A parallel plate capacitor has each plate area A and separation d. Charge Q is given to positive plate of capacitor. The force on each plate of capacitor is
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Correct option is
A
4C
Q
2
Initial capacitance C
1
=
d
Aϵ
0
Final capacitance C
2
=
1.5d
Aϵ
0
=
3d
2Aϵ
0
Since the capacitor was isolated, the charge stored remains the same.
Thus work done=
2C
2
Q
2
−
2C
1
Q
2
=
4C
Q
2
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