A parallel plate capacitor is of area \(6 \mathrm{~cm}^{2}\) and a separation 3 mm. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants \(\mathrm{K}_{1}=10\), \(\mathrm{K}_{2}=12\) and \(\mathrm{K}_{3}=14\). The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be:

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Capacitance

A parallel plate capacitor is of area

6 cm^{2}

and a separation 3 mm. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants

K_{1} = 10

K_{2} = 12

and

K_{3} = 14

. The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be:

(A)4
(B)14
(C)12✓ Correct
(D)36

Answer: (C)

Solution

The three dielectric slabs each fill one third of the plate area but span the whole gap, so they act as three capacitors in parallel, not in series. Each is

C_{i} = \frac{K _{i} ( A /3 ) ϵ _{0}}{d}

, and in parallel they add:

C_{n e t} = \frac{( K _{1} + K _{2} + K _{3} ) A ϵ _{0}}{3 d}

. A single material filling the gap would give

\frac{K _{e q} A ϵ _{0}}{d}

, so matching the two needs

K_{e q} = \frac{K _{1} + K _{2} + K _{3}}{3} = \frac{10 + 12 + 14}{3} = 12

. Because the slabs share the area rather than the gap, the equivalent constant is just the plain average.

JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Capacitance

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