Question

When x = $\frac{d}{3}$ then capacitance is C1 = 2 μF then if x = $\frac{2d}{3}$ then capacitance C2 will be (in μF).

Answer 1

$\frac{1}{C_{eq}}=\frac{1}{C_{air}}+\frac{1}{C_{di}}$
$\frac{1}{C_{eq}}=\frac{2d}{3(\epsilon _{0}A)}+\frac{d}{(3)4\epsilon _{0}A}=\frac{3d}{4A\epsilon _{0}}$
$C_{eq}=\frac{4A\epsilon _{0}}{3d}=2\mu F$
$\frac{A\epsilon _{0}}{d}=1.5\mu F$
$\frac{1}{C'_{eq}}=\frac{d}{3(\epsilon _{0}A)}+\frac{2d}{(3)4\epsilon _{0}A}$= \frac{(4+2)d}{12\epsilon _{0}{A}}$$=\frac{6d}{12\epsilon _{0}{A}}$$\Rightarrow C'_{eq}=2[\frac{\epsilon _{0}A}{d}]=3\mu F

So, the correct answer is 3 μF