91Ó°ÊÓ

Coulomb's law states that the magnitude F of the force exerted by two-point charges q₁ and q₂ separated by 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, r is the distance between the charges q₁ and q₂.

Consider the given data as below.

The charge,${\mathrm{q}}_2=-3.00\mathrm{nC}=-3.00×{10}^{-9}\mathrm{C}$

The charge, ${\mathrm{q}}_3=+5.00\mathrm{nC}=+5.00×{10}^{-9}\mathrm{C}$

The force, F = 0.600 N

The distance between two point charges,

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

Draw the free body diagram as below.

The charge exerts a force on charge q₃, the net force acting on q₃ is zero.

As q₃ is affected by two forces F₁₃ and F₂₃, thus, the net force is;

${\mathrm{F}}_{\mathrm{net}}={\mathrm{F}}_{13}+{\mathrm{F}}_{23}$

Thus, the charge q₁ towards the positive x-direction, which is a repulsive force between q₃ and q₁, so the net force is;

$$\begin{gathered} {\mathrm{F}}_{13}={\mathrm{F}}_{23} \\ \mathrm{k}\frac{\left|{\mathrm{q}}_1{\mathrm{q}}_3\right|}{{\mathrm{r}}_1^2}=\mathrm{k}\frac{\left|{\mathrm{q}}_2{\mathrm{q}}_3\right|}{{\mathrm{r}}_2^2} \\ \frac{\left|{\mathrm{q}}_1\right|}{{\mathrm{r}}_1^2}=\frac{\left|{\mathrm{q}}_2\right|}{{\mathrm{r}}_2^2} \\ \begin{array}{rl}\left|{\mathrm{q}}_1\right|& =\left|{\mathrm{q}}_2\right|\left(\frac{{\mathrm{r}}_1^2}{{\mathrm{r}}_2^2}\right)\end{array} \end{gathered}$$

Substitute known values ion the above equation, and you have

$$\begin{gathered} \begin{array}{rl}\left|q_1\right|& =âˆ\pounds −3.00\mathrm{nCl}{\left(\frac{2.00\mathrm{cm}}{4.00\mathrm{cm}}\right)}^2\end{array} \\ =\frac{+3.00\mathrm{nC}}{4.00} \\ =+0.75\mathrm{nC} \end{gathered}$$

Hence, the magnitude and sign of q₁ is $+0.75\mathrm{nC}$.