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JEE Main 202116 MarchMorning Shift
Exam
JEE Main
Year
2021
Date
16 March
Shift
Morning Shift
Subject
Physics
Chapter
Capacitance

JEE Main 2021 · 16 March · Morning Shift · Physics · Capacitance

For changing the capacitance of a given paralle plate capacitor, a...

For changing the capacitance of a given paralle plate capacitor, a dielectric material of dielectric constant KK is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is 34d\dfrac{3}{4}d, where dd is the separation between the plates of parallel plate capacitor. The new capacitance C′{C}^{'} in terms of original capacitance C0{C}_{0} is given by the following relation :For changing the capacitance of a given paralle plate capacitor, a dielectric material of dielectric constant K is

Options

  1. AC′=3+K4KC0{C}^{'} = \dfrac{3 + K}{4K}{C}_{0}
  2. BC′=4+K3C0{C}^{'} = \dfrac{4 + K}{3}{C}_{0}
  3. CC′=4KK+3C0{C}^{'} = \dfrac{4K}{K + 3}{C}_{0}
  4. DC′=43+KC0{C}^{'} = \dfrac{4}{3 + K}{C}_{0}

Correct answer

C. C′=4KK+3C0{C}^{'} = \dfrac{4K}{K + 3}{C}_{0}

Step-by-step solution

For changing the capacitance of a given paralle plate capacitor, a dielectric material of dielectric constant K is

C0=ϵ0Ad{C}_{0} = \dfrac{{ϵ}_{0}A}{d}

C′=C1{C}^{'} = {C}_{1} and C2{C}_{2} in series.

i.e. 1C′=1C1+1C2\dfrac{1}{{C}^{'}} = \dfrac{1}{{C}_{1}} + \dfrac{1}{{C}_{2}}

1C′=(3d/4)ϵ0KA+d/4ϵ0A\dfrac{1}{{C}^{'}} = \dfrac{\left(3d / 4\right)}{{ϵ}_{0}KA} + \dfrac{d / 4}{{ϵ}_{0}A}

1C′=d4ϵ0A(3+KK)\dfrac{1}{{C}^{'}} = \dfrac{d}{4{ϵ}_{0}A}\left(\dfrac{3 + K}{K}\right)

C′=4KC0(3+K){C}^{'} = \dfrac{4K{C}_{0}}{\left(3 + K\right)}

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