Ncert Exercise with Solution Chapter 2 Electrostatic Potential and Capacitance

NCERT EXERCISES WITH SOLUTION

 

CHAPTER 2 ELECTROSTATIC POTENTIAL AND CAPACITANCE

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

Ans. Zero of electric potential for two charges. As shown in Fig., suppose the two charges are placed on X-axis with the positive charge located at the origin O.

Let the potential be zero at the point P and OP = x. For x < 0 (i.e., to the left of O), the potentials of the two charges cannot add up to zero. Clearly, x must be positive. If x lies between O and A, then

V₁ + V₂ = 0

or 

or 

or x = 0.10 m = 10 cm.

The other possibility is that x may also lie on OA produced, as shown in Fig. 

As V₁ + V₂ = 0

∴

or 

or x = 0.40 m = 40 cm.

Thus the electric potential is zero at 10 cm and 40 cm away from the positive charge on the side of the negative charge.

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

Ans. Clearly, distance of each charge from the centre O is

r = 10 cm = 0.10 m

Magnitude of each charge is

q = 5 μC = 5 × 10^-6C

∴ Potential at the centre O is

V = 6.  = 2.7 × 10⁶ V.

2.3. Two charges + 2 μC and – 2 μC are placed at points A and B, 6 cm apart, (i) Identify an ecjuipotential surface of the system (ii) What is the direction of the electric field at every point on the surface ?

Ans. (i) The equipotential surface will be a plane normal to AB and passing through its midpoint O, as shown in Fig. It has zero potential everywhere.

(ii) The direction of electric field is normal to the plane in the direction AB i.e., from positive to negative charge.

 

2.4. A spherical conductor of radius 12 cm has a charge of 1.6 × 10^-7C 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 ?

Ans

2.5. A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1 pF = 10^-12 F). What will be the capacitance if the distance between the plates be reduced by half, the space between them is filled with a substance of dielectric constant, κ = 6 ?

Ans. Capacitance of the capacitor with air between its plates,

=  = 8 pF

When the capacitor is filled with dielectric (κ = 6) between its plates and the distance between the plates is reduced by half, capacitance becomes,

C’ =  = 12 × 8 = 96 pF.

2.6. Three capacitors each of capacitance 9 pF are connected in series, (a) what is the total capacitance of the combination? (b) What is the potential difference across each capacitor when the combination is connected to a 120 V supply?

Ans. (a) If C is the equivalent capacitance of the series combination, then

or C = 3 pF.

(b) As all the capacitors have equal capacitance, so potential drop ΔV would be same across each capacitor.

V = ΔV₁ + ΔV₂ + ΔV₃

= ∆V + ΔV + ΔV = 3∆V

or ΔV =  =  = 40 V.

2.7. Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel, (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.

Ans. (a) For the parallel combination, total capacitance is given by

C = C₁ + C₂ + C₂ = 2 + 3 + 4 = 9pF.

(b) When the combination is connected to 100 V supply, charges on the capacitors will be

q₁ = C₁V = 2 × 10^-12 × 100 = 2 × 10^-10C

q₂ = C₂V = 3 × 10^-12 × 100 = 3 × 10^-10C

q₂ = C₃V = 4 × 10^-12 × 100 = 4 × 10^-10 C.

2.8. In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10^-3 m² and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If the capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

Ans. Capacitance of capacitor with air between its plates is

C₀ =  = 

= 1.8 × 10^-11F =18 pF.

Charge,

q = C₀V = 1.8 × 10^-11 × 100 = 1.8 × 10^-9 C.

2.9. Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (i) while the voltage supply remains connected (ii) after the supply was disconnected.

[CBSE Sample Paper 98]

Ans. From the above question, we have

C₀ = 1.8 × 10^-11 F = 18pF, q₀ = 1.8 × 10^-9 C

Also, κ = 6

(i) When the voltage supply remains connected, the potential difference between capacitor plates remains same i.e., 100 V. The capacitance increases κ times.

∴ C = κC₀ = 6 × 18 = 108 pF.

