Two insulated identically sized charged copper spheres A and B have their centers separated by a distance of 50 cm. A third sphere of the same size but uncharged is brought in contact with the first, then in contact with the second and finally removed from both. What is the new force of repulsion between A and B?

Explanation

To solve the problem step by step, we will follow these procedures:

### Step 1: Calculate the initial force of repulsion between spheres A and B

Given:
- Charge on each sphere (A and B), \( Q = 6.5 \times 10^{-7} \, \text{C} \)
- Distance between centers of A and B, \( r = 50 \, \text{cm} = 0.5 \, \text{m} \)
- Coulomb's constant, \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
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