Two capacitors of capacitance $2C$ and $C$ are joined in parallel and charged to potential $V$. The battery is now removed and the capacitor $C$ is filled with a medium of dielectric constant $K$. The potential difference across each capacitor will be
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$\mathrm{Q}_{\text {total }}=\mathrm{CV}+2 \mathrm{CV}=3 \mathrm{CV}$
$\mathrm{V}_{\text {common }}=\frac{\mathrm{Q}_{\text {total }}}{\mathrm{C}_{\text {total }}}=\frac{3 \mathrm{CV}}{2 \mathrm{C}+\mathrm{KC}}=\frac{3 \mathrm{V}}{\mathrm{K}+2}$
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