[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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16/02/2022 12:03 pm
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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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16/02/2022 12:05 pm
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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