Question
A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in
✓
Answer
Due to slab.
$\mathrm{c} \rightarrow \mathrm{K C}, \quad \mathrm{E}=\frac{1}{2} \mathrm{C V}^{2}$
$\mathrm{V} \rightarrow \mathrm{V} / \mathrm{K}, \quad \mathrm{E}=\mathrm{E} / \mathrm{K}$
$\mathrm{Q}=\mathrm{CV}=$ constant
$\mathrm{V} \rightarrow$ Decrease, Energy decrease.
$\mathrm{Q} \rightarrow$ Remain constant
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