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+2 votes
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in Physics by (69.6k points)

 A parallel plate air capacitor with its plates spaced 2cm apart is charged to a potential of 300 volts. What will be the electric field intensity inside the capacitor, if the plates are moved apart to a distance of 5cm without disconnecting the power source? Calculate the change in energy of the capacitor. Area of the plates is equal to A = 100cm2 . Also solve the problem assuming the entire operation was done after disconnecting the power source. Account for the change in energy in both the cases.(∈0 = 9 x 10 -12 SI units)

1 Answer

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Best answer

Now in 2nd case , Q constant

We have Q= CV

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