Question

Two small plastic spheres are given positive electric charges. When they are 15.0 cm apart, the repulsive force between them has magnitude 0.220 N. What is the charge on each sphere (a) if the two charges are equal and (b) if one sphere has four times the charge of the other?

Short answer

1. The charge on each sphere when the two charges are equal is $7.42\times {10}^{-7}C$.
2. The charge on each sphere when one sphere has four times the charge of the other is $3.7\times {10}^{-7}C$.

Step-by-step solution

Step 1: Concept of Coulomb’s Law

According to the Coulomb’s law, the force between two charges spaced at a distance is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them.

Mathematically,

$\mathbf{F}\mathbf{=}\mathbf{k}\frac{\left|q_1{q}_2\right|}{{\mathbf{r}}^{\mathbf{2}}}$ ….. (1)

Here, F is the force,${\mathrm{q}}_1$ and${\mathrm{q}}_2$ are the charges, k is the Coulomb’s charge, and is the distance between two charges.

(a) Determination of the charge on each sphere when the two charges are equal.

Consider the given data as below.

The repulsive force,$\mathrm{F}=0.220\mathrm{N}$

The Coulomb’s constant, $\mathrm{k}=9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2$

The distance between charges,$\mathrm{r}=15\mathrm{cm}=0.15\mathrm{m}$

Coulomb’s law of force between two charges is,

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

If q₁=q₂, then,

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

Solve for the charge magnitude,

$$\begin{gathered} \mathrm{q}=\mathrm{r}\sqrt{\frac{\mathrm{F}}{\mathrm{k}}} \\ =0.150\mathrm{m}\sqrt{\frac{0.220\mathrm{N}}{9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2}} \\ =7.42\times {10}^{-7}\mathrm{C} \end{gathered}$$

(b) Determination of the charge on each sphere when one sphere has four times the charge of the other.

Coulomb’s law of force between two charges is,

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

If 4q₁=q₂, then,

$\mathrm{F}=\mathrm{k}\frac{4{\mathrm{q}}^2}{{\mathrm{r}}^2}$

Solve for the charge magnitude,

$$\begin{gathered} \mathrm{q}=\mathrm{r}\sqrt{\frac{\mathrm{F}}{\mathrm{k}4}} \\ =\frac{1}{2}\left(7.42\times {10}^{-7}\mathrm{C}\right) \\ =3.7\times {10}^{-7}\mathrm{C} \end{gathered}$$

Hence, charge on each sphere when one sphere has four times the charge of the other is $3.7\times {10}^{-7}\mathrm{C}$.