Question

Two metal plates are separated by a distance $d$ in a parallel plate condenser. A metal plate of thickness $t$ and of the same area is inserted between the condenser plates. The value of capacitance increases by a factor of :

• A. $\frac{d-t}{d}$
• B. $(2-\frac{t}{d})$
• C. $(t-\frac{t}{d})$
• D. $\frac{1}{(1-\frac{t}{d})}$

Answer 1

The correct option is D $\frac{1}{(1-\frac{t}{d})}$

The capacitance after inserting dielectric slab of thickness $t$ and dielectric constant $K$ is: $\frac{{ϵ}_{0}A}{d-t(1-\frac{1}{K})}$

For metal plate $K=\infty$
$\Rightarrow C_{eff}=\frac{{ϵ}_{0}A}{d-t}$
or, $C_{eff}=\frac{{ϵ}_{0}A}{d(1-\frac{t}{d})}$
$\therefore$ C increases by a factor of $\frac{1}{(1-\frac{t}{d})}$