Electrostatics Questions and Answers

Answer: C

The correct statement is \(E_{1}(r_{0}/2) = 2E_{2}(r_{0}/2)\), option (C).

The three fields are

\[ \begin{aligned} E_{1} &= \frac{Q}{4\pi\varepsilon_{0}r^{2}} \\ E_{2} &= \frac{\lambda}{2\pi\varepsilon_{0}r} \\ E_{3} &= \frac{\sigma}{2\varepsilon_{0}} \end{aligned} \]

Setting all three equal at \(r_{0}\) gives \(Q = 2\pi\sigma r_{0}^{2}\) and \(r_{0} = \lambda/(\pi\sigma)\), so (A), with \(4\sigma\pi r_{0}^{2}\), and (B), with \(2\pi\sigma\), are both off by a factor of 2.

Let the common value at \(r_{0}\) be E. Halving the distance multiplies \(E_{1}\) by 4 (inverse square) and \(E_{2}\) by 2 (inverse distance), while \(E_{3}\) does not change. So at \(r_{0}/2\): \(E_{1} = 4E\), \(E_{2} = 2E\), \(E_{3} = E\). Then \(E_{1} = 2E_{2}\), which is (C), while \(E_{2} = 2E_{3}\), not \(4E_{3}\), so (D) is false.

Answer: A

(a) is an observation while (b) and (c) are inferences.

An observation is what can be perceived directly through the senses. One can see the leaves of the electroscope move further apart (a).

That the two leaves carry similar charge (b) and that negative charges move on to the electroscope (c) cannot be seen; they are explanations reasoned from the observed movement, so they are inferences. Options calling b or c observations are wrong.

Answer: E

The net charge is \(-3 \times 10^{4}\varepsilon_{0}\) C.

Outward flux counts as positive and inward flux as negative, so the net flux is

\[ \begin{aligned} \Phi &= 3 \times 10^{4} - 6 \times 10^{4} = -3 \times 10^{4}\ \text{N m}^{2}\text{C}^{-1} \\ q &= \varepsilon_{0}\Phi = -3 \times 10^{4}\varepsilon_{0} \end{aligned} \]

by Gauss's law, \(\Phi = \frac{q}{\varepsilon_{0}}\). More flux entering than leaving means a net negative charge inside. \(9 \times 10^{4}\varepsilon_{0}\) adds the fluxes, and \(\pm 6 \times 10^{4}\varepsilon_{0}\) uses only the inward flux.

Answer: D

It can be reduced by connecting another capacitor in series with it.

In series, \(\frac{1}{C_{\text{eff}}} = \frac{1}{C} + \frac{1}{C'}\), so \(C_{\text{eff}}\) is smaller than either capacitance.

A capacitor in parallel adds its capacitance; a dielectric slab multiplies \(C\) by \(K\); a metal plate between the plates effectively reduces the gap and increases \(C\). Changing the potential difference does not change capacitance at all, since \(C = \frac{\varepsilon_{0}A}{d}\) depends only on the geometry and the medium.

Answer: E

The force is maximum for \(\frac{Q}{2}\) each.

With parts \(q\) and \(Q - q\), the force is \(F = \frac{kq(Q - q)}{d^{2}}\). Maximise \(q(Q - q)\):

\[ \begin{aligned} \frac{d}{dq}\left(qQ - q^{2}\right) &= Q - 2q = 0 \\ q &= \frac{Q}{2} \end{aligned} \]

The product of two numbers with a fixed sum is greatest when they are equal. For the unequal splits the products are \(\frac{3Q^{2}}{16}\), \(\frac{7Q^{2}}{64}\), \(\frac{2Q^{2}}{9}\) and \(\frac{5Q^{2}}{36}\), all less than \(\frac{Q^{2}}{4}\).

Answer: B

The field on the other side is directed towards the plane perpendicularly.

The charge −q induces positive charge on the near face of the conductor and leaves an equal negative charge on the far face. The electric field just outside a conductor is always perpendicular to its surface, and next to a negatively charged surface it points towards that surface. So at points on the far side the field is normal to the plane and directed into it.

Radial directions centred on the point charge would ignore the screening by the conductor, a field parallel to the surface is impossible at a conductor in equilibrium, and 'away from the plane' would need positive charge on the far face.

Answer: C

The charge drawn is \(48\ \mu C\).

Three equal capacitors in series have \(\frac{1}{C} = \frac{3}{12}\), so \(C = 4\ \mu F\). The same charge flows onto each capacitor and from the battery:

\[ Q = CV = 4\ \mu F \times 12\ V = 48\ \mu C \]

\(144\ \mu C\) (using 12 µF) and \(432\ \mu C\) (parallel, 36 µF) are not offered; the listed 12, 16, 24 and 32 µC do not come from the series formula.

Answer: C

The new force is \(\frac{4}{9}F\).

Moving one charge away by \(d/2\) increases the separation from \(d\) to \(d + d/2 = 3d/2\). Coulomb's force varies as the inverse square of the separation:

\[ \begin{aligned} \frac{F'}{F} &= \left(\frac{d}{3d/2}\right)^{2} = \left(\frac{2}{3}\right)^{2} \\ F' &= \frac{4}{9}F \end{aligned} \]

\(\frac{2}{3}F\) forgets to square the ratio, and \(\frac{9}{4}F\) and \(\frac{3}{2}F\) invert it, as if the charges had moved closer. The force must decrease, because the charges are farther apart.

Answer: D

The new torque is \(\sqrt{2}\tau\).

The torque on a dipole is \(pE\sin\theta\), so

\[ \begin{aligned} \frac{\tau'}{\tau} \\ &= \frac{\sin 45^{\circ}}{\sin 30^{\circ}} \\ &= \frac{1/\sqrt{2}}{1/2} \\ &= \sqrt{2} \end{aligned} \]

\(\frac{1}{\sqrt{2}}\tau\) inverts the ratio, and \(2\tau\) or \(\frac{1}{2}\tau\) use the angles in proportion instead of their sines.

The torque is greatest at \(90^{\circ}\), where it equals \(pE = 2\tau\).

Answer: E

The ratio of potentials is 1 : 3.

After the change, A holds \(4 - 2 = 2\) C and B holds \(4 + 2 = 6\) C. For identical capacitors \(V = \frac{Q}{C}\) with the same \(C\), so

\[ V_{A} : V_{B} = 2 : 6 = 1 : 3 \]

1 : 1 ignores the added charges, and 3 : 1 inverts the ratio.

Equivalently, potential is proportional to charge for a fixed capacitance, so the capacitor holding three times the charge sits at three times the potential.