Dielectrics and Capacitance Questions and Answers

Answer: B

The permittivity should be high.

For a parallel-plate capacitor,

\[ C = \frac{\varepsilon_{0}\varepsilon_{r}A}{d} \]

so the capacitance is directly proportional to the permittivity of the dielectric. For a given plate area and spacing, a dielectric with a high relative permittivity, such as a ceramic, gives a much larger capacitance than air.

A low or medium permittivity, or the permittivity of air (\(\varepsilon_{r} \approx 1\)), gives a smaller capacitance, so none of those options achieves a high value.

Answer: A

The field form of Ohm's law is \(\overline{J} = \sigma\overline{E}\).

In electromagnetic theory, Ohm's law is written at each point of a conductor: the current density is proportional to the electric field, with the conductivity \(\sigma\) as the constant of proportionality. Integrating it over a uniform wire of length L and cross-section A gives the circuit form \(V = IR\), with \(R = L/\sigma A\).

V = IR is the circuit law, not the expression for electromagnetic fields the question asks for. \(\sigma = \overline{J}\overline{E}\) multiplies where it should divide, and \(\overline{E} = \overline{J}\sigma\) puts the conductivity where the resistivity \(1/\sigma\) belongs.

Answer: C

The entropy of the universe increases by \(\frac{CV^{2}}{2T}\).

The battery moves charge \(Q = CV\) through emf V, so it does work \(W = QV = CV^{2}\). Half of it, \(\frac{1}{2}CV^{2}\), is stored in the capacitor's field, which is ordered energy carrying no entropy. The other half is dissipated as heat into the surroundings at temperature T:

\[ \begin{aligned} Q_{heat} &= CV^{2} - \frac{1}{2}CV^{2} = \frac{1}{2}CV^{2} \\ \Delta S &= \frac{Q_{heat}}{T} = \frac{CV^{2}}{2T} \end{aligned} \]

\(CV^{2}/T\) counts the whole work as heat, and \(CV^{2}/4T\) and \(CV^{2}/8T\) halve the heat again.