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
$C=\frac{{\epsilon }_{0}A}{d}$ ....(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,
$C^{\prime }=\frac{{\epsilon }_{0}A}{d-t\left(1-\frac{1}{K}\right)}$
Since a metal sheet of thickness $d/2$ is introduced, hence here, $t=d/2,K=\infty$ (for metals)
or $\frac{1}{K}=0$
$\therefore C^{\prime }=\frac{{\epsilon }_{0}A}{d-\frac{d}{2}}=\frac{2{\epsilon }_{0}A}{d}$ .....(ii)
Hence, from eqs. (i) and (ii), we get
$\therefore \frac{C^{\prime }}{C}=\frac{\frac{2{\epsilon }_{0}A}{d}}{\frac{{\epsilon }_{0}A}{d}}=\frac{2}{1}=2:1$