Question

Two parallel conducting plates of area A = 2.5 $m^2$ each are placed 6 mm apart and are both earthed. A third plate, identical with the first two, is placed at a distance of 2 mm from one of the earthed plates and is given a charge of 1 C. The potential of the central plate is

• A. $6\times {10}^{7}V$
• B. $3\times {10}^{7}V$
• C. $4\times {10}^{7}V$
• D. $2\times {10}^{7}V$

Answer 1

The correct option is A $6\times {10}^{7}V$

we know $V=\frac{Q}{C_{eq}}$

Here $Q=1$ Coulomb

Hence $V=\frac{1}{C}$

Now, $C_{eq}=\frac{3}{2}\frac{A{\in }_0}{d}$

$\therefore C=\frac{3}{2}\times \frac{2.5\times 8.85\times {10}^{–12}}{(2\times {10}^{–3})}⟹C=16.59\times {10}^{–9}F$

$\therefore V=\frac{1}{(16.59\times {10}^{–9})}⟹V=6.02\times {10}^{7}V\approx 6.0\times {10}^{7}V$

Hence, option $\left(A\right)$ is correct.