Question

12mC and 6mC charges are given to the two conducting plates having same cross-sectional area and placed face to face close to each other as shown in the figure. The resulting charge distribution in mC on surface A,B,C and D are respectively,

Answer 1

9, 3, -3, 9

Answer 2

When two conducting plates are placed face to face, the charges redistribute themselves on the surfaces due to electrostatic induction. Let the two plates be Plate 1 and Plate 2, with total charges $Q_1 = 12 \text{ mC}$ and $Q_2 = 6 \text{ mC}$ respectively. Let the surfaces be labeled as A, B, C, and D as shown in the figure.

Surface A: Outer surface of Plate 1. Surface B: Inner surface of Plate 1. Surface C: Inner surface of Plate 2. Surface D: Outer surface of Plate 2.

Let the charges on these surfaces be $q_A, q_B, q_C,$ and $q_D$ respectively. The total charge on Plate 1 is $Q_1 = q_A + q_B = 12 \text{ mC}$. The total charge on Plate 2 is $Q_2 = q_C + q_D = 6 \text{ mC}$.

For parallel conducting plates placed close to each other, the following principles apply:

1. The charges on the inner surfaces (B and C) are equal in magnitude and opposite in sign due to electrostatic induction. This is because the electric field lines originating from one inner surface must terminate on the other inner surface. Thus, $q_B = -q_C$.
2. The charges on the outer surfaces (A and D) are equal and are equal to half of the total charge of the system. The total charge of the system is $Q_{total} = Q_1 + Q_2$. Thus, $q_A = q_D = \frac{Q_1 + Q_2}{2}$.

Let's calculate the total charge: $Q_{total} = 12 \text{ mC} + 6 \text{ mC} = 18 \text{ mC}$.

Now, calculate the charge on the outer surfaces: $q_A = q_D = \frac{18 \text{ mC}}{2} = 9 \text{ mC}$.

Now, use the total charge on each plate to find the charges on the inner surfaces: For Plate 1: $q_A + q_B = 12 \text{ mC}$. Substitute $q_A = 9 \text{ mC}$: $9 \text{ mC} + q_B = 12 \text{ mC}$ $q_B = 12 \text{ mC} - 9 \text{ mC} = 3 \text{ mC}$.

For Plate 2: $q_C + q_D = 6 \text{ mC}$. Substitute $q_D = 9 \text{ mC}$: $q_C + 9 \text{ mC} = 6 \text{ mC}$ $q_C = 6 \text{ mC} - 9 \text{ mC} = -3 \text{ mC}$.

Let's verify the condition $q_B = -q_C$: $3 \text{ mC} = -(-3 \text{ mC})$ $3 \text{ mC} = 3 \text{ mC}$. The condition is satisfied.

So, the resulting charge distribution on surfaces A, B, C, and D respectively is: $q_A = 9 \text{ mC}$ $q_B = 3 \text{ mC}$ $q_C = -3 \text{ mC}$ $q_D = 9 \text{ mC}$

The charge distribution in mC on surfaces A, B, C and D are 9, 3, -3, and 9 respectively.