A parallel - plate capacitor with plate area A has separation d between the plates. Two dielectric slabs of dielectric constant K 1 and K 2 of same area A2 and thickness d2 are inserted in the space between the plates. The capacitance of the capacitor will be given by:

(B) \frac{\varepsilon_{0} {A}}{{d}}\left(\frac{1}{2}+\frac{{K}_{1} {K}_{2}}{2\left({K}_{1}+{K}_{2}\right)}\right)

(D) \frac{\varepsilon_{0} {A}}{{d}}\left(\frac{1}{2}+\frac{2\left({K}_{1}+{K}_{2}\right)}{{K}_{1} {K}_{2}}\right)

JEE Mains · Physics · STD 12 - Electric potential and capacitance

A parallel - plate capacitor with plate area \(A\) has separation \(d\) between the plates. Two dielectric slabs of dielectric constant \({K}_{1}\) and \({K}_{2}\) of same area \(\frac A2\) and thickness \(\frac d2\) are inserted in the space between the plates. The capacitance of the capacitor will be given by :

Pick an option to check your answer.

Verified Solution

Correct Answer

(A) \(\frac{\varepsilon_{0} {A}}{{d}}\left(\frac{1}{2}+\frac{{K}_{1} {K}_{2}}{{K}_{1}+{K}_{2}}\right)\)

Detailed explanation

\(C_{eq}=\frac{\frac{A}{2} \varepsilon_{0}}{d}+\frac{A \varepsilon_{0}}{d} \frac{K_{1} K_{2}}{K_{1}+K_{2}}\)

\(=\frac{A \varepsilon_{0}}{d}\left(\frac{1}{2}+\frac{K_{1} K_{2}}{K_{1}+K_{2}}\right)\)

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