JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Capacitance

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In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be: (Given area of plate

= A

)

(A) $\frac{15}{34} \frac{K ε _{0} A}{d}$ ✓ Correct
(B) $\frac{15}{6} \frac{K ε _{0} A}{d}$
(C) $\frac{25}{6} \frac{K ε _{0} A}{d}$
(D) $\frac{9}{6} \frac{K ε _{0} A}{d}$

Answer: (A)

Solution

The three slabs sit one behind the other between the same pair of plates, so the same charge crosses each and they combine in series. With thicknesses

d

2 d

3 d

and dielectric constants

K

3 K

5 K

C_{1} = \frac{K ε _{0} A}{d}, C_{2} = \frac{3 K ε _{0} A}{2 d}, C_{3} = \frac{5 K ε _{0} A}{3 d}

\frac{1}{C} = \frac{d}{K ε _{0} A} (1 + \frac{2}{3} + \frac{3}{5}) = \frac{d}{K ε _{0} A} \cdot \frac{34}{15}

The combination has capacitance

\frac{15}{34} \frac{K ε _{0} A}{d}

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Capacitance

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