Question

Two identical conducting spheres B and C are equally charged and there is a force of repulsion F between them when they are placed at some distance. The third spherical conductor is also identical to them but is uncharged. It is first kept in contact with B and then with C and is then separated. Determine the new force of repulsion between B and C.

Answer 1

Let the spherical conductors Band C have same charge as q. The electric force between them is

$F=\frac{1}{4\pi {\epsilon }_0}\times \left(\frac{q^2}{r^2}\right)$

Here, r being the distance between them, When third uncharged conductor A is brought in contact with B, then charge on each conductor

$q_A=q_B=\frac{q_A+q_b}{2}=\frac{0+q}{2}=\frac{q}{2}$

When this conductor A is now being in contact with C, then charge on each conductor

$q_A=q_C=\frac{q_A+q_C}{2}=\frac{\frac{q}{2}+q}{2}=\frac{3q}{4}$

Hence, the electric force acting between B and C is

$F^{\prime }=\frac{1}{4\pi {\epsilon }_0}\times \frac{q_B\times q_C}{r^2}$

$F^{\prime }=\frac{1}{4\pi {\epsilon }_0}\times \frac{\frac{q}{2}\times \frac{3q}{4}}{r^2}$

solve the above, we get

$F^{\prime }=\frac{3F}{8}$