Solveeit Logo

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

A point charge at point 'P will experience electric force due to capacitor

B

The potential difference between the plates will be Q2C\frac{Q}{2C}

C

The energy stored in the electric field in the region between the plates is 9Q24C\frac{9Q^2}{4C}

D

The force on one plate due to the other plate is Q22πϵ0d2\frac{Q^2}{2\pi \epsilon_0 d^2}

Answer

A point charge at point 'P will experience electric force due to capacitor

Explanation

Solution

Let the charges on the plates be Q1=2QQ_1 = 2Q and Q2=QQ_2 = -Q. The area of the plates is AA. The capacitance is C=ϵ0AdC = \frac{\epsilon_0 A}{d}.

The electric field outside the capacitor (at point P) is given by Eout=Q1+Q22Aϵ0E_{out} = \frac{Q_1 + Q_2}{2A\epsilon_0}. Substituting the charges: Eout=2Q+(Q)2Aϵ0=Q2Aϵ0E_{out} = \frac{2Q + (-Q)}{2A\epsilon_0} = \frac{Q}{2A\epsilon_0}. Since Eout0E_{out} \neq 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=Q1Q22Aϵ0=2Q(Q)2Aϵ0=3Q2Aϵ0E_{in} = \frac{Q_1 - Q_2}{2A\epsilon_0} = \frac{2Q - (-Q)}{2A\epsilon_0} = \frac{3Q}{2A\epsilon_0}. The potential difference between the plates is V=Eind=3Q2Aϵ0dV = E_{in} \cdot d = \frac{3Q}{2A\epsilon_0} \cdot d. Using C=ϵ0AdC = \frac{\epsilon_0 A}{d}, we get Aϵ0=CdA\epsilon_0 = Cd. So, V=3Qd2Cd=3Q2CV = \frac{3Qd}{2Cd} = \frac{3Q}{2C}. Statement (2) is incorrect.

The energy stored in the electric field between the plates is U=12CV2=12C(3Q2C)2=12C9Q24C2=9Q28CU = \frac{1}{2} C V^2 = \frac{1}{2} C \left(\frac{3Q}{2C}\right)^2 = \frac{1}{2} C \frac{9Q^2}{4C^2} = \frac{9Q^2}{8C}. Statement (3) is incorrect.

The force on one plate (say, Q1Q_1) due to the other plate (Q2Q_2) is F=Q1EQ2F = |Q_1 E_{Q_2}|. The electric field due to a single plate Q2Q_2 is EQ2=Q22Aϵ0=Q2Aϵ0E_{Q_2} = \frac{|Q_2|}{2A\epsilon_0} = \frac{Q}{2A\epsilon_0}. So, F=2QQ2Aϵ0=Q2Aϵ0=Q2CdF = |2Q \cdot \frac{Q}{2A\epsilon_0}| = \frac{Q^2}{A\epsilon_0} = \frac{Q^2}{Cd}. Statement (4) is incorrect.