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

Let the two conducting plates be P1 and P2.

Plate P1 has a total charge of $Q_1 = 12 \, \text{mC}$.

Plate P2 has a total charge of $Q_2 = 6 \, \text{mC}$.

The plates are placed face to face and close to each other.

Surface A is the outer surface of plate P1.

Surface B is the inner surface of plate P1, facing plate P2.

Surface C is the inner surface of plate P2, facing plate P1.

Surface D is the outer surface of plate P2.

Let the charge on surface A be $q_A$, on surface B be $q_B$, on surface C be $q_C$, and on surface D be $q_D$.

The total charge on plate P1 is distributed between surfaces A and B: $q_A + q_B = Q_1 = 12 \, \text{mC}$ (1)

The total charge on plate P2 is distributed between surfaces C and D: $q_C + q_D = Q_2 = 6 \, \text{mC}$ (2)

For two conducting plates placed parallel and close to each other, the charges on the inner surfaces facing each other are equal in magnitude and opposite in sign. This is due to electrostatic induction, where the positive charge on one inner surface attracts an equal amount of negative charge on the opposite inner surface. $q_B = -q_C$ (3)

Also, for two parallel conducting plates, the charges on the outer surfaces are equal. $q_A = q_D$ (4)

Now we have a system of four linear equations with four variables:

1. $q_A + q_B = 12$
2. $q_C + q_D = 6$
3. $q_B = -q_C$
4. $q_A = q_D$

Substitute (3) into (2): $-q_B + q_D = 6$ (5)

Now we have three equations:

1. $q_A + q_B = 12$
2. $q_A - q_D = 0$
3. $-q_B + q_D = 6$

From (4), $q_D = q_A$. Substitute this into (5): $-q_B + q_A = 6$ (6)

Now we have a system of two equations with two variables $q_A$ and $q_B$:

1. $q_A + q_B = 12$
2. $q_A - q_B = 6$

Add equation (1) and equation (6): $(q_A + q_B) + (q_A - q_B) = 12 + 6$ $2q_A = 18$ $q_A = 9 \, \text{mC}$

Substitute the value of $q_A$ into equation (1): $9 + q_B = 12$ $q_B = 12 - 9$ $q_B = 3 \, \text{mC}$

Now we can find $q_C$ and $q_D$ using equations (3) and (4): From (3), $q_C = -q_B = -3 \, \text{mC}$ From (4), $q_D = q_A = 9 \, \text{mC}$

So, the charge distribution on surfaces A, B, C, and D are:

$q_A = 9 \, \text{mC}$

$q_B = 3 \, \text{mC}$

$q_C = -3 \, \text{mC}$

$q_D = 9 \, \text{mC}$

Let's check if these values satisfy the original total charges:

Charge on plate P1: $q_A + q_B = 9 + 3 = 12 \, \text{mC}$ (Correct)

Charge on plate P2: $q_C + q_D = -3 + 9 = 6 \, \text{mC}$ (Correct)

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