Two small similar metal sphere A and B having charge 4 q and - 4q when placed at a certain distance apart exert and electric force F on each other then another identical uncharged sphere C first touched with A then with B and then removed infinity, the force of attraction between A and B for the same separation will be (a) F/2 (b) F/8 (c) F/16 (d) F/32?

Question

Answer

Solution:

Given: Two small similar metal sphere A and B having charge 4q and -4q respectively.

Electric force between A and B = F

When sphere C is touched with sphere A, it acquires a charge of +4q due to induction. Then, when sphere C is touched with sphere B, it acquires a charge of -4q due to induction.

Now, sphere A has a net charge of 4q + 4q = 8q, and sphere B has a net charge of -4q - 4q = -8q.

Therefore, the electric force between A and B will be:

F' = (1/4πε₀) * |(8q) * (-8q)| / d²

where ε₀ is the permittivity of free space and d is the distance between A and B.

Simplifying, we get:

F' = F/16

Hence, the correct answer is option (c) F/16.

Explanation:

To solve this problem, we can use the principle of superposition of charges. When sphere C is touched with A and B, it induces charges on C due to the presence of A and B. The net charge on C is zero, but it redistributes the charges on A and B.

When A and B are brought close to each other, they experience a force of attraction due to their opposite charges. The magnitude of this force is given by Coulomb's law.

When C is touched with A and B, the charges on A and B are redistributed in such a way that the net charge on each of them remains the same, but their positions change. This does not affect the force of attraction between them, as it depends only on their net charges and the distance between them.

Therefore, the force of attraction between A and B remains the same as before, but with a different configuration of charges. The new force can be calculated using Coulomb's law and the principle of superposition of charges.

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