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

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Aicharya · Accepted Answer

$C=\frac{\varepsilon_{0} A}{d}, \quad \frac{4 C}{3}=\frac{\varepsilon_{0} A}{\left[d-\frac{d}{2}+\frac{d}{2 k}\right]}, \frac{4}{3}=\frac{d}{\frac{d}{2}+\frac{d}{2 k}}, 2 d+\frac{2 d}{k}=3 d, \quad k=2$ .

Capacitance of a capacitor becomes $\frac{4}{3}$ times its original value if a dielectric slab of thickness…

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