Q1.25 An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10^4 NC^−1 (Millikan’s oil drop experiment). The density of the oil is 1.26 g cm^−3. Estimate the radius of the drop. (g = 9.81 m s^−2; e = 1.60 × 10^−19 C).

A1.25 9.81 × 10^−4 mm.

Q1.26 Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?

Fig 1.35 (a)

Fig 1.35 (b)

Fig 1.35 (c)

Fig 1.35 (d)

Fig 1.35 (e)

A1.26 Only (c) is right; the rest cannot represent electrostatic field lines,
(a) is wrong because field lines must be normal to a conductor,
(b) is wrong because field lines cannot start from a negative charge,
(d) is wrong because field lines cannot intersect each other,
(e) is wrong because electrostatic field lines cannot form closed loops.

Q1.27 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 10^5 NC^−1 per metre. What are the force and torque experienced by a system having a total dipole moment equal to 10^−7 Cm in the negative z-direction ?

A1.27 The force is 10^−2 N in the negative z-direction, that is, in the direction of decreasing electric field. You can check that this is also the direction of decreasing potential energy of the dipole; torque is zero.

Q1.28
(a) A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor.

Fig. 1.36 (a)

(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 [Fig. 1.36(b)].

Fig. 1.36(b)

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

A1.28 (a) Hint: Choose a Gaussian surface lying wholly within the conductor and enclosing the cavity.
(b) Gauss’s law on the same surface as in
(a) shows that q must induce −q on the inner surface of the conductor.
(c) Enclose the instrument fully by a metallic surface.

Q1.29 A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ/2ε0) n cap, where n cap is the unit vector in the outward normal direction, and σ is the surface charge density near the hole.

A1.29 Hint: Consider the conductor with the hole filled up. Then the field just outside is (σ/ε0) n cap and is zero inside. View this field as a superposition of the field due to the filled up hole plus the field due to the rest of the charged conductor. Inside the conductor, these fields are equal and opposite. Outside they are equal both in magnitude and direction. Hence, the field due to the rest of the conductor is (σ/2ε0) n cap.

Q1.30 Obtain the formula for the electric field due to a long thin wire of uniform linear charge density E without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]

Q1.31 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.

A1.31 p;uud; n;udd.

Q1.32 (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E vector = 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 a certain distance apart.

A1.32 (a) Hint: Prove it by contradiction. Suppose the equilibrium is stable; then the test charge displaced slightly in any direction will experience a restoring force towards the null-point. That is, all field lines near the null point should be directed inwards towards the null-point. That is, there is a net inward flux of electric field through a closed surface around the null-point. But by Gauss’s law, the flux of electric field through a surface, not enclosing any charge, must be zero. Hence, the equilibrium cannot be stable.
(b) The mid-point of the line joining the two charges is a null-point. Displace a test charge from the null-point slightly along the line. There is a restoring force. But displace it, say, normal to the line. You will see that the net force takes it away from the null-point. Remember, stability of equilibrium needs restoring force in all directions.

Q1.33 A particle of mass m and charge (−q) enters the region between the two charged plates initially moving along x-axis with speed vx (like particle 1 in Fig. 1.33). The length of 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 qEL^2/(2m vx^2).

Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class 11 Textbook of Physics.

Q1.34 Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vx = 2.0 × 10^6 m s^−1. If E between the plates separated by 0.5 cm is 9.1 × 10^2 N/C, where will the electron strike the upper plate? (|e|=1.6 × 10^−19 C, me = 9.1 × 10^−31 kg.)

A1.34 1.6 cm

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End of Chapter 1 ELECTRIC CHARGES AND FIELDS