91Ó°ÊÓ

Coulomb's law states that the magnitude F of the force exerted by two-point charges ${\mathbf{q}}_{\mathbf{1}}$ and ${\mathbf{q}}_{\mathbf{2}}$ separated by r is directly proportional to the product of the charges $\left(q_1×q_2\right)$ and inversely proportional to the square of the distance between them.

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

Here; F is the force on each point charge exerting on each other, k is the proportionality constant, is the distance between the charges ${\mathrm{q}}_1$ and ${\mathrm{q}}_2$.

Consider the given data as below.

The charge, ${\mathrm{q}}_1=-0.550\mathrm{μ°ä}=-0.550×{10}^{-6}\mathrm{C}$

The force, F = 0.600 N

The distance between two point charges, $\mathrm{r}=0.300\mathrm{m}$

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

Rewrite equation (1) for charge as below.

${\mathrm{q}}_2=\frac{{\mathrm{Fr}}^2}{{\mathrm{kq}}_1}$

Substitute known value of the unknown charge.

$$\begin{gathered} {\mathrm{q}}_2=\frac{\left(0.600\mathrm{N}\right)×{\left(0.300\mathrm{m}\right)}^2}{\left(9×{10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(-0.550×{10}^{-6}\mathrm{C}\right)} \\ =1.091×{10}^{-5}\mathrm{C} \\ =10.91×{10}^{-6}\mathrm{C} \\ =10.91\mathrm{μ°ä} \end{gathered}$$

Hence, the value of the unknown charge is $10.91\mathrm{μ°ä}$.

To calculate the magnitude and direction of the forces exerted by the unknown charge on the negative charge q₁ . The magnitude of the force exerted on the negative charge is,

F = 0.600 N

And because the force is attractive, it is directed downward as shown the figure below.

Hence, F = 0.600 N is the force exerted by the unknown charge on the negative charge, and the direction is downward.