Manipal Entrance Test 2015 Physics Paper

Charge acquired by the plates of the capacito
q0=CV=(10µF)×(50V) =500µC
Now, let the charge distribution is as follows.
Total charge on positive plate has now become 700 µC while that in negative plate is still - 500 µC, Here, charges are in µC.
Net electric field at point P is zero.
∴

(700−q)

2Aε0

+

q

2Aε0

+(

500−q

2Aε0

)=

q

2Aε0

⇒q=600µC
∴ Potential difference between the plates is
ΔV=

q

C

=

600µC

10µF

=60V