Question

Separation between the plates of a parallel plate capacitor is $d$ and the area of each plate is $A$. When a slab of material of dielectric constant $k$ and thickness $t(t < d)$ is introduced between the plates, its capacitance becomes

• A. $\frac{\varepsilon_{0} A}{d+t\left(1-\frac{1}{k}\right)}$
• B. $\frac{\varepsilon_{0} A}{d+t\left(1+\frac{1}{k}\right)}$
• C. $\frac{\varepsilon_{0} A}{d-t\left(1-\frac{1}{k}\right)}$
• D. $\frac{\varepsilon_{0} A}{d-t\left(1+\frac{1}{k}\right)}$

Answer 1

Answer: (C)

Potential difference between the plates,
$V=V_{\text {air }}+V_{\text {medium }}=\frac{\sigma}{\varepsilon_{0}} \times(d-t)+\frac{\sigma}{k \varepsilon_{0}} \times t$
$\Rightarrow V=\frac{\sigma}{\varepsilon_{0}}\left(d-t+\frac{t}{k}\right)$
$=\frac{Q}{A \varepsilon_{0}}\left(d-t+\frac{t}{k}\right)$

Hence, capacitance, $C=\frac{Q}{V}$
$=\frac{Q}{\frac{Q}{A \varepsilon_{0}}\left(d-t+\frac{t}{k}\right)}$
$=\frac{\varepsilon_{0} A}{\left(d-t+\frac{t}{k}\right)}$
$=\frac{\varepsilon_{0} A}{d-t\left(1-\frac{1}{k}\right)}$