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:
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Correct option is D)
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=r2KQ2....(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=2Q
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=2Q+2Q=43Q
Step 3: Calculation of new force
New coulomb force between B and C:
From equation (1), we get: F′=83F
Hence, Option D is correct
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