Three conducting plates are given 3Q, –2Q and Q charge as shown in figure. After distribution of charge find value of q1, q2, q3, q4, q5 and q6.

To solve the problem of charge distribution among three conducting plates given the charges $3Q$, $-2Q$, and $Q$, we can follow these steps:

1. Break Down the Problem

We need to analyze the charge interactions between the plates. Let's denote the plates as:

• Plate A: $3Q$
• Plate B: $-2Q$
• Plate C: $Q$

We will be finding the induced charges on the surfaces of each plate in response to the given charges, denoting them as $q_1, q_2, q_3, q_4, q_5,$ and $q_6$ for the respective surfaces of the plates.

2. Relevant Concepts

The principle of charge conservation and the behavior of conductors in electrostatic equilibrium will be important:

• The total charge in the system must be conserved.
• The induced charge on the inner surface of a conductor must equal the negative of the charge on the outer surface of the adjacent conductor.

3. Analysis and Detail

Let's assume:

• $q_1$: Charge on the surface of Plate A that faces Plate B.
• $q_2$: Charge on the outer surface of Plate A.
• $q_3$: Charge on the inner surface of Plate B facing Plate A.
• $q_4$: Charge on the outer surface of Plate B.
• $q_5$: Charge on the surface of Plate B that faces Plate C.
• $q_6$: Charge on the outer surface of Plate C.

Step 1: Charge Distribution

1. For Plate A with charge $3Q$:

   • $q_1 + q_2 = 3Q$ (Total charge on Plate A)

2. For Plate B with charge $-2Q$:

   • $q_3 + q_4 = -2Q$
   • Since Plate A induces $-q_1$ on Plate B, we have $q_3 = -q_1$.

3. For Plate C with charge $Q$:

   • The surface facing Plate B will have $-q_5$ induced by Plate B, hence $q_6 + q_5 = Q$.

4. Verify and Summarize

Using our relationships:

1. $q_1 + q_2 = 3Q$ (Equation 1)
2. $-q_1 + q_4 = -2Q$ (Substituting $q_3$ into Equation 2)
3. $q_2 + q_6 = Q$ (Equation 3)

From Equation 1:

• Rearrange it to find $q_2 = 3Q - q_1$
  • Substitute into Equation 3:
  • $3Q - q_1 + q_6 = Q$
  • Simplifying gives: $q_6 = Q - 3Q + q_1 = -2Q + q_1$

From $-q_1 + q_4 = -2Q$:

• Rearranging gives us $q_4 = -2Q + q_1$

Solving the System of Equations

1. From $q_2$ and $q_4$: We need numerical values for $q_1, q_2, q_3=q_1, q_4, q_5, q_6$.
2. Set constraints for $q_5$ as the other plate's total charge.

Final Answer

Assuming symmetry and using charge conservation:

1. Assuming $q_1 = Q$

2. $q_2 = 2Q$

3. $q_3 = -Q$

4. $q_4 = -Q$

5. Assume charge on Plate C edges balances so $q_5 = 0$

6. Thus $q_6 = 0$

7. $q_1 = Q$

8. $q_2 = 2Q$

9. $q_3 = -Q$

10. $q_4 = -Q$

11. $q_5 = 0$

12. $q_6 = 0$

Thus we have the configuration based on charge symmetry among the plates.