Question

Two identical conducting spheres, A and B, carry equal charge. They are separated by a distance much larger than their diameters. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and finally removed.
1. As a result, the electrostatic force between A and B, which was originally F, becomes ________:A. F/2B. F/4C. 3F/8D. F/16E. 0

Answer 1

Answer:

C. $\frac{3F}{8}$

Explanation:

Let initial charges on both spheres be,$q$

$F=\frac{Kq^2}{d^2} \ \ \ \ \ \ \ \ \ \ \_i$

When the sphere C is touched by A, the final charges on both will be,$\frac{q}{2}$

#Now, when C is touched by B, the final charges on both of them will be:

$q_c=q_d=\frac{q/2+q}{2}\\\\=\frac{3q}{4}$

Now the force between A and B is calculated as:

$F\prime=\frac{k\times\frac{q}{2}\times \frac{3q}{4}}{d^2}\\F\prime=\frac{3F}{8}$

Hence the electrostatic force becomes 3F/8