JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A parallel plate capacitor of capacitance \(12.5 \mathrm{pF}\) is charged by a battery connected between its plates to potential difference of \(12.0 \mathrm{~V}\). The battery is now disconnected and a dielectric slab \(\left(\epsilon_{\mathrm{r}}=6\right)\) is inserted between the plates. The change in its potential energy after inserting the dielectric slab is _______.\(\times 10^{-12} \mathrm{~J}\).
- A \(720\)
- B \(730\)
- C \(750\)
- D \(770\)
Answer & Solution
Correct Answer
(C) \(750\)
Step-by-step Solution
Detailed explanation
Before inserting dielectric capacitance is given \(\mathrm{C}_0=12.5 \mathrm{pF}\) and charge on the capacitor \(\mathrm{Q}=\mathrm{C}_0 \mathrm{~V}\) After inserting dielectric capacitance will become \(\epsilon_{\mathrm{s}} \mathrm{C}_0\). Change in potential energy of the…
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