Question

Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is F. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres A and B (radii of A and B are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as:

पर्याय

• `(3 F)/5`

• `(2F)/3`

• `F/2`

• `(3F)/8`

Answer 1

`(3F)/8`

Explanation:

Let the separation between the centres of the two spheres be d.

`F = (kq^2)/(d^2)`

When the third identical uncharged sphere C is brought in contact with sphere A, both share equal charges

⇒ V_A = V_C

`q_A = q_C = q/2`

Now when the third sphere (with charg `q/2`) is brought into contact with sphere B, again both share equal charges

⇒ V_B = V_C

`q'_B = q'_C`

= `(q + q/2)/(2)`

= `(3q)/4`

The value of F' between spheres A and B is given by,

`F' = (k(q/2)((3q)/4))/(d^2)`

= `k((3q^2)/8)/(d^2)`

= `3/8 F`