The charge on the capacitor plates will be

q = CV = 108 × 10^-12 × 100 = 1.08 × 10^-8 C.

(ii) After the supply is disconnected, the charge on the capacitor plates remains same i.e., q₀ = 18 × 10^-9 C

The capacitance increases κ times.

C = κC₀ = 108 pF.

The potential difference between the capacitor plates becomes

V =  = 16.6 V.

2.10. A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor ?

Ans. Here C = 12 pF = 12 × 10^-12 F, V = 50 V

Energy stored,

U = .

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

Ans. Here C₁ = 600 pF, V₁ = 200 V,

C₂ = 600 pF, V₂ = 0

Common potential,

V =  = 100 V

Initial energy stored,

=

Final energy stored,

=

= 6 × 10^-6J

Electrostatic energy lost,

ΔU = U_i – U_f = 12 × 10^-6 – 6 × 10^-6 = 6 × 10^-6 J

2.12. A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of-2 × 10^-9Cfrom a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0) via a point R(0, 6 an, 9 cm).

Ans. As the work done in taking a charge from one point to another is independent of the path followed, therefore

W = 

= 

Here q = 8 mC = 8 × 10 ³C, q₀ = – 2 × 10 ⁹ C

r₁ = 3cm = 3 × 10^-2 m, r₂ = 4cm = 4 × 10^-2m

∴ W = -2 × 10^-9 × 8 × 10^-3 × 9 × 10⁹ × 

= 1.2 J.

2.13. 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.

Ans. Length of longest diagonal of the cube

=  =  b

Distance of each charge (placed at vertex) from the centre of the cube is

r =  b

∴ Potential at the centre of the cube is

V = .

Electric fields at the centre due to any pair of charges at the opposite comers will be equal and opposite thus cancelling out in pairs. Hence resultant electric field at the centre will be zero.

2.14. 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 midpoint 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 midpoint.

Ans. (a)

Fig. (a)

Electric potential at the midpoint O,

 V.

Electric field at the midpoint O,

 

= 4.0 × 10⁵ Vm^-1, from charge 2.5 μC to 1.5 μC

(b) PA = PB =  cm

=  cm – 18 cm = 0.18 m

Fig. (b)

Electric potential at point P,

 .

Electric field at  due to charge 

=

Electric field at  due to charge 

=== 

∠APO = ∠BOP = θ/2

In right ΔAOP,

∴ 

The resultant field at P will be

= Vm^-1

= 

[cos112.6 = -0.3843]

 Vm^-1

Let the field E make angle α with E₁. Then,

=

[sin 112.6° = sin 67.4° = 0.9239]

∴ α = tan^-1(4.3) = 76.9°

Let E make angle β with the direction from B to A Then from right ΔPOC,

β+ α –  = 90°

β = 90°- α +  = 90° – 76.9°+ 56.3°= 69.4°

Hence, the resultant field E makes an angle of 69.4° with the line joining charge 2.5μC to 1.5μC.

2.15. 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.

Ans. (a) The charge q placed at the centre of the shell induces a charge – q on the inner surface of the shell and charge + q on its outer surface.

∴ Surface charge density on the inner surface of the shell

=  = – 

Surface charge density on the outer surface of the shell

= .

(b) Even if the shell is not spherical, the entire charge resides on its outer surface. The net charge on the inner surface enclosing the cavity is zero. From Gauss’s theorem, electric field vanishes at all points inside the cavity. For a cavity of arbitrary shape, this is not enough to claim that electric field inside must be zero. The cavity surface may have positive and negative charges with total charge zero.

Electric field vanishes inside a cavity of any shape.

To overrule this possibility, consider a closed loop PQRSP, such that part PQR is inside the cavity along a line of force and the part RSP is inside the conductor. Since the field inside a conductor is zero, this gives a network done by the field (in part RSP) in carrying a test charge over a closed loop. But this is not possible for a conservative field like the electrostatic field. Hence there are no lines of force (i.e., no field), and no charge on the inner surface of the conductor, whatever be its shape.

