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[Solved] A parallel plate capacitor of capacitance 200 µF is connected to a battery of 200 V. A dielectric slab of dielectric constant 2 is now inserted

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A parallel plate capacitor of capacitance 200 µF 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.

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ΔU = \(\frac{1}{2}(ΔC)V^2\)

ΔU = \(\frac{1}{2}(KC -C)V^2\)

ΔU = \(\frac{1}{2}(2 - 1)CV^2\)

ΔU = \(\frac{1}{2}\)x 200 x 10-6 x 200 x 200

ΔU = 4 J

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