Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with their diameters (Fig. 21-22a). The electrostatic force acting on sphere 2 due to sphere 1 is$\overrightarrow{F}$.Suppose now that a third identical sphere 3, having an insulating handle and initially neutral, is touched first to sphere 1 (Fig. 21-22b), then to sphere 2 (Fig. 21-22c), and finally removed (Fig. 21-22d). The electrostatic force that now acts on sphere 2 has magnitude$\mathit{F}\mathbf{\text{'}}$. What is the ratio$\mathit{F}\mathbf{\text{'}}\mathbf{/}\mathit{F}$?

1. Two identical spheres 1 and 2 are separated by a distance$\left(r\right)\gg diameter$.
2. Electrostatic force on sphere 2 due to sphere 1 is $\stackrel{\rightharpoonup }{F}$.
3. The new electrostatic force on sphere 2 due to sphere 1 is$F\text{'}$ .

Conduction is the process in which an uncharged body is charged by bringing it in contact with the charged body. Here, when two identical spheres are brought in together, the net charge of the system is equally distributed among them. Using this theory, the charge on each sphere is calculated, and then using the electrostatic force formula, we can get the magnitude of the forces in both cases.

The electrostatic force of attraction or repulsion on a body by another charged body according to Coulomb’s law,

$\mathbf{F}\mathbf{=}\frac{{\mathbf{kq}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}$ (i)

Here, q1 and q₂ are the charges on each sphere and is the separation between them.

Let,q be the charge on sphere 1 and 2.

For the first case, the electrostatic force on sphere 2 by sphere 1 can be given using equation (i) as follows:

$F=\frac{k{q}^2}{r^2}.....................\left(a\right)$

Now, for the second case, when the neutral sphere 3 is brought into contact with sphere 1 having charge q, they both acquire charge q/2 as per the concept. Now, when the sphere 3 with charge q/2 is brought into contact with sphere 2 with charge q , the charge acquired by sphere 2 and 3 will be given as:

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

Now, the force on sphere 2 by sphere 1 can be given using equation (i) as follows:

role="math" $$\begin{gathered} F\text{'}=\frac{k\left(q/2\right)\left(3q/4\right)}{r^2} \\ =\frac{3k{q}^2}{8r^2}...................\left(b\right) \end{gathered}$$

Thus, using values from equations (a) and (b), the ratio of the forces can be given as:

role="math" localid="1661870975624" $$\begin{gathered} \frac{F\text{'}}{F}=\frac{\left(\frac{3k{q}^2}{8r^2}\right)}{\frac{k{q}^2}{r^2}} \\ =\frac{3}{8} \\ =0.375 \end{gathered}$$

Hence, the required ratio is 0.375.