JEE Main · 2008 · 27 Apr 2008 · Offline · Capacitance

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A parallel plate capacitor with air between the plates has a capacitance of 9 pF. The separation between its plates is 'd'. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant

k_{1} = 3

and thickness

\frac{d}{3}

while the other one has dielectric constant

k_{2} = 6

and thickness

\frac{2 d}{3}

. Capacitance of the capacitor is now

(A)1.8 pF
(B)45 pF
(C)40.5 pF✓ Correct
(D)20.25 pF

Answer: (C)

Solution

With two dielectric slabs stacked between the plates the arrangement is two capacitors in series, so the reciprocals of their capacitances add. Each slab contributes

\frac{k ε _{0} A}{d _{i}}

, giving

\frac{1}{C ^{'}} = \frac{d /3}{3 ε _{0} A} + \frac{2 d /3}{6 ε _{0} A} = \frac{d}{9 ε _{0} A} + \frac{d}{9 ε _{0} A} = \frac{2 d}{9 ε _{0} A}

. Hence

C^{'} = \frac{9}{2} \cdot \frac{ε _{0} A}{d}

, and since the empty capacitor had

\frac{ε _{0} A}{d} = 9 pF

, the filled capacitor is

\frac{9}{2} \times 9 = 40.5 pF

JEE Main · 2008 · 27 Apr 2008 · Offline · Single correct option

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JEE Main · 2008 · 27 Apr 2008 · Offline · Capacitance

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