A parallel plate capacitor is charged and source is then disconnected. If the plates are now moved f

Question

A parallel plate capacitor is charged and source is then disconnected. If the plates are now moved further apart by insulating handles, the\

Professor PC Thomas Classes · Answer

1. The source is removed. Therefore the charge remains constant\2. $\mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$. Plates are moved apart. Therefore $d$ increases $C$ decreases \3. $ \mathrm{V}=\frac{\mathrm{q}}{\mathrm{C}}$ increases \4. Surface density of charge $\sigma=\frac{\mathrm{q}}{\mathrm{A}}$ remains same \5. Electric field $\mathrm{E}=\frac{\sigma}{\varepsilon_{0}}$ remains constant. \6. Potential energy $U=\frac{q^{2}}{2 C}$ increases.