A parallel plate capacitor remains connected to a battery. If the plates are moved apart\

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Professor PC Thomas Classes · Answer

(i) Source remains connected. So V remains constant \(ii) $\mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$. Plates are moved apart \(iii) $\mathrm{q}=\mathrm{CV}$ decreases. ie charge flows from the capacitor to the battery.\(iv) Charge density $\sigma=\frac{q}{A}$ decreases \(v) Electric field $\mathrm{E}=\frac{\sigma}{\varepsilon_{0}}$ decreases\vi) Potential energy $\mathrm{U}=\frac{1}{2} \mathrm{CV}^{2}$ decreases