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

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

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

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

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

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Capacitance

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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

A /2

and thickness

d /2

are inserted in the space between the plates. The capacitance of the capacitor will be given by:

(A) $\frac{ε _{0} A}{d} (\frac{1}{2} + \frac{K _{1} + K _{2}}{K _{1} K _{2}})$
(B) $\frac{ε _{0} A}{d} (\frac{1}{2} + \frac{2 ( K _{1} + K _{2} )}{K _{1} K _{2}})$
(C) $\frac{ε _{0} A}{d} (\frac{1}{2} + \frac{K _{1} K _{2}}{K _{1} + K _{2}})$ ✓ Correct
(D) $\frac{ε _{0} A}{d} (\frac{1}{2} + \frac{K _{1} K _{2}}{2 ( K _{1} + K _{2} )})$

Answer: (C)

Solution

Half of the plate area is left empty, and over that half the plates are a distance

d

apart:

C_{1} = \frac{ε _{0} ( A /2 )}{d} = \frac{ε _{0} A}{2 d}

Over the other half the two slabs sit one above the other, each of area

\frac{A}{2}

and thickness

\frac{d}{2}

, so they carry the same charge and combine in series:

C_{K_{1}} = \frac{K _{1} ε _{0} ( A /2 )}{d /2} = \frac{K _{1} ε _{0} A}{d}, C_{K_{2}} = \frac{K _{2} ε _{0} A}{d}

C_{2} = \frac{C _{K_{1}} C _{K_{2}}}{C _{K_{1}} + C _{K_{2}}} = \frac{ε _{0} A}{d} \cdot \frac{K _{1} K _{2}}{K _{1} + K _{2}}

The two halves share the same pair of plates and so stand in parallel:

C = C_{1} + C_{2} = \frac{ε _{0} A}{d} (\frac{1}{2} + \frac{K _{1} K _{2}}{K _{1} + K _{2}})

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Capacitance

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