Question

Two spherical conduction $B$ and $C$ having equal radii and carrying equal charges on 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, in 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

• A. $F/4$
• B. $3F/4$
• C. $F/8$
• D. $3F/8$

Answer 1

The correct option is D $3F/8$

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, 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

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