JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Capacitance

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For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is

\frac{3}{4} d

, where '

d

' is the separation between the plates of parallel plate capacitor. The new capacitance (

C^{'}

) in terms of original capacitance (

C_{0}

) is given by the following relation:

(A) $C^{'} = \frac{4 + K}{3} C_{0}$
(B) $C^{'} = \frac{3 + K}{4 K} C_{0}$
(C) $C^{'} = \frac{4}{3 + K} C_{0}$
(D) $C^{'} = \frac{4 K}{K + 3} C_{0}$ ✓ Correct

Answer: (D)

Solution

The slab occupies

\frac{3 d}{4}

of the gap and leaves

\frac{d}{4}

of air, and both sections carry the same charge, so the arrangement behaves as two capacitors in series over the full plate area. Writing

C_{0} = \frac{ε _{0} A}{d}

C_{air} = \frac{ε _{0} A}{d /4} = 4 C_{0}, C_{slab} = \frac{K ε _{0} A}{3 d /4} = \frac{4 K}{3} C_{0}

\frac{1}{C ^{'}} = \frac{1}{4 C _{0}} + \frac{3}{4 K C _{0}} = \frac{K + 3}{4 K C _{0}}

The new capacitance is

C^{'} = \frac{4 K}{K + 3} C_{0}

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Capacitance

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