91Ó°ÊÓ

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.

Consider the given data as below.

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

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

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,

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

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,

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

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