JEE MAIN - Physics (2021 - 31st August Evening Shift - No. 21)
A parallel plate capacitor of capacitance 200 $$\mu$$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
Explanation
$$\Delta U = {1 \over 2}(\Delta C){V^2}$$
$$\Delta U = {1 \over 2}(KC - C){V^2}$$
$$\Delta U = {1 \over 2}(2 - 1)C{V^2}$$
$$\Delta U = {1 \over 2} \times 200 \times {10^{ - 6}} \times 200 \times 200$$
$$\Delta U = 4$$ J
$$\Delta U = {1 \over 2}(KC - C){V^2}$$
$$\Delta U = {1 \over 2}(2 - 1)C{V^2}$$
$$\Delta U = {1 \over 2} \times 200 \times {10^{ - 6}} \times 200 \times 200$$
$$\Delta U = 4$$ J
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