Electrostatics Questions and Answers

Answer: E

The work done is zero.

The work done in rotating a dipole from angle \(\theta_1\) to \(\theta_2\) in a uniform field is \(W = pE(\cos\theta_1 - \cos\theta_2)\).

\[ W = pE[\cos(-45^\circ) - \cos 45^\circ] = 0 \]

The dipole's potential energy, \(-pE\cos\theta\), is the same at -45° and +45°, so no net work is needed. Any non-zero option would need start and end angles with different cosines.

Answer: E

The flux is \(6 \times 10^{6}\) N m² C^-1.

The cube has 6 faces of 1 cm², so its total charge is \(q = 8.85\ \mu\text{C cm}^{-2} \times 6\ \text{cm}^{2} = 53.1\ \mu\text{C}\). The uncharged metal sphere does not change the net enclosed charge, so by Gauss's law:

\[ \begin{aligned} \Phi &= \frac{q}{\varepsilon_0} = \frac{53.1 \times 10^{-6}}{8.85 \times 10^{-12}} \\ &= 6 \times 10^{6} \end{aligned} \]

Using one face only gives \(1 \times 10^{6}\); the other values do not follow from the cube's six faces.

Answer: D

The incorrect formula is \(\frac{1}{2}C^{2}V\).

The energy stored is \(U = \frac{1}{2}QV\). With Q = CV it becomes \(\frac{1}{2}CV^{2}\) and \(\frac{Q^{2}}{2C}\). In terms of the field, the energy density \(\frac{1}{2}\varepsilon_0E^{2}\) times the volume Ad gives \(\frac{1}{2}\varepsilon_0E^{2}Ad\).

\(\frac{1}{2}C^{2}V\) has the wrong dimensions (it equals \(\frac{1}{2}CQ\)), so it cannot be an energy.

Answer: E

The work done is \(3 \times 10^{-2}\) J.

Outside them, charged spheres act as point charges at their centres. The work equals the increase in potential energy.

\[ \begin{aligned} W &= kq^{2}\left(\frac{1}{0.3} - \frac{1}{0.4}\right) \\ &= 9 \times 10^{9} \times 4 \times 10^{-12} \times \frac{5}{6} \\ &= 0.036 \times \tfrac{5}{6} = 0.03\ \text{J} \end{aligned} \]

The radius of the spheres does not matter as long as they do not overlap. \(12 \times 10^{-2}\) J is the final energy \(\frac{kq^{2}}{0.3}\), not the change.

Answer: D

An electric field can deflect α-particles.

An electric field exerts a force only on charged particles. Alpha particles are helium nuclei carrying a charge of +2e, so they are deflected towards the negative plate.

X-rays and gamma rays are electromagnetic radiation with no charge, and neutrons are uncharged particles, so all three pass through an electric field undeflected.

Answer: A

The work done is about 5.8 J.

The work equals the increase in electrostatic potential energy:

\[ \begin{aligned} W &= k q_1 q_2 \left( \frac{1}{r_2} - \frac{1}{r_1} \right) \\ &= 9 \times 10^{9} \times 96 \times 10^{-12} \\ &\quad \times \left( \frac{1}{0.06} - \frac{1}{0.10} \right) \\ &= 0.864 \times 6.67 = 5.76\ \text{J} \end{aligned} \]

Because both charges are positive, work must be done against their repulsion to bring them closer, so it cannot be zero.

Answer: D

The final potential difference is Zero.

Each capacitor holds charge Q = CV. Connecting them in parallel with opposite polarity joins the positive plate of one to the negative plate of the other, so the total charge on each joined pair is Q - Q = 0. With no net charge on the combined capacitance of 2C:

\[ V_{f} = \frac{Q - Q}{2C} = 0 \]

V would be the result with the same polarity, and V/2 is what one charged and one uncharged capacitor share. The charges cancel and the stored energy is lost as heat in the connecting wires.

Answer: A

The electric field inside a charged conductor is Zero.

A conductor has free electrons. If there were any field inside, they would move under it, so in electrostatic equilibrium the charges redistribute themselves on the outer surface until the field inside is exactly zero. This is the principle behind electrostatic shielding.

The field inside is not uniform, maximum or dependent on the charge; the charge only affects the field outside, which is perpendicular to the surface.