91Ó°ÊÓ

The given data can be listed below as:

• The number of the excess electrons is, $\mathrm{Q}=1×{10}^{10}$.
• The distance amongst the centers of the first ball and the uncharged identical metal ball is,$\text{°ù=20 c³¾Ã—}\left(\frac{{\text{10}}^{\text{-2}}\text{″¾}}{\text{1 c³¾}}\right){\text{=20×10}}^{\text{-2}}\text{″¾}$ .

The magnitude of the force of two charged spheres is directly proportional to the charge of the spheres. Moreover, the force is also inversely proportional to the square of the distance between the center of the spheres.

As there are excess electrons, hence the electrons get equally distributed on the ball. Hence, the charges on both the spheres are

$\begin{array}{c}\frac{\mathrm{Q}}{2}={\mathrm{q}}_1\\ ={\mathrm{q}}_2\\ =0.5×{10}^{10}\mathrm{e}\end{array}$

The equation of the electric force exerted by one ball over another ball is expressed as:

$\mathrm{F}=\mathrm{k}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}$

Here,$\mathrm{F}$is the electric force exerted by one ball over another ball, $\mathrm{k}$is the electric force constant, $q_1$and $q_2$are the charges of the first and the second spheres and $\mathrm{r}$is the distance between the sphere’s center.

Substitute the values in the above equation.

$\begin{array}{c}\mathrm{F}=(9×{10}^9\text{ N}â‹\dots {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}})×\frac{{(0.5×{10}^{10}\mathrm{e})}^2}{{(20×{10}^{−2}\text{″¾})}^2}\\ =(9×{10}^9\text{ N}â‹\dots {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}})×\frac{(2.5×{10}^{19}{(1.602×{10}^{−19}\text{ C})}^2)}{\left(0.04{\text{″¾}}^2\right)}\\ =(9×{10}^9\text{ N}â‹\dots {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}})×1.604×{10}^{−17}{\text{ C}}^2{\text{/m}}^{\text{2}}\\ =1.44×{10}^{−7}\text{ N}\end{array}$

The assumptions made in this part is the force exerted on the second sphere by the first sphere and force exerted on the first sphere by the second sphere is equal in sign but opposite in the direction. Hence, the force is repulsive.

Thus, the electric force one ball exerts on the other is $1.44×{10}^{−7}\text{ N}$and the force is repulsive.

The assumptions made in this part is the force exerted on the second sphere by the first sphere and force exerted on the first sphere by the second sphere is equal in sign but opposite in the direction.

The diagram of the forces on the sphere has been provided below:

According to the above figure, there is non-uniform charge distribution is going on between the spheres because of the polarization. As the negative charge is concentrated on the sphere's farther side, then because of the farther sides, the electric field gets generated that are${\mathrm{E}}_{\mathrm{f}1}$ and${\mathrm{E}}_{\mathrm{f}2}$ respectively.

Because of the polarization, basically two forces act on both of the sphere. The first force acting on the first sphere mainly moves the sphere in the right direction because of the non-uniform charge distribution. The second force acting on the first sphere mainly moves the sphere in the left direction because of the repulsion of the like charges. The same forces act on the second sphere and moves the sphere in both right and the left direction.

Thus, due to the polarization and the movement of the spheres in both the directions, the force exerted by one sphere over another sphere decreases.