Question: In Fig. 21-40, four particles are fixed along anxaxis, separated bydistances -2.00. The charges are$q_1=+2e,q_2=-e,q_3=+e,and{q}_4=+4e$, with$e=1.60\times {10}^{-19}C$. In unit-vector notation, what is the net electrostatic force on (a) particle 1 and (b) particle 2 due to the other particles?

• a)Four charged particles are on x-axis, separated by distance d=2.00cm.
• b) The values of the charges,$q_1=+2e,q_2=-e,q_3=+eand{q}_4=+4e.$

Using the concept of Coulomb's law, we can find the value of the net electrostatic force in the unit-vector notation for particle 1 and particle 2.

Formula:The electrostatic force on a particle 1 due to particle 2 in unit-vector notationis given by

$F=k\frac{q_1{q}_2}{{\left|\overline{r}\right|}^3}\hat{r}$

Considering the net force on $q_1$, we have the force $\overrightarrow{F_{12}}$ pointing in the +x direction since $q_1$is attracted to $q_2$and the forces, and $\overrightarrow{F_{13}}$ and $\overrightarrow{F_{14}}$both point in the –x direction since $q_1$ is repelled by $q_3$ and $q_4$. As all the particles are on x-axis here, the magnitude $\left|\overrightarrow{r}\right|=(\hat{r}forx-c0mponent)=d$.

Thus, the net force on the particle 1 is given by

role="math" localid="1663156273051" $$\begin{gathered} \overrightarrow{F_1}=\overrightarrow{F_{12}}+\overrightarrow{F_{13}}+\overrightarrow{F_{14}} \\ =\frac{2e\left|-e\right|}{4\pi {\epsilon }_0{d}^2}-\frac{\left(2e\right)\left(e\right)}{4\pi {\epsilon }_0{\left(2d\right)}^2}-\frac{\left(2e\right)\left(4e\right)}{4\pi {\epsilon }_0{\left(3d\right)}^2} \\ =\frac{11}{18}\frac{e^2}{4\pi {\epsilon }_0{d}^2} \\ =\frac{11}{18}\frac{\left(9.00\times {10}^{9}N{m}^2/C^2\right){\left(1.60\times {10}^{-19}C\right)}^2}{{\left(2.00\times {10}^{-2}C\right)}^2} \\ =3.52\times {10}^{-25}N \end{gathered}$$

Hence, the value of the force is$3.52\times {10}^{-25}N$.

Considering the net force on$q_2$, we have the force $\overrightarrow{F_{21}}=-\overrightarrow{F_{12}}$pointing in the

–x direction, and both forces $\overrightarrow{F_{23}}and\overrightarrow{F_{24}}$point in the +x direction. As all the particles are on the x-axis here, the magnitude $\left|\overrightarrow{r}\right|=(\hat{r}forx-component)=d$, .

Thus, the net force on the particle 2 is givenby

$$\begin{gathered} \overrightarrow{F_2}=\overrightarrow{F_{24}}+\overrightarrow{F_{23}}-\overrightarrow{F_{12}} \\ =\frac{4e\left|-e\right|}{4\pi {\epsilon }_0{\left(2d\right)}^2}+\frac{e\left|-e\right|}{4\pi {\epsilon }_0{d}^2}-\frac{2e\left|-e\right|}{4\pi {\epsilon }_0{d}^2} \\ =0 \end{gathered}$$

Hence, the net electrostatic force on particle 2 is 0 .