

B CB

A CA

1 515

directed in direction of ECB i.e from C to A

(b) At given point P on perpendicular bisector of line joining two

charges, electric potential is

V P = V PA + V PB or V P = 41

0πε

q r

A PA

B PB

and horizontal component of net electric field at point P is

E x = E PA cos θ– E PB cos θ = (E PA – E PB) cos θ

or E x = 1

q r

A PA

B PB

2 518

6 4

−

− × 1518

or E x = –2.3 × 105V m–1

whereas vertical component of net electric field at point p is

E y = E PA sinθ+ E PB sinθ = [E PA + E PB] sinθ

or E y = 1

q r

A PA

B PB

2 518

6 4

−

− × 1018

×

− × = –2.6956 or θ = – 69.6°

with the horizontal in –ve x-direction i.e at 69.6° with BA.

15 A spherical conducting shell of inner radius r1 and outer radius r2 has a

charge Q.

(a) A charge q is placed at the centre of the shell What is the surface charge

density on the inner and outer surfaces of the shell?

(b) Is the electric field inside a cavity (with no charge)

zero, even if the shell is not spherical, but has any

irregular shape? Explain.

Soln. (a) Surface charge density on the inner surface

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and on the outer surface of shell is

σout = Q q

r

+

4π22(b) The electric flux linked with any closed surface

S inside the conductor is zero as electric field inside

conductor is zero

So, by Gauss’s theorem net charge enclosed by

closed surface S is also zero i.e., qnet = 0

So, if there is no charge inside the cavity, then there cannot be any charge

on the inner surface of the shell and hence electric field inside the cavity will be zero, even though the shell may not be spherical

16 Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

= − σε

^and on right side (side 2), E2 n

02

= σε

^Discontinuity in the normal component from one side to the other is

σε

0

σε

^.(b) To show that the tangential component of electrostatic field is continuous from one side of a charged surface to another, we use the fact that work done by electrostatic field on a closed loop is zero

17 A long charged cylinder of linear charge density λ is surrounded by a

hollow co-axial conducting cylinder What is the electric field in the space between the two cylinders?

r2

r1 +Q S

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Soln. In Figure, A is a long charged cylinder of linear charge density λ, length l and radius a A hollow co-axial conducting cylinder B of of length

l and radius b surrounds A.

The charge q = λl spreads uniformly on the outer

surface of A It induces – q charge on the cylinder B,

which spreads on the inner surface of B An electric

field E is produced in the space between the two

cylinders, which is directed radially outwards Let

us consider a co-axial cylindrical Gaussian surface

of radius r The electric flux through the cylindrical

(c) What are the answers to (a) and (b) above if the zero of potential energy

∴ Total energy E = P.E + K.E = –27.2 + 13.6 = –13.6 eV

Now W = ∆U = 0 – (–13.6) or W = +13.6 eV

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For (a): Taking –13.585 eV as zero of P.E., then

P.E of the system = –27.17 – (–13.585) = –13.585 eV = –13.6 eV

For (b): DU = – 13.6 eV – (–13.6 eV) = 0

∴ W = 0

19 If one of the two electrons of a H 2 molecule is removed, we get a hydrogen molecular ion H 2 + In the ground state of an H 2 + , the two protons are separated by roughly 1.5 Å, and the electron is roughly 1 Å from each proton Determine the potential energy of the system Specify your choice

of the zero of potential energy.

Soln. It is a system of three point charges and the potential energy stored

is this system of charges is

U′ = 4πε1

0

1 2 12

1 3 13

2 3 23

q q r

with zero potential energy at infinity

20 Two charged conducting spheres of radii a and b are connected to each

other by a wire What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.

Soln. As the two spheres are connected to each other by wire, so they have

same electric potential i.e., V a = V b

or 41

0πε

q q

b a

a b

b a

=

⋅

14

2 2

2 2πε

πε

E

b a

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24 NCERT Solutions (Class XII) i.e sphere with smaller radius produces more electric field on its surface

Hence, the charge density on the sharp and pointed ends of conductor is higher than on its flatter portions

21 Two charges –q and +q are located at points (0, 0, –a) and (0, 0, a),

respectively.

(a) What is the electrostatic potential at the points (0, 0, z) and (x, y, 0)?

(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> 1.

(c) How much work is done in moving a small test charge from the point

(5, 0, 0) to (–7, 0, 0) along the x-axis? Does the answer change if the path of the test charge between the some points is not along the x-axis?

Soln. (a) The two point charges form an electric dipole of moment p = q ⋅ 2a directed along + z-axis Point A (0, 0, z) lies on the axis of electric dipole,

so electric potential at point A is

(b) Electric potential at any point on the axis of electric dipole at distance r

from its centre is

so in that case V ∝ 12

r (c) As both the points C (5, 0, 0) and D (–7, 0, 0) lie on the perpendicular

bisector of electric dipole, so electric potential at both the points is zero

Hence work done in moving the charge from C to D is

W CD = q o [V D – V C ] = q o × 0 or W CD = 0

This work done will remain equal to zero even if the path of the test charge between the same points is changed, as electric field is conservative field and work done in moving a charge between the two points in electric field

is independent of the path chosen to move the charge

22 Figure shows a charge array known as an

electric quadrupole For a point on the axis of

quadrupole, obtain the dependence of potential

on r for r/a >> 1, and contrast your results with that due to an electric dipole,

and an electric monopole (i.e., a single charge).

