91Ó°ÊÓ

1. Point charges of charges,${\mathrm{q}}_1=+6.0\mathrm{μ°ä}$,${\mathrm{q}}_2=−4.0\mathrm{μ°ä}$and$q_3$are on x-axis at distances,$r_1=8m$,$r_2=16m$and${\mathrm{r}}_3=24\mathrm{m}$.
2. A point to be placed at origin for which net force on it is zero.

Using the concept of Coulomb's law, we can find the magnitude of the required force between the particles. Now, equating this net force to zero, we can get the value of the charge.

Formula:

The magnitude of the electrostatic force between any two particles, $\mathrm{F}=\mathrm{k}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}$ (i)

Let, us consider the charge at origin be $\mathrm{q}$, a positive charge. Now, considering all the charges, we can see that force due to ${\mathrm{q}}_1$ is repulsive towards-x-axis, force due to ${\mathrm{q}}_2$ is attractive towards +x-axis and force due to ${\mathrm{q}}_3$can be either attractive $({\mathrm{q}}_3>0)$or repulsive $({\mathrm{q}}_3<0)$. Considering the distance, we get that the force is towards +x-axis.

Thus, the net force on the charge $\mathrm{q}$can be given using equation (i) and the given data as:

$\begin{array}{c}\stackrel{→}{{\mathrm{F}}_{\mathrm{net}}}=\stackrel{→}{{\mathrm{F}}_1}+\stackrel{→}{{\mathrm{F}}_2}+\stackrel{→}{{\mathrm{F}}_3}\\ 0=−\mathrm{k}\frac{\left|{\mathrm{q}}_1\right|\mathrm{q}}{{\mathrm{r}}_1^2}+\mathrm{k}\frac{\left|{\mathrm{q}}_2\right|\mathrm{q}}{{\mathrm{r}}_2^2}−\mathrm{k}\frac{\left|{\mathrm{q}}_3\right|\mathrm{q}}{{\mathrm{r}}_3^2}\\ 0=−\frac{6×{10}^{−6}}{8^2}+\frac{4×{10}^{−6}}{{16}^2}+\frac{{\mathrm{q}}_3}{{24}^2}\\ {\mathrm{q}}_3=–45\mathrm{μ°ä}.\end{array}$

Hence, the value of the charge of the third particle is $–45\mathrm{μ}C.$