Electric Potential and Energy Questions and Answers

Answer: A

The quadrupole term is \(-\frac{\lambda R^{3}}{4\varepsilon_{0}d^{3}}\).

On the axis the exact potential of the ring (charge \(Q = 2\pi R\lambda\)) is expanded in powers of \(\frac{R}{d}\):

\[ \begin{aligned} V &= \frac{Q}{4\pi\varepsilon_{0}\sqrt{d^{2} + R^{2}}} \\ &= \frac{\lambda R}{2\varepsilon_{0}d}\left(1 - \frac{R^{2}}{2d^{2}} + \cdots\right) \end{aligned} \]

The first term is the monopole, there is no dipole term (the ring's dipole moment is zero by symmetry), and the next term, falling as \(\frac{1}{d^{3}}\), is the quadrupole contribution:

\[ \begin{aligned} V_{\text{quad}} &= -\frac{\lambda R}{2\varepsilon_{0}d}\cdot\frac{R^{2}}{2d^{2}} \\ &= -\frac{\lambda R^{3}}{4\varepsilon_{0}d^{3}} \end{aligned} \]

The positive version has the wrong sign, the \(\frac{1}{2\varepsilon_{0}}\) version drops the factor \(\frac{1}{2}\) of the binomial expansion, and 0 would be the dipole term, not the quadrupole.

Answer: D

The heat released is \(CV^{2}/2\).

The battery moves a charge \(Q = CV\) through a constant emf V, so it does work \(W = QV = CV^{2}\). The energy stored in the capacitor is \(\frac{1}{2}CV^{2}\), so the rest is dissipated in the resistance of the circuit:

\[ H = CV^{2} - \frac{1}{2}CV^{2} = \frac{1}{2}CV^{2} \]

Remarkably, this half is lost whatever the resistance. The option \(CV^{2}\) is the battery's total work, not the heat, and zero would require charging with no loss, which is impossible from a fixed-voltage source.

Answer: C

The energy will increase eight times.

The energy stored in the field of a charged conducting sphere is:

\[ \begin{aligned} U &= \frac{Q^{2}}{8\pi\varepsilon_{0}R} \\ U' &= \frac{(2Q)^{2}}{8\pi\varepsilon_{0}(R/2)} = 8U \end{aligned} \]

Doubling the charge multiplies the energy by 4, and halving the radius doubles it again. 'Four times' accounts only for the charge.