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NEET
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Physics
Chapter
Capacitance

NEET · Physics · Capacitance

An air-filled parallel plate capacitor is charged by a battery so...

An air-filled parallel plate capacitor is charged by a battery so that it stores an electrostatic energy U0U_{0}. The battery is then disconnected. A dielectric slab of dielectric constant K=2K = 2 and thickness equal to half the plate separation (d2\dfrac{d}{2}) is inserted between the plates. The new electrostatic energy stored in the capacitor will be:

Options

  1. A43U0\dfrac{4}{3}U_{0}
  2. B12U0\dfrac{1}{2}U_{0}
  3. C2U02U_{0}
  4. D34U0\dfrac{3}{4}U_{0}

Correct answer

D. 34U0\dfrac{3}{4}U_{0}

Step-by-step solution

Let the initial capacitance be C0=ϵ0AdC_{0} = \dfrac{\epsilon_{0}A}{d}.
The initial energy stored is U0=Q22C0U_{0} = \dfrac{Q^{2}}{2C_{0}}, where QQ is the charge on the plates.

When the battery is disconnected, the charge QQ remains constant.

A dielectric slab of thickness t=d2t = \dfrac{d}{2} and K=2K = 2 is inserted. The new capacitance CC is:
C=ϵ0Ad−t+tK=ϵ0Ad−d2+d4C = \dfrac{\epsilon_{0}A}{d - t + \dfrac{t}{K}} = \dfrac{\epsilon_{0}A}{d - \dfrac{d}{2} + \dfrac{d}{4}}

C=ϵ0A3d4=43C0C = \dfrac{\epsilon_{0}A}{\dfrac{3d}{4}} = \dfrac{4}{3}C_{0}

The new energy stored in the capacitor is:
U=Q22C=Q22(43C0)U = \dfrac{Q^{2}}{2C} = \dfrac{Q^{2}}{2 \left(\dfrac{4}{3}C_{0}\right)}

U=34(Q22C0)=34U0U = \dfrac{3}{4} \left(\dfrac{Q^{2}}{2C_{0}}\right) = \dfrac{3}{4}U_{0}

Answer: 34U0\dfrac{3}{4}U_{0}
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