When 30μC charge is given to an isolated conductor of capacitance 5μF . Find energy stored in the electric field of conductor.
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Solution
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The energy stored in the capacitor is given
as follows
E=q22C
By substituting the values in the above given
equation we get
E=(30×10−6)2C2×5×10−6F
E=9×10−5J
Hence the energy stored in the electric field
of conductor isE=9×10−5J
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