Capacitance of a parallel plate capacitor becomes $\frac{4}{3}$ times its original value, if a dielectric slab of thickness $t = \frac{d}{2}$ is inserted between the plates [$d$ is the separation between the plates]. The dielectric constant of the slab is

Question

Dr. Abubacker · Accepted Answer

(c) The capacitance of air capacitor
$C = \frac{\varepsilon_{0}A}{d}$
When a dielectric slab of thickness $t = \frac{d}{2}$ is inserted between plates, the capacity becomes
$C^{'} = \frac{A\varepsilon_{0}}{d - \frac{d}{2}\left( 1 - \frac{1}{K} \right)}$
$\frac{4}{3\ }\frac{A\varepsilon_{0}}{d} = \frac{\varepsilon_{0}A}{d - \frac{d}{2}\left( 1 - \frac{1}{K} \right)}$
$3d = 4d\left( 1 - \frac{1}{2} + \frac{1}{2K}\  \right)$
$3 = 4\ \left( \frac{1}{2} + \frac{1}{2K} \right)$
or $\ \ \ \frac{4}{\ \ 2K} = 3 - 2$
or $\ \ \ K = 2$

Q. Capacitance of a parallel plate capacitor becomes $\frac{4}{3}$ times its original value, if a dielectric slab of thickness $t = \frac{d}{2}$ is inserted between the plates [$d$ is the separation between the plates]. The dielectric constant of the slab is

Explanation

Reviewed by Dr. Abubacker

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