If the distance between parallel plates of a capacitor is halved and dielectric constant is doubled then the capacitance will become
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$C=\frac{\varepsilon_{0} \varepsilon_{\mathrm{r}} A}{d}$
$C^{\prime}=\frac{\varepsilon_{0} \times 2 \varepsilon_{\mathrm{r}} \mathrm{A}}{\frac{\mathrm{d}}{2}}=4 \times \frac{\varepsilon_{0} \varepsilon_{\mathrm{r}} \mathrm{A}}{\mathrm{d}}=4 \mathrm{\,C}$
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