P D

BC a A +q –q

a –q +q r

Trang 30

Soln. V P = V PA + V PB + V PC + V PD

or V P = 41

0πε

pa

r or V P ∝ r13 .However, electric potential at any point on axis of electric dipole is

23 An electrical technician requires a capacitance of 2 µF in a circuit across

a potential difference of 1 kV A large number of 1 µF capacitors are available

to him each of which can withstand a potential difference of not more than

400 V Suggest a possible arrangement that requires the minimum number

C′ = 13mF

Minimum number of rows of 3 capacitors each to be connected in parallel

to obtain net capacitance of 2 mF are

m = 21

3

mm

F

So minimum number of capacitors required are m × n = 6 × 3 = 18

Trang 31

24 What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm?

This is why ordinary capacitors are in the range of mF or less.

25 Obtain the equivalent capacitance of the network in figure For a 300 V supply, determine the charge and voltage across each capacitor.

Soln. Q C2 and C3 are in series, so, C′ = 100 pF

Q C1 and C′ are parallel,

1 22004

C C= ′′+C = + = + or C2 200

3

= pF= 66.7 pFNet charge stored on the combination is

×

−

−C

F = 50 V.

Trang 32

26 The plates of a parallel plate capacitor have an area of 90 cm 2 each and are separated by2.5 mm The capacitor is charged by connecting it to a 400

V supply

(a) How much electrostatic energy is stored by the capacitor?

(b) View this energy as stored in the electrostatic field between the plates,

and obtain the energy per unit volume u Hence arrive at a relation between

u and the magnitude of electric field E between the plates.

2

CV = × 32 × 10–12 × 4002 or U = 2.56 mJ (b) V = A × d = 2.25 × 10–4 m3 = Energy

2 .JOIN ON TELEGRAM @iitjeeadv

Trang 33

Soln. Magnitude of electric field between the plates of charged capacitor

12

29 A spherical capacitor consists of two

concentric spherical conductors held in

position by suitable insulating supports

Show that the capacitance of a spherical

capacitor is given by C = 4 0 1 2

πε r r

r r− .

Where r1 and r2 are the radii of outer and

inner spheres, respectively.

Soln. Let + Q be the charge on outer spherical shell A of radius r1 and + Q

be the charge on inner spherical shell B of radius r2 Then electric potential

Q

r or V A = 0 and electric potential on shell B is

V B = V BA + V BB = 1

Q r

r1

Q Q

r2 r1

Charge Q

Charge –Q

Trang 34

30 A spherical capacitor has an inner sphere of radius 12 cm and outer sphere of radius 13 cm The outer sphere is earthed and the inner sphere is given a charge of 2.5 µC The space between the concentric spheres is filled with liquid of dielectric constant 32.

(a) Determine the capacitance of the capacitor.

(b) What is the potential of the inner sphere.

(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm Explain why the latter is much smaller.

(b) Electric potential of inner sphere is V B = V BB + V BA

= 1

Q r

Q k

Here C > C0, because a single conductor A can be charged to a electric

potential till it reaches the breakdown value of surroundings But when

another earthed metallic conductor B is brought near it, negative charge induced on it decreases the electric potential on A, hence more charge can not be stored on A.

31 Answer carefully:

(a) Two large conducting spheres carrying charges Q1 and Q2 are brought close to each other Is the magnitude of electrostatic force between them exactly given by Q Q

r

1 2

0 2 4πε , where r is the distance between their centres?

(b) If coulomb’s law involved 1/r3 dependence (instead of 1/r2 ), would Gauss’s law be still true?

(c) A small test charge is released at rest at a point in an electrostatic field configuration Will it travel along the field line passing through that point? (d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?

(e) We know that electric field is discontinuous across the surface of a charged conductor Is electric potential also discontinuous there?

Trang 35

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

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

Soln. (a) No, because coulomb’s law holds good only for point charges.(b) No, because in that case electric flux linked with the closed surface

will also become dependent on ‘r’ other than charge enclosed by it.

(c) No, it will travel along the field line only if it is a straight line.(d) Zero, whatever may be the shape of orbit may be It is because work done in moving a charge in closed path in electric field is zero, as electric field is a conservative field

(e) No, electric potential is continuous there As E = 0, so dV

32 A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm The outer cylinder is earthed and the inner cylinder

is given a charge of 3.5 µC Determine the capacitance of the system and the potential of the inner cylinder Neglect end effects (i.e bending of the field lines at the ends).

Soln. Capacitance of cylindrical capacitor is

2 9

10

πε L b a

Soln. V = 1 × 103 V, K = 3, C = 50 × 10–12 F

Given E = 10% of dielectric strength

or E = 10010 × 107 V m–1 = 106 V m–1

Trang 36

–3 m2 = 19 cm2

34 Describe schematically the equipotential surfaces corresponding to

(a) A constant electric field in the z-direction.

