Question

Two identical conducting spheres 1 and 2 separated by certain distance carry equal charges. They repel each other with a force F. Now a third identical conductor 3, initially uncharged is first brought in contact with 1, then 2 in turn, and is finally removed. What is the new force between 1 and 2?

• A. F
• B. $\frac{F}{2}$
• C. 2F
• D. $\frac{3}{8}F$

Answer 1

The correct option is D $\frac{3}{8}F$

Let 1 and 2 carry a charge q each.
The force between I and 2 is $F_1=\frac{1}{4\pi {ϵ}_0}\frac{qq}{r^2}$
The charge left on 1 after its contact with 3 is $\frac{q+0}{2}=\frac{q}{2}$
The charge left on 2 after its contact with 3 is $\frac{q+\frac{q}{2}}{2}=\frac{3q}{4}$
Therefore, the new force between the spheres is $F=\frac{1}{4\pi {ϵ}_0}\frac{\left(\frac{q}{2}\right)\left(\frac{3q}{4}\right)}{r^2}=\frac{3}{8}F$