Charge acquired by the plates of the capacitor q0=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.
∴(700−q)2Aε0+q2Aε0+(500−q2Aε0)=q2Aε0
⇒q=600μC
∴ Potential difference between the plates is
ΔV=qC=600μC10μF=60V