(b) a field that uniformly increases in magnitude but remains in a constant

(say, z) direction.

(c) a single positive charge at the origin, and

(d) a uniform grid consisting of long equally spaced parallel charged wires

in a plane.

Soln. (a) Planes parallel to x-y plane or normal to the electric field in z-direction.

(b) Planes parallel to x-y plane or normal to the electric field in z-direction,

but the planes having different fixed potentials will become closer with increase in electric field intensity

(c) Concentric spherical surfaces with their centres at origin

(d) A time dependent changing shape nearer to grid, and at far off distances from the grid, it slowly becomes planar and parallel to the grid

35 In a Van-de-Graff type generator, a spherical metal shell is to be a 15 ×

10 6 V electrode The dielectric strength of the gas surrounding the electrode

is 5 × 10 7 V m –1 What is the minimum radius of the spherical shell required?

Soln. V = 15 × 106 V, E = 5 × 107 V m–1

As V = 41

0πε

q

R and E = 4πε10 2

q R

36 A small sphere of radius r1 and charge q1 is enclosed by a spherical shell

of radius r2 and charge q2 Show that if q1 is positive, charge will necessary flow from the sphere to the shell (when the two are connected by a wire) no

matter what the charge q2 on the shell is.

Soln. The potential on inner small sphere is

V A = V AA + V AB

or V A = 1

q r

Trang 37

32 NCERT Solutions (Class XII) whereas the potential on the outer shell B is

V B = V BA + V BB = 41

0

1 1

2 2πε

q r

+ + +++ + +

++

+ + + A

r2

q1

q2O

r1

So, V A – V B = 1

q r

V A – V B = 0 or q1 = 0 [As r1 ≠ r2]

So, charge q1 given to sphere A will flow on the shell B, no matter what the charge on the shell B is.

37 Answer the following :

(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude Near the surface of the earth, the field is about 100 V m –1 Why then

do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage, so there is no field inside).

(b) A man fixes outside house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1 m 2 Will he get an electric shock if he touches the metal sheet next morning?

(c) The discharging current in the atmosphere due to the small conductivity

of air is known to be 1800 A on an average over the globe Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged? (d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning?

Soln. (a) Our body and the ground are at same electric potential As we step out into the open, the original equipotential surfaces of open air change, keeping our head and the ground at the same potential

(b) Yes, it is because the aluminium sheet gets charged due to discharging current and raises to the extent depending on the capacitor formed by the sheet and the ground

(c) The atmosphere of earth gets continuously charged due to lightning, thunderstorms but simultaneously it gets discharged through normal weather zones This keeps the system balanced

(d) Light, sound and heat energy Light energy in lightning and heat and sound energy in the accompanying thunder

a b

Trang 38

Current Electricity 33

1 The storage battery of a car has an emf of 12 V If the internal resistance

of the battery is 0.4 Ω, what is the maximum current that can be drawn from

the battery?

Soln. The maximum current will be obtained

when no external resistance is offered by wire

joining the two terminals

2 A battery of emf 10 V and internal resistance 3 Ω is connected to a

resistor If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Soln. I = E

R r+ ⇒ 0.5 =

103

R +

R + 3 = 20 ⇒ R = 17 Ω

Terminal voltage of the battery

V = E – Ir = 10 – 0.5 × 3 = 8.5 V

3 (a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series What is

the total resistance of the combination?

(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

Soln. (a) Total resistance of the combination in series, Req = R1 + R2 + R3

Trang 39

4 (a) Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel What is the

total resistance of the combination?

(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance Determine the current through each resistor, and the total current drawn from the battery.

Soln. (a) Total resistance of the combination in parallel

I1 = V

R1

202

= = 10 A

I2 = V

R2

204

= = 5 A

I3 = V

R3

205

= = 4 A

Total current drawn from the cell, I = I1 + I2 + I3 or I = 19 A.

5 At room temperature (27.0°C) the resistance of a heating element is

100 Ω What is the temperature of the element if the resistance is found

to be 117 Ω, given that the temperature coefficient of the material of the

7 A silver wire has a resistance of 2.1 Ω at 27.5°C and a resistance of 2.7 Ω

at 100°C Determine the temperature coefficient of resistivity of silver.

Soln. T1 = 27.5°C, R1 = 2.1 Ω, T2 = 100°C and R2 = 2.7 Ω

Using the relation, R2 = R1 [1 + α(T2 – T1)], we have

Temperature coefficient of resistivity of silver,

Trang 40

Current Electricity 35

8 A heating element using nichrome connected to a 230 V supply draws

an initial current of 3.2 A which settles after a few seconds to a steady value

of 2.8 A What is the steady temperature of the heating element if the room temperature is 27.0 °C ? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10 –4 °C –1

Soln.At room temperature 27 °C, the resistance of the heating element

Soln. Let us first distribute the current in different branches

Now, equations for different loops using Kirchhoff’s II law,

Tiêu đề	Ncert Solutions Physics Class 12
Tác giả	MTG Learning Media (P) Ltd.
Trường học	mtg learning media (p) ltd.
Chuyên ngành	Physics
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