Solution:
Once disconnected, charge \(Q\) remains constant. Introducing a slab increases capacitance \(C\). Since potential difference \(V = Q/C\) and stored energy \(U = \frac{Q^2}{2C}\), both potential difference and stored energy decrease.
A parallel plate capacitor is connected across a \(2\text{ V}\) battery and charged. The battery is then disconnected and a glass slab is introduced between the plates. Which of the following pairs of quantities decrease?
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