Q. 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) $\frac{\varepsilon_{0}A}{d - t\left( 1 + \frac{1}{k} \right)}$
Explanation
(c) Potential difference between the plates $V = V_{air} + V_{medium}$
$= \frac{\sigma}{\varepsilon_{0}} \times (d - t) + \frac{\sigma}{K\varepsilon_{0}} \times t$
$\Rightarrow V = \frac{\sigma}{\varepsilon_{0}}(d - t + \frac{t}{K})$
$= \frac{Q}{A\varepsilon_{0}}(d - t + \frac{t}{K})$
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Hence capacitance $C = \frac{Q}{V} = \frac{Q}{\frac{Q}{A\varepsilon_{0}}(d - t + \frac{t}{K})}$
$= \frac{\varepsilon_{0}A}{(d - t + \frac{t}{K})} = \frac{\varepsilon_{0}A}{d - t\left( 1 - \frac{1}{K} \right)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.