Answer
(a) increase
(b) $0.071mm$
Work Step by Step
(a) We know that $C=\frac{K\epsilon_{\circ}A}{d}$. This equation shows that the capacitance is inversely proportional to the separation between the plates. Thus, if the key is pressed then the separation decreases and hence the capacitance increases.
(b) We can find the required distance as follows:
$\Delta C=C_f-C_i$
$\Delta C=\frac{K\epsilon_{\circ}A}{d_f}-\frac{K\epsilon_{\circ}A}{d_i}$
$\implies 0.425\times 10^{-12}F=K\epsilon_{\circ}A[\frac{1}{d_f}-\frac{1}{d_i}]$
This can be rearranged as:
$\frac{1}{d_f}-\frac{1}{d_i}=\frac{0.425\times 10^{-12}F}{3.75(8.85\times 10^{-12}C^2/N.m^2)(47.5\times 10^{-6}m^2)}$
$\implies \frac{1}{d_f}=\frac{0.425\times 10^{-12}F}{3.75(8.85\times 10^{-12}C^2/N.m^2)(47.5\times 10^{-6}m^2)}+\frac{1}{0.55\times 10^{-3}m}$
$\frac{1}{d_f}=2087.8m^{-1}$
$d_f=4.79\times 10^{-4}m$
Now $d=d_f-d_i$
$d=0.479mm-0.55mm$
$d=0.071mm$
