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A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in
Reduction of charge on the plates and increase of potential difference across the plates
Increase in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates.
Decrease in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates
None of the above
Solution
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