2.16. (a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by (₂ – ₁).  where  is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of  is from side 1 to side 2)

Hence show that just outside a conductor, the electric field is σ / ε₀.

(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.

[Hint : For (a), Use Gauss’s law. For (b), use the fact that work done by electrostatic field on a closed loop is zero.]

Ans. (a) Electric field near a plane sheet of charge is given by

E = 

If  is a unit vector normal to the sheet from side 1 to side 2, then electric field on side 2

₂ = 

in the direction of the outward normal to the side 2.

Similarly, electric field on side 1 is

₁ = – 

in the direction of the outward normal to the side 1.

∴ (₂ – ₁). =  = 

As ₁ and ₂ act in opposite directions, there must be discontinuity at the sheet of charge. Now electric field vanishes inside a conductor, therefore

₁ = 0

Hence outside the conductor, the electric field is

 = ₂ =  

(b) Let XY be the charged surface of a dielectric and ₁ and ₂ be the electric fields on the two sides of the charged surface as shown in.

Consider a rectangular loop ABCD with length 1 and negligibly small breadth. Line integral along the closed path ABCD will be

∫ 

or E₁l cos θ₁ – E₂l cos θ₂ = 0

(E₁ cos θ₁ – E₂ cos θ₂) l = 0

where  and  are the tangential components of  and , respectively. Thus,

 =   (∵ l ≠ 0)

Hence the tangential component of the electrostatic field is continuous across the surface.

2.17. A long charged cylinder of linear charged density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders ?

Ans. 

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

(i) Estimate the potential energy of the system in eV, taking the zero of potential energy at infinite separation of the electron from proton.

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

(iii) What are the answers to (i) and (ii) above if the zero of potential energy is taken at 1.06 A separation ?

Ans. (i) q₁ = -1.6 × 10^-19 C, q₂ = + 1.6 × 10^-19 C,

r = 0.53 Å = 0.53 × 10^-10 m

P.E. of the electron-proton system will be

U = 

= 9 × 10⁹ × 

=  – .

(ii) K.E. of the electron in the orbit

=  P.E. =  ×27.2eV = 13.6eV 2 2

∴ Total energy of the electron

= P.E. + K.E.

= (- 27.2 + 13.6) eV = -13.6 eV.

As minimum energy of the free electron is zero, so minimum work required to free the electron

= 0 – (-13.6) = 13.6 eV.

(iii) When the zero of potential energy is not taken at infinity, the potential energy of the system is

 

This indicates that the K.E. of 13.6 eV of case (i) is used up in increasing the P.E. from -27.2 eV to – 13.6 eV as the electron is carried from 0.53 Å to 1.06 Å position. K.E. in this situation should be zero. As the total energy in this case is zero, therefore, minimum work required to free the electron

= 0 – (-13.6eV)=13.6 eV.

2.19. If one of the two electrons of a H₂ molecule is removed, we get a hydrogen molecular ion (H₂⁺). In the ground state of a H₂ ion, 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.

Ans. The system of charges is shown in Fig. 

Fig. Charge on an electron,

q₁ = – e = – 1.6 × 10^-19 C

Charge on each proton,

cj₂ = q₃ = + e = + 1.6 × 10^-19 C

If the zero of potential energy is taken at infinity, then potential energy of the system is

U = 

= 

=  [q = e]

= 

= – 

[∵ 1 eV = 1.6 × 10^-19 J]

2.20. 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 that on its flatter portions.

Ans. The charges will flow between the two spheres till their potentials become equal. Then the charges on the two spheres would be

But 

∴ 

The ratio of the electric fields at the surface of the two spheres will be

Also, 

∴ 

Thus the surface charge densities are inversely proportional to the radii of the spheres. Since the flat portion may be considered as a spherical surface of large radius and a pointed portion as that of small radius, that is why, the surface charge density on the sharp and pointed ends of a conductor is much higher than that on its flatter portion.

2.21. Two charges-q and + q are located at points (0, 0, -a) and (0, 0, α) respectively.

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

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

(iii) 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 same points is not along the x-axis ?

