JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Capacitance

The two dielectrics occupy half the plate area each, with the full separation $d$ between the plates, so the two halves act as capacitors in PARALLEL.

Air capacitance of each half $=\frac{C}{2}=\frac{10}{2}=5\mu \mathrm{F}$

$C_1=K_1\times 5=2\times 5=10\mu \mathrm{F}$
$C_2=K_2\times 5=3\times 5=15\mu \mathrm{F}$

$C_{\mathrm{eq}}=C_1+C_2=25\mu \mathrm{F}$

Force between the plates: at a fixed supply voltage $V$, the attractive force is

$F=\frac{1}{2}{V}^2\left|\frac{dC}{dd}\right|$

Since $C=\frac{K{\epsilon }_{0}A}{d}\propto \frac{1}{d}$, we get $\left|\frac{dC}{dd}\right|=\frac{C}{d}$, so

$F=\frac{C_{\mathrm{eq}}{V}^2}{2d}$

Substituting $F=8$ N, $d=10\mathrm{mm}=10^{-2}$ m, $C_{\mathrm{eq}}=25\mu \mathrm{F}$:

$V^2=\frac{2Fd}{C_{\mathrm{eq}}}=\frac{2\times 8\times 10^{-2}}{25\times 10^{-6}}=\frac{0.16}{25\times 10^{-6}}=6400$

$V=80\mathrm{V}$

Note: the plate area 4 cm${}^2$ is NOT needed. It is in fact inconsistent with the stated 10 $\mu$F at $d=10$ mm (that separation and area give only about 0.35 pF). Use the given capacitance, not the geometry.