Question

Two metal plate form a parallel plate capacitor. The distance between the plates is $d$. A metal sheet of thickness $d/2$ and of the same area is introduced between the plates. What is the ratio of the capacitances in the two cases?

• A. 4 : 1
• B. 2 : 1
• C. 3 : 1
• D. 5 : 1

Answer 1

Answer: (B)

The capacitance of the air filled parallel plate capacitor is given by
$C = \frac{\varepsilon_0 A}{d} $ ....(i)
When a slab of dielectric constant $K$, and thickness $t$ is introduced in between the plates of the capacitor, its new capacitance is given by,
$C' = \frac{\varepsilon_0 A}{d - t \left( 1 - \frac{1}{K} \right)}$
Since a metal sheet of thickness $d/2$ is introduced, hence here, $t = d/2, K = ∞$ (for metals)
or $\frac{1}{K} = 0$
$\therefore \:\:\: C' = \frac{ \varepsilon_0 A}{d - \frac{d}{2}} = \frac{ 2\varepsilon_0 A}{d}$ .....(ii)
Hence, from eqs. (i) and (ii), we get
$\therefore \:\:\:\:\frac{C'}{C} = \frac{\frac{2 \varepsilon_{0}A}{d}}{\frac{ \varepsilon_{0}A}{d}} = \frac{2}{1} =2 :1$