Question
Question: In the figure shown the plates of a parallel plate capacitor have unequal charges. Its capacitance i...
In the figure shown the plates of a parallel plate capacitor have unequal charges. Its capacitance is 'C. P is a point outside the capacitor and close to the plate of charge -Q. The distance between the plates is 'd' then which statement is correct

A point charge at point 'P will experience electric force due to capacitor
The potential difference between the plates will be 2CQ
The energy stored in the electric field in the region between the plates is 4C9Q2
The force on one plate due to the other plate is 2πϵ0d2Q2
A point charge at point 'P will experience electric force due to capacitor
Solution
Let the charges on the plates be Q1=2Q and Q2=−Q. The area of the plates is A. The capacitance is C=dϵ0A.
The electric field outside the capacitor (at point P) is given by Eout=2Aϵ0Q1+Q2. Substituting the charges: Eout=2Aϵ02Q+(−Q)=2Aϵ0Q. Since Eout=0, a point charge placed at P will experience an electric force. Thus, statement (1) is correct.
The electric field between the plates is Ein=2Aϵ0Q1−Q2=2Aϵ02Q−(−Q)=2Aϵ03Q. The potential difference between the plates is V=Ein⋅d=2Aϵ03Q⋅d. Using C=dϵ0A, we get Aϵ0=Cd. So, V=2Cd3Qd=2C3Q. Statement (2) is incorrect.
The energy stored in the electric field between the plates is U=21CV2=21C(2C3Q)2=21C4C29Q2=8C9Q2. Statement (3) is incorrect.
The force on one plate (say, Q1) due to the other plate (Q2) is F=∣Q1EQ2∣. The electric field due to a single plate Q2 is EQ2=2Aϵ0∣Q2∣=2Aϵ0Q. So, F=∣2Q⋅2Aϵ0Q∣=Aϵ0Q2=CdQ2. Statement (4) is incorrect.