the plates of a 0.6uf uncharged capacitor are connected to a 10v battery. the charge on the capacitor after a long time is
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Answer:
Initial energy:
U
1
=
2
1
CV
2
=
2
1
×100×10
−6
×(50)
2
=12.5×10
−2
J
When the battery is kept connected, voltage across the capacitor does not change.
We know: C=
d
ϵ
0
A
=100 μF
And when distance is halved:
C
′
=
d
2ϵ
0
A
=200 μF
So, the final energy:
U
2
=
2
1
C
′
V
2
=
2
1
×200×10
−6
×(50)
2
=25×10
−2
J
∴ additional energy given = U
2
−U
1
=(25×10
−2
−12.5×10
−2
) J=12.5×10
−2
J
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