Question #42799

This question tests the student's ability to couple electrostatics with steady-state current electricity in a leaky dielectric medium.

A common trap is assuming that 'steady state' for a capacitor implies zero current, leading students to incorrectly conclude that the potential difference across the shells is simply the battery's emf $\mathcal{E}$ (yielding $Q = 6\pi \epsilon a \mathcal{E}$).

However, because the medium is weakly conducting, a steady radial current flows, creating a voltage drop across the external resistive wire.

Students must apply Kirchhoff's Loop Rule to find the true potential difference across the shells before applying the capacitance formula $Q = CV$.

Another trap is calculating the voltage drop across the external resistor instead of the medium, which yields an incorrect potential difference of $3\mathcal{E}/4$.