JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Capacitance

A parallel plate capacitor of capacitance

200 μ F

is connected to a battery of 200 V. A dielectric slab of dielectric constant 2 is now inserted into the space between plates of capacitor while the battery remain connected. The change in the electrostatic energy in the capacitor will be ______ J.

Answer: 4

Solution

When a dielectric is inserted with the battery connected, the voltage remains constant at

V = 200

V.

Initial capacitance

C_{0} = 200 μ F = 200 \times 1 0^{- 6}

F.

Initial energy

U_{0} = \frac{1}{2} C_{0} V^{2} = \frac{1}{2} \times 200 \times 1 0^{- 6} \times (200)^{2} = \frac{1}{2} \times 200 \times 1 0^{- 6} \times 40000 = 4

J.

After inserting dielectric, new capacitance

C = K C_{0} = 2 \times 200 \times 1 0^{- 6} = 400 \times 1 0^{- 6}

F.

New energy

U = \frac{1}{2} C V^{2} = \frac{1}{2} \times 400 \times 1 0^{- 6} \times 40000 = 8

J.

Change in energy

Δ U = U - U_{0} = 8 - 4 = 4

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Integer answer

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JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Capacitance

Solution

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