2.19 ADDITIONAL EXERCISES

Q2.12: A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of


−2 × 10^−9 C from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).

A2.12: 1.2 J; the point R is irrelevant to the answer.

Q2.13: 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.

A2.13: Potential = 4q/ {(√3π) ε0b); field is zero, as expected by symmetry.

Q2.14: Two tiny spheres carrying charges 1.5 μ C and 2.5 μ C 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 mid-point in a plane normal to the line and passing through the mid-point.

A2.14:


(a) 2.4 × 10^5 V; 4.0 × 10^5 Vm^−1 from charge 2.5 μC to 1.5 μC.


(b) 2.0 × 10^5 V; 6.6 × 10^5 Vm^−1 in the direction that makes an angle of about 69° to the line joining charge 2.5 μC to 1.5 μC.

Q2.15: 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.

A2.15:


(a) −q/ (4π r1^2), (Q + q) / (4π r2^2)


(b) By Gauss’s law, the net charge on the inner surface enclosing the cavity (not having any charge) must be zero. For a cavity of arbitrary shape, this is not enough to claim that the electric field inside must be zero. The cavity may have positive and negative charges with total charge zero. To dispose of this possibility, take a closed loop, part of which is inside the cavity along a field line and the rest inside the conductor. Since field inside the conductor is zero, this gives a net work done by the field in carrying a test charge over a closed loop. We know this is impossible for an electrostatic field. Hence, there are no field lines inside the cavity (i.e., no field), and no charge on the inner surface of the conductor, whatever be its shape.

Q2.16:


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


(E2 vector − E1 vector). n cap = σ/ε0


where n cap is a unit vector normal to the surface at a point and sigma is the surface charge density at that point. (The direction of n cap is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ n cap/ 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.]

Q2.17: 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?

A2.17: λ/ (2πε0r), where r is the distance of the point from the common axis of the cylinders. The field is radial, perpendicular to the axis.

Q2.18: In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Angstrom:


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


(b) 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 (a)?


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

A2.18:


(a) −27.2 eV


(b) 13.6 eV


(c) − 13.6 eV, 13.6 eV.


Note in the later choice the total energy of the hydrogen atom is zero.

Q2.19: If one of the two electrons of a H2 molecule is removed, we get a hydrogen molecular ion H2^+. In the ground state of an H2^+, 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.

A2.19: − 19.2 eV; the zero of potential energy is taken to be at infinity.

Q2.20: 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.

A2.20: The ratio of electric field of the first to the second is (b/a). A flat portion may be equated to a spherical surface of large radius, and a pointed portion to one of small radius.

Q2.21: 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 same points is not along the x-axis?

A2.21:


(a) On the axis of the dipole, potential is (±1/4πε0) {p/(x^2 − a^2)} where p = 2q a is the magnitude of the dipole moment; the + sign when the point is closer to +q and − sign when it is closer to −q. Normal to the axis, at points (x, y, 0), potential is zero.


(b) The dependence on r is 1/r^2 type.


(c) Zero. No, because work done by electrostatic field between two points is independent of the path connecting the two points.

Q2.22: Figure 2.32 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 >> 1, and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).

Fig.2.32

A2.22: For large r, quadrupole potential goes like 1/r^3, dipole potential goes like 1/r^2, monopole potential goes like 1/r.

Q2.23: An electrical technician requires a capacitance of 2 μ F in a circuit across a potential difference of 1 kV. A large number of 1 μ F 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 of capacitors.

A2.23: Eighteen 1 μ F capacitors arranged in 6 parallel rows, each row consisting of 3 capacitors in series.

Q2.24: 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? [You will realize from your answer why ordinary capacitors are in the range of μ F or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]

A2.24: 1130 km^2

Q2.25: Obtain the equivalent capacitance of the network in Fig.2.33. For a 300 V supply, determine the charge and voltage across each capacitor.