Ans. (i) When the point P lies closer to the charge + q as shown in Fig. (a), the potential at this point P will be

V = 

= 

or V =   [∵ p = q × 2a]

When the point P lies closer to charge – q, as shown in Fig. (b), it can be easily seen that

V = – 

Again, any point (x, y, 0) lies in XY-plane which is perpendicular bisector of Z-axis. Such a point will be at equal distances from the charges – q and + q. Hence potential at point (x, y, 0) will be zero.

(ii) If the distance of point P from the origin O is r, then from the results of part (i), we get

V = ±  [Put z = r]

If r >> a, we can neglect a² compared to r², so

V = ± 

∴ For r >> a, the dependence of potential V on r is 1/ r² type.

(iii) (5, 0, 0) and (- 7, 0, 0) are the points on the X-axis i.e., these points lie on the perpendicular bisector of the dipole. Each point is at the same distance from the two charges. Hence electric potential at each of these points is zero.

Work done in moving the test charge q₀ from the point (5, 0, 0) to (-7, 0, 0) is

W = q(V₁ – V₂) = q(0 – 0) = 0.

No, the answer will not change if the path of the test charge between the same two points is not along X-axis. This is because the work done by the electrostatic field between two points is independent of the path connecting the two points.

2.22. Figure below shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on r for r >> a. Contrast your result with that due to an electric dipole and an electric monopole (i.e. a single charge).

Ans. Potential at point P is

V = 

 = 

 = 

= 

where Q = 2q a² is the quadrupole moment of the given charge distribution. As r >> a, so we can write

V = 

Hence for large r, quadrupole potential varies as 1 / r³, whereas dipole potential varies as 1 / r² and monopole potential varies as 1 / r.

2.23. An electrical technician requires a capacitance of 2 μf in a circuit across a potential difference of 1 kV. A large number of 1 μP 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 a minimum number of capacitors.

Ans. Let this arrangement require n capacitors of 1 μF each in series and m such series combinations to be connected in parallel.

P.D. across each capacitor of a series combination

=  = 400 or n =  = 2.5

But number of capacitors cannot be a fraction,

∴ n = 3

Equivalent capacitance of the combination is

 . m = 2 or m = 2n = 6

∴ Total number of capacitors required

= 3 × 6 = 18

So six series combinations, each of three 1 μF capacitors, should be connected in parallel as shown in Fig. 

 

2.24. 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

Ans. Here C = 2F, d = 0.5 cm = 5 × 10^-3 m

As C = 

∴ A =  m²

≃ 1130 ×10⁶ m² =1130 km².

2.25. Obtain the equivalent capacitance of the network shown in Fig.. For a 300 V supply, determine the charge and voltage across each capacitor.              [CBSE OD 08]

Ans. As C₂ and C₃ are in series, their equivalent capacitance

= 

Series combination of C₂ and C₃ is in parallel with C₁, their equivalent capacitance

= 100 pF+ 100 pF = 200 pF

The combination of C₁, C₂ and C₃ is in series with C₄, equivalent capacitance of the network

= 

Total charge on the network is

q = cV =  × 10^-12 × 300 = 2 × 10^-8 C

This must be equal to charge on C₄ and also to the sum of the charges on the combination of C₁, C₂ and C₃.

∴ q₄ = q = 2 × 10^-8 C

V₄ = 

P.D. between points A and B

= V – V₄ = (300 – 200) V = 100 V

∴ V₁ = 100 V

q₁ = C_lV₁= 100 × 10^-12 × 100 = 10^-8 C

Also the P.D. across the series combination of C₂ and C₃

= 100 V

Now since C₂ = C₃

∴ V₂ = V₃ =  = 50 V

and q₂ = q₃ = 200 × 10^-12 × 50 = 10^-8 C.

2.26. The plates of a parallel plate capacitor have an area of 90 cm² each and are separated by 2.5 mm The capacitor is charged by connecting it to a 400 V supply.

