A capacitor is connected to a battery of voltage $V$. Now a di electric slab of dielectric constant $k$ is completely inserted between the plates, then the final charge on the capacitor will be
(If initial charge is $q_{0}$ )
AIIMS 2019, Medium
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The initial charge on the capacitor is,
$q_{0}=C_{0} V$
$=\frac{\varepsilon_{0} A}{d} V$
The charge on the capacitor when di$-$electric is inserted is,
$q=k q_{0}$
$=\frac{k \varepsilon_{0} A}{d} V$
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