A dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → The charge on the capacitor
The capacitor is isolated, so its charge has nowhere to go: Q stays constant. Inserting a dielectric of dielectric constant K increases the capacitance K times, and the other quantities follow from Q.
\( Q = \text{constant} \)
\( C' = KC \)
\( \displaystyle V' = \frac{Q}{KC} = \frac{V}{K} \)
\( \displaystyle E' = \frac{V'}{d} = \frac{E}{K} \)
\( \displaystyle U' = \frac{Q^{2}}{2KC} = \frac{U}{K} \)