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Question

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An air-filled capacitor stores a potential energy of 15 mJ due to its charge. It is accidentally filled with water in such a way as not to discharge its plates. How much energy does it continue to store after it is filled? (The dielectric constant for water is 78 and for air it is 1.0006.)

Explanation / Answer

initialy U = C*V^2/2

C = 15*10^(-3)*2/V^2

capacitance after filling water between plates

C' = k*C

so U' = k*C*V^2/2

U' = k*2*15*V^2 / 2*V^2

U' = k*15 = 78*15*10^(-3)

U' = 1170 mJ

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