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 .

Question

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 .,

MARKS App · Accepted Answer

Initially C = 200 μ F E i = 1 2 C V 2 = 1 2 × 200 × 10 - 6 × 200 2 Finally C ' = K C = 400 μ F E f = 1 2 C ' V 2 = 1 2 × 400 × 10 - 6 × ( 200 ) 2 ΔE = 1 2 × 400 - 200 × 10 - 6 × 4 × 10 4 = 4 J

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