Capacitance of a parallel plate capacitor becomes $4/3$ times its original value if a dielectric slab of thickness $t = d/2$ is inserted between the plates ($d$ is the separation between the plates). The dielectric constant of the slab is

Question

Dr. Abubacker · Accepted Answer

(d) $C_{air} = \frac{\varepsilon_{0}A}{d},$ with dielectric slab $C^{'} = \frac{\varepsilon_{0}A}{\left( d - t + \frac{t}{K} ight)}$
Given $C^{'} = \frac{4}{3}C \Rightarrow \frac{\varepsilon_{0}A}{\left( d - t + \frac{t}{K} ight)} = \frac{4}{3} 	imes \frac{\varepsilon_{0}A}{d}$
$\Rightarrow K = \frac{4t}{4t - d} = \frac{4(d/2)}{4\lbrack(d/2) - dbrack} = 2$

Q. Capacitance of a parallel plate capacitor becomes $4/3$ times its original value if a dielectric slab of thickness $t = d/2$ 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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