A slab of dielectric constant $\mathrm{K}$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4} \mathrm{~d}$, where $\mathrm{d}$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be : (Given $\mathrm{C}_{0}$ = capacitance of capacitor with air as medium between plates.)

Question

A slab of dielectric constant $\mathrm{K}$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4} \mathrm{~d}$, where $\mathrm{d}$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be :

(Given $\mathrm{C}_{0}$ = capacitance of capacitor with air as medium between plates.)

$\frac{4 K C_{0}}{3+K}$

$\frac{3 K C_{0}}{3+K}$

$\frac{3+K}{4 K C_{0}}$

$\frac{K}{4+K}$

Solution

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