Fig.2.33

“Image description by Dr T K Bansal starts


Four capacitors, C1, C2, C3, C4, & a battery of 300 V is connected in a circuit. The positive terminal of the battery is connected to a parallel combination of two branches; branch 1 contains capacitor C1 of capacitance of 100 pF, & branch 2 contains a series combination of two capacitors C2 & C3 of capacitance 200pF each.


This combination is then connected to a capacitor C4 = 100 pF in series, which is in turn connected to the negative terminal of the battery.


Image description ends”

A2.25: Equivalent capacitance = (200/3) pF.

Q1 = 10^−8 C, V1 = 100 V ; Q2 = Q3 = 10^−8 C


V2 = V3 = 50 V


Q4 = 2.55 × 10^−8 C, V4 = 200 V

Q2.26: The plates of a parallel plate capacitor have an area of 90 cm^2 each and are separated by 2.5 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.

A2.26:


(a) 2.55 × 10^−6 J


(b) u = 0.113 J m^−3, u = (1/2) ε0 E^2

Q2.27: A 4 μ F capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 μ F capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

A2.27: 2.67 × 10^−2 J

Q2.28: Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1/2) Q E, 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 1/2.

A2.28: Hint: Suppose we increase the separation of the plates by Δx Work done (by external agency) = F Δx. This goes to increase the potential energy of the capacitor by uaΔx where u is energy density. Therefore, F = ua which is easily seen to be (1/2) QE, using u = (1/2) ε0 E^2. The physical origin of the factor 1/2 in the force formula lies in the fact that just outside the conductor, field is E, and inside it is zero. So, the average value E/2 contributes to the force.

Q2.29: A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig.2.34). Show that the capacitance of a spherical capacitor is given by


C = 4πε0 r1r2/ (r1 − r2)


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

Fig.2.34

Q2.30: 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.

A2.30:


(a) 5.5 × 10^−9 F


(b) 4.5 × 10^2 V


(c) 1.3 × 10^11 F

Q2.31: 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/ (4πε0r^2), where r is the distance between their centres?


(b) If Coulomb’s law involved 1/r^3 dependence (instead of 1/r^2), 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).

A2.31:


(a) No, because charge distributions on the spheres will not be uniform.


(b) No.


(c) Not necessarily. (True only if the field line is a straight line.) The field line gives the direction of acceleration, not that of velocity, in general.


(d) Zero, no matter what the shape of the complete orbit.


(e) No, potential is continuous.


(f) A single conductor is a capacitor with one of the ‘plates’ at infinity.


(g) A water molecule has permanent dipole moment. However, detailed explanation of the value of dielectric constant requires microscopic theory and is beyond the scope of the book.

Q2.32: 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).

A2.32: 1.2 × 10^−10 F, 2.9 × 10^4 V

Q2.33: 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 10^7 Vm^−1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionization.) 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?

A2.33: 19 cm^2

Q2.34: 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.

A2.34:


(a) Planes parallel to x-y plane.


(b) Same as in (a), except that planes differing by a fixed potential get closer as field increases.


(c) Concentric spherical shells centred at the origin.


(d) A periodically varying shape near the grid which gradually reaches the shape of planes parallel to the grid at far distances.

Q2.35: 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.

A2.35: Hint: By Gauss’s law, field between the sphere and the shell is determined by q1 alone. Hence, potential difference between the sphere and the shell is independent of q2. If q1 is positive, this potential difference is always positive.

Q2.36: 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 aluminum 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?


(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 Cm^−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.)

A2.36:


(a) Our body and the ground form an equipotential surface. 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. The steady discharging current in the atmosphere charges up the aluminum sheet gradually and raises its voltage to an extent depending on the capacitance of the capacitor (formed by the sheet, slab and the ground).


(c) The atmosphere is continually being charged by thunderstorms and lightning all over the globe and discharged through regions of ordinary weather. The two opposing currents are, on an average, in equilibrium.


(d) Light energy involved in lightning; heat and sound energy in the accompanying thunder.

End of Class 12 Physics Chapter 2