Question

A metal ball with diameter of a half a centimeter and hanging from an insulating thread is charged up with $\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}$excess electrons. An initially uncharged identical metal ball hanging from an insulating thread is brought in contact with the first ball, then moved away, and they hang so that the distance between their centers is $\mathbf{20}\mathbf{}\mathbf{cm}$.

(a) Calculate the electric force one ball exerts on the other, and state whether it is attractive or repulsive. If you have to make any simplifying assumptions, state them explicitly and justify them.

(b) Now the balls are moved so that as they hang, the distance between their centers is only $\mathbf{5}\mathbf{}\mathbf{cm}$. Naively one would expect the force that one ball exerts on the other to increase by a factor of ${\mathbf{\text{4}}}^{\mathbf{2}}\mathbf{=}\mathbf{16}$, but in real life the increase is a bit less than a factor of role="math" localid="1661330186132" $\mathbf{16}$. Explain why, including a diagram. (Nothing but the distance between centers is changed—the charge on each ball is unchanged, and no other objects are around.)

Short answer

(a) The electric force one ball exerts on the other is$1.44\times {10}^{-7}\text{\hspace{0.17em}N}$ 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.

(b) Due to the polarization and the movement of the spheres in both the directions, the force exerted by one sphere over another sphere decreases.

Step-by-step solution

Identification of the given data

The given data can be listed below as:

• The number of the excess electrons is, $\mathrm{Q}=1\times {10}^{10}$.
• The distance amongst the centers of the first ball and the uncharged identical metal ball is,$\text{r=20\hspace{0.17em}cm×}\left(\frac{{\text{10}}^{\text{-2}}\text{\hspace{0.17em}m}}{\text{1\hspace{0.17em}cm}}\right){\text{=20×10}}^{\text{-2}}\text{\hspace{0.17em}m}$ .

Significance of the magnitude of the force of two charged spheres

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.

 Step 3: (a) Determination of the electric force exerted by one ball over another ball

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

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

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.

$\begin{array}{c}\mathrm{F}=(9\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}})\times \frac{{(0.5\times {10}^{10}\mathrm{e})}^2}{{(20\times {10}^{-2}\text{\hspace{0.17em}m})}^2}\\ =(9\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}})\times \frac{(2.5\times {10}^{19}{(1.602\times {10}^{-19}\text{\hspace{0.17em}C})}^2)}{\left(0.04{\text{\hspace{0.17em}m}}^2\right)}\\ =(9\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}})\times 1.604\times {10}^{-17}{\text{\hspace{0.17em}C}}^2{\text{/m}}^{\text{2}}\\ =1.44\times {10}^{-7}\text{\hspace{0.17em}N}\end{array}$

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 $1.44\times {10}^{-7}\text{\hspace{0.17em}N}$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.

(b) Determination of the increase of the force

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.