a capacitor of 10 micro farad is charged to a potential 50 v the battery is now disconnected and addition charge 20 is given to the plates of capacitor new potential difference of the capacitor will be
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Answer:
Charge acquired by the plates of the capacitor q
0
=CV=(10μF)×(50V)=500μC
Now, let the charge distribution is as follows.
Total charge on positive plate has now become 700μC while that in negative plate is still −500μC.
Here, charges are in μC.
Net electric field at point P is zero.
∴
2Aε
0
(700−q)
+
2Aε
0
q
+(
2Aε
0
500−q
)=
2Aε
0
q
⇒q=600μC
∴ Potential difference between the plates is
ΔV=
C
q
=
10μF
600μC
=60V
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