# Parallel Plate Capacitor with different Charges

Parallel Plate Capacitor with different Charges :

Two identical plates of parallel plate capacitor are given unequal charges Q1 and Q2 . Let the charge appearing on the inner surface be +q and -q, then the charges appearing on outer surfaces are as shown in the figure.

If we take a point P inside the plate 1, then electric field at P should be zero. Suppose surface area of the each surface is A.

Using the equation
$\displaystyle E = \frac{\sigma}{2 \epsilon_0}$

$\displaystyle \vec{E_p} = \vec{E_1} + \vec{E_2} + \vec{E_3} + \vec{E_4}$

Where E1 , E2 , E3 , E4 are the electric fields due to surfaces 1, 2, 3 and 4 at the inner point P .

$\displaystyle \vec{E_p} = \frac{Q_1 – q}{2 \epsilon_0 A}\hat{i} – \frac{ q}{2 \epsilon_0 A}\hat{i} + \frac{ q}{2 \epsilon_0 A}\hat{i} – \frac{Q_2 + q}{2 \epsilon_0 A}\hat{i}$

$\displaystyle \vec{E_p} = (\frac{Q_1 – q}{2 \epsilon_0 A}- \frac{Q_2 + q}{2 \epsilon_0 A})\hat{i}$

But |E| = 0 Which gives ,

$\displaystyle q = \frac{Q_1 – Q_2}{2}$

Hence, charge distribution on each surfaces are shown in figure

Note:
The charge on inner surface of plate A is the half of the difference of the charge on plate A and B (Q1 and Q2 respectively) i.e. (Q1 – Q2)/2 , and on inner surface of B is opposite of the charge appearing on inner surface of A i.e. -(Q1 – Q2)/2 ; while the charge on outer surfaces of A and B is half of the sum of charge on A and B i.e. (Q1 + Q2)/2

Illustration : Three very large metal plates are given charges as shown. Find out the final charge distribution. We can assume the cross-sectional area of each plate to be A.

Solution : Let the distribution be as shown

Let P1 , P2 and P3 be the three points

Inside the plates

EP1 = EP2 = EP3 = 0

$\displaystyle E_{P_1} = \frac{1}{2 \epsilon_0 A} (Q-q_1-q_1 + (5Q + q_2)-q_2 -10Q + q_3 -q_3) = 0$

q1 = – 2 Q

EP2 = 0

q2 = – 7 Q

EP3 = 0

q3 = 3 Q

The final distribution is :