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Question

Two spherical conductors B and C having equal radii and carrying equal charges in them repel each other with a force F, when kept apart at some distance. A third spherical conductor having same radius as that of B but uncharged, is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is:
  1. F4
  2. 3F4
  3. F8
  4. 3F8

A
F4
B
F8
C
3F4
D
3F8
Solution
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Step 1: Calculation of initial force
Let B and C have charge Q each initially and are seperated by r distance.
From Coulomb's law, force between them F=KQ2r2 ....(1)

Step 2: Distribution of charges
When an identical uncharged spherical conductor A is brought in contact with charged sphere B.
By symmetry, the total charge will be shared equally among them, as both have equal radii.
Charge on A and B =Q2

Now, when the conductor A is brought in contact with C.
They will also share equal charge among themselves, as both have equal radii.
Charge on A and C =Q+Q22=3Q4

Step 3: Calculation of new force
New coulomb force between B and C:
F=KQ2×3Q4r2=38KQ2r2
From equation (1), we get: F=38F

Hence, Option D is correct

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