(i) How much energy is stored by the capacitor ?

(ii) View this energy 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.

Ans. (i) Here A = 90 cm² = 90 × 10^-4 m² = 9 × 10^-3m²

d = 2.5 mm = 2.5 × 10^-3m,

ε₀ = 8.85 × 10^-12Fm^-1,

V = 400 V

Capacitance of the parallel plate capacitor is

C = 

= 31.86 × 10^-12 F = 31.86 pF.

Electrostatic energy stored by the capacitor,

U =  CV² =  × 31.86 × 10^-12 × (400)² J

= 25488 × 10^-8 J = 2.55 × 10^-6 J.

(ii) Energy stored per unit volume or energy density of the capacitor is

 Jm 

 Jm 

The relation between  and  can be arrived at as follows:

or  

2.27. A  capacitor is charged by a 200  supply. It is then disconnected from the supply and is connected to another uncharged 2  capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation 7 [CBSE OD 05]

 Ans. Initial electrostatic energy of the 4  capacitor is

Charge on 4 μF capacitor

When the 4 μF and 2μF capacitors are connected together both attain a common potential V. Thus

Final electrostatic energy of the combination,

= 5.33 × 10^-2 J

Electrostatic energy of the first capacitor lost in the form of heat and electromagnetic radiation is

∆U = U_i – U_f = (8 – 5.33) × 10^-2 J

= 2.67 × 10^-2 J.

2.28. Show that the force on each plate of a parallel plate capacitor has a magnitude equal to  qE, where q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor .

Ans. Let A be the plate area and σ, the surface charge density of the capacitor. Then

q = σA

Suppose we increase the separation of the capacitor plates by small distance ∆x against the force F. Then work done by the external agency = F. ∆x

If u be the energy stored per unit volume or the energy density of the capacitor, then increase in potential energy of the capacitor

= u × increase in volume = u. A. ∆x

∴ F.∆x = u.A.∆x

or F = uA =  ε₀ E² . A =  (ε₀E) A E

= .σA.E =  qE

The physical origin of the factor  in the force formula lies in the fact that just inside the capacitor, field is E, and outside it is zero. So the average value E / 2 contributes to the force.

2.29. A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig.). Show that the capacitance of a spherical capacitor is given by

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

Ans. 

2.30. A spherical capacitor has an inner sphere of radius 12 an and an 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 co-centric spheres is filled with a 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.

Ans. Here a = 12 cm = 12 × 10^-2m,

b = 13 cm = 13 × 10^-2cm,

q = 2.5 μC = 2.5 × 10^-6C,  = 32

(a) Capacitance of the spherical capacitor is

C = 

=  F

=  × 10^-11 F = 5.5 × 10^-9 F.

(b) Potential of the inner sphere is

V =  = 0.45 × 10³ V = 4.5 ×10²

(c) Capacitance of the isolated sphere of radius 12 cm is

C = 4πε₀R =  = 1.3 × 10^-11 F.

When an earthed conductor is placed near a charged conductor, the capacitance of the latter increases. The two conductors form a capacitor. But the capacitance of an isolated conductor is always small.

2.31. Answer carefully :

(i) 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 , where r is the distance between their centres ?

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

(iii) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the line of force passing through that point ?

(iv) 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 ?

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

(vi) What meaning would you give to the capacity of a single conductor ?

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

Ans. (i) No. When the two spheres are brought close to each other, their charge distributions do not remain uniform and they will not act as point charges.

(ii) No. Gauss’s law will not hold if Coulomb’s law involved 1/r³ or any other power of r (except 2). In that case the electric flux will depend upon r also.

(iii) Not necessarily. The small test charge will move along the line of force only if it is a straight line. The line of force gives the direction of acceleration, and not that of velocity.

(iv) Zero. But when the orbit is elliptical, work is done in moving the electron from one point to the other. However, net work done over a complete cycle is zero, (z;) No, potential is everywhere constant as it is a scalar quantity.

(vi) A single conductor is a capacitor with one plate at infinity. It also possesses capacitance.

