JEE MAIN - Physics (2021 - 17th March Morning Shift - No. 22)
A parallel plate capacitor whose capacitance C is 14 pF is charged by a battery to a potential difference V = 12 V between its plates. The charging battery is now disconnected and a porcelin plate with k = 7 is inserted between the plates, then the plate would oscillate back and forth between the plates with a constant mechanical energy of _____________ pJ. (Assume no friction)
Answer
864
Explanation
$${U_i} = {1 \over 2}c{v^2}$$
$$ = {1 \over 2} \times 14 \times {(12)^2}$$ pJ
= 1008 pJ
$${U_f} = {{{Q^2}} \over {2kC}}$$
$$ = {{{{(14 \times 12)}^2}} \over {2 \times 7 \times 14}}$$
= 144 pJ
oscillating energy = Ui $$-$$ Uf
= 1008 $$-$$ 144
= 864 pJ
$$ = {1 \over 2} \times 14 \times {(12)^2}$$ pJ
= 1008 pJ
$${U_f} = {{{Q^2}} \over {2kC}}$$
$$ = {{{{(14 \times 12)}^2}} \over {2 \times 7 \times 14}}$$
= 144 pJ
oscillating energy = Ui $$-$$ Uf
= 1008 $$-$$ 144
= 864 pJ
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