12/ph/c2/37
Question 2.37: 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 Vm−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 his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m2. 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? (Hint: The earth has an electric field of about 100 Vm−1 at its surface in the downward direction, corresponding to a surface charge density = −10−9 C m−2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)
Answer :
(a) We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential.
(b) Yes, the man will get an electric shock if he touches the metal slab next morning. The steady discharging current in the atmosphere charges up the aluminium sheet. As a result, its voltage rises gradually. The raise in the voltage depends on the capacitance of the capacitor formed by the aluminium slab and the ground.
(c) The occurrence of thunderstorms and lightning charges the atmosphere continuously. Hence, even with the presence of discharging current of 1800 A, the atmosphere is not discharged completely. The two opposing currents are in equilibrium and the atmosphere remains electrically neutral.
(d) During lightning and thunderstorm, light energy, heat energy, and sound energy are dissipated in the atmosphere.
12/ph/c2/36
Question 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.
Answer :
According to Gauss’s law, the electric field between a sphere and a shell is determined by the charge q1on a small sphere. Hence, the potential difference, V, between the sphere and the shell is independent of charge q2. For positive charge q1, potential difference V is always positive.
12/ph/c2/35
Question 2.35: In a Van de Graaff type generator a spherical metal shell is to be a 15 × 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? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)
Answer :
Potential difference, V = 15 × 106 V
Dielectric strength of the surrounding gas = 5 × 107 V/m
Electric field intensity, E = Dielectric strength = 5 × 107 V/m
Minimum radius of the spherical shell required for the purpose is given by,
Hence, the minimum radius of the spherical shell required is 30 cm.
12/ph/c2/34
Question 2.34:
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.
Answer :
(a) Equidistant planes parallel to the x-y plane are the equipotential surfaces.
(b) Planes parallel to the x-y plane are the equipotential surfaces with the exception that when the planes get closer, the field increases.
(c) Concentric spheres centered at the origin are equipotential surfaces.
(d) A periodically varying shape near the given grid is the equipotential surface. This shape gradually reaches the shape of planes parallel to the grid at a larger distance.
12/ph/c2/33
Question 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. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should 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?
Answer :
Potential rating of a parallel plate capacitor, V = 1 kV = 1000 V
Dielectric constant of a material, 
=3
=3
Dielectric strength = 107 V/m
For safety, the field intensity never exceeds 10% of the dielectric strength.
Hence, electric field intensity, E = 10% of 107 = 106 V/m
Capacitance of the parallel plate capacitor, C = 50 pF = 50 × 10−12 F
Distance between the plates is given by,
Where,
A = Area of each plate
= Permittivity of free space =
Hence, the area of each plate is about 19 cm2.
12/ph/c2/32
Question 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).
Answer :
Length of a co-axial cylinder, l = 15 cm = 0.15 m
Radius of outer cylinder, r1 = 1.5 cm = 0.015 m
Radius of inner cylinder, r2 = 1.4 cm = 0.014 m
Charge on the inner cylinder, q = 3.5 µC = 3.5 × 10−6 C
Where,
= Permittivity of free space = 
Potential difference of the inner cylinder is given by,
12/ph/c2/31
Question 2.31:
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 Q1 Q2/4π
r 2, where r is the distance between their centres?
r 2, 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?
(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) than say, mica (= 6).
Answer :
(a) The force between two conducting spheres is not exactly given by the expression, Q1 Q2/4πr2, because there is a non-uniform charge distribution on the spheres.
r2, because there is a non-uniform charge distribution on the spheres.
(b) Gauss’s law will not be true, if Coulomb’s law involved 1/r3 dependence, instead of1/r2, on r.
(c) 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 are straight. This is because the field lines give the direction of acceleration and not of velocity.
(d) Whenever the electron completes an orbit, either circular or elliptical, the work done by the field of a nucleus is zero.
(e) No
Electric field is discontinuous across the surface of a charged conductor. However, electric potential is continuous.
(f) The capacitance of a single conductor is considered as a parallel plate capacitor with one of its two plates at infinity.
(g) Water has an unsymmetrical space as compared to mica. Since it has a permanent dipole moment, it has a greater dielectric constant than mica.
12/ph/c2/30
Question 2.30: A spherical capacitor has an inner sphere of radius 12 cm 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 concentric 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.
Answer :
Radius of the inner sphere, 
= 12 cm = 0.12 m
= 12 cm = 0.12 m
Radius of the outer sphere, 
= 13 cm = 0.13 m
= 13 cm = 0.13 m
Charge on the inner sphere,
Dielectric constant of a liquid, 
(a) 
Where,
= Permittivity of free space = 
Hence, the capacitance of the capacitor is approximately 
.
.
(b) Potential of the inner sphere is given by,
Hence, the potential of the inner sphere is 
.
.
(c) Radius of an isolated sphere, r = 12 × 10−2 m
Capacitance of the sphere is given by the relation,
The capacitance of the isolated sphere is less in comparison to the concentric spheres. This is because the outer sphere of the concentric spheres is earthed. Hence, the potential difference is less and the capacitance is more than the isolated sphere.
12/ph/c2/29
Question 2.29:
A spherical capacitor consists of two concentric spherical conductors, held in position by suitable 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,
= Permittivity of free space
Hence, proved.
12/ph/c2/28
Question 2.28: Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) 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 ½.
Answer :
Let F be the force applied to separate the plates of a parallel plate capacitor by a distance of Δx. Hence, work done by the force to do so = FΔx
As a result, the potential energy of the capacitor increases by an amount given as uAΔx.
Where,
u = Energy density
A = Area of each plate
d = Distance between the plates
V = Potential difference across the plates
The work done will be equal to the increase in the potential energy i.e.,
Electric intensity is given by,
However, capacitance, 
Charge on the capacitor is given by,
Q = CV
The physical origin of the factor, 
, in the force formula lies in the fact that just outside the conductor, field is E and inside it is zero. Hence, it is the average value, 
, of the field that contributes to the force.
, in the force formula lies in the fact that just outside the conductor, field is E and inside it is zero. Hence, it is the average value, , of the field that contributes to the force.
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