Question

A slab of copper of thickness $d / 2$ is introduced between the plates of a parallel plate capacitor where $d$ is the seperation between its two plates. If the capacitance of the capacitor without copper slab is $C$ and with copper slab is $C'$ then $\frac{C'}{C}$ is :

• A. $\sqrt{2}$
• B. $2$
• C. $1$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

Answer: (B)

Without copper $C =\frac{\epsilon_{0} A}{d}$
with copper $C ^{1}=\frac{\epsilon_{0} A}{d-t\left(1-\frac{1}{\epsilon_{r}}\right)}$
for copper $\epsilon_{r}=\infty$ and $t=\frac{d}{2}$
Hence $C^{1}=\frac{\epsilon_{0} A}{d-\frac{d}{2}}=2 \frac{\epsilon_{0} A}{d}$
$\frac{C^{1}}{C}=2$