(vii) Because of its bent shape and the presence of two highly polar O – H bonds, a water molecule possesses a large permanent dipole moment about 0.6 × 10^-29 Cm. Hence water has a large dielectric constant.

2.32. 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 field lines at the ends).

Ans. Here L = 15 cm = 0.15 m, q = 3.5 μC = 3.5 × 10^-6 C, a = 1.4 cm = 0.014 m, b =1.5 cm =0.015 m

Capacitance of a cylindrical capacitor is given by

C = 

=  F

=  F

= 0.1206 × 10^-9 F = 1.2 × 10^-10 F

Potential,

V =  V = 2.9 ×10⁴ V.

2.33. A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 10⁷ Vm^-1. For safety, we would like the field never to exceed say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF ?              [CBSE OD 05]

Ans. Maximum permissible voltage

= 1 kV = 10³ V

Maximum permissible electric field

= 10% of 10⁷ Vm^-1 = 10⁶ Vm^-1

∴ Minimum separation d required between the plates is given by

E =  or d = 

Capacitance of a parallel plate capacitor is

C = 

∴ A = 

= 18.8 × 10^-4m² ≃ 19 cm².

2.34. Describe schematically the equipotential surfaces corresponding to

(i) a constant electric field in the Z-direction.

(ii) a field that uniformly increases in magnitude but remains in a constant (say, Z) directions.

(iii) a single positive charge at the origin.

(iv) a uniform grid consisting of long equally spaced parallel charged wires in a plane.

Ans. (i) For a constant electric field in Z-direction, equipotential surfaces will be planes parallel to XY-planes, as shown in Fig. 

 (ii) In this case also, the equipotential surfaces will be planes parallel to XY-plane. However, as field increases, such planes will get closer.

(iii) For a single positive charge at the origin, the equipotential surfaces will be concentric spheres having origin as their common centre, as shown in Fig. 2.25. The separation between the equipotentials differing by a constant potential increases with increase in distance from the origin.

(iv) Near the grid the equipotential surfaces will have varying shapes. At far off distances, the equipotential surfaces will be planes parallel to the grid.

2.35. In a Van de Graaff type generator, a spherical metal shell is to be a 1.5 × 10⁶ V electrode. The dielectric strength of the gas surrounding the electrode is 5 × 10⁷ Vm^-1. What is the minimum radius of the spherical shell required ? [CBSE OD 08]

Ans. Maximum permissible potential, V = 1.5 × 10⁶ V

For safety, the maximum permissible electric field is E = 10% of dielectric strength

= 10% of 5 × 10⁷ Vm^-1 = 5 × 10⁶ Vm^-1

Now for a spherical shell,

V = 

E = 

∴ Minimum radius required is

r =  = 3 × 10^-1 m = 30 cm.

2.36. 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 necessarily 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.

Ans. 

2.37. Answer the following :

(i) 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 Vm^-1. Why do then 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.)

(ii) A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1 m². Will he get an electric shock if he touches the metal sheet next morning ?

(iii) 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 zoords, what keeps the atmosphere charged ?

(iv) What are the forms of energy into which the electric energy of the atmosphere is dissipated during a lightning ?

Ans. (i) Normally the equipotential surfaces are parallel to the surface of the earth as shown in Fig. 2.201. Now our body is a good conductor. So as we step out into the open, the original equipotential surfaces of open air get modified, but keeping our head and the ground at the same potential and we do not get any electric shock.

 (ii) Yes. The aluminium sheet and the ground form a capacitor with insulating slab as dielectric. The discharging current in the atmosphere will charge the capacitor steadily and raise its voltage. Next morning, if the man touches the metal sheet, he will receive shock to the extent depending upon the capacitance of the capacitor formed.

(iii) The atmosphere is continuously being charged by thunder storms and lightning bolts all over globe and maintains an equilibrium with the discharge of the atmosphere in ordinary weather conditions.

(iv) The electrical energy is lost as (i) light energy involved in lightning (ii) heat and sound energy in the accompanying thunder.