$\therefore D=\frac{(2 \varepsilon)_{0 V}}{\sigma}$
$=2 \frac{\left(8.86 \times 10^{-12}\right)(5)}{10^{-7}}$
$=0.88 \times 10^{-3} \mathrm{m}$
$=0.88 \mathrm{mm}$

$A$. the charge stored in it, increases.
$B$. the energy stored in it, decreases.
$C$. its capacitance increases.
$D$. the ratio of charge to its potential remains the same.
$E$. the product of charge and voltage increases.
Choose the most appropriate answer from the options given below:
