A parallel plate capacitor of capacitance 200 μFis connected to a battery of 200V. A dielectric slab of dielectric constant 2 is now inserted into the space between plates of capacitor while the batt…

Question

A parallel plate capacitor of capacitance 200 μFis connected to a battery of 200V. 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.

Accepted Answer

Correct answer: 4. (4) U= 12 ( C )V^2 U= 12( KC-C)V^2 U= 12(2-1)CV^2 U= 12 200 10^-6 200 200 U=4J

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

Solution

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