Question

A charged particle $q_1$ is at position $\left(2,-1,3\right)$ .The electrostatic force on another charge particle $q_2$ at $\left(0,0,0\right)$ is

• A. $\frac{q_1{q}_2}{56\pi {\left(ϵ\right)}_0}\left(2\hat{i}-\hat{j}+3\hat{k}\right)$
• B. $\frac{q_1{q}_2}{56\sqrt{14}{\left(ϵ\right)}_0}\left(2\hat{i}-\hat{j}-3\hat{k}\right)$
• C. $\frac{q_1{q}_2}{56\pi {\left(ϵ\right)}_0}\left(\hat{i}-2\hat{j}-3\hat{k}\right)$
• D. $\frac{q_1{q}_2}{56\sqrt{14}\pi {\left(ϵ\right)}_0}\left(2\hat{i}-\hat{j}+3\hat{k}\right)$

Answer 1

Answer: (D)

$\mathrm{A}\left(q_1\right)\to (2,-1,3)$
$\mathrm{B}\left(q_2\right)\to (0,0,0)$
${\vec{F}}_{\mathrm{BA}}=\frac{k{q}_{\mathrm{A}}{q}_{\mathrm{B}}}{\mid \overrightarrow{r_{\mathrm{B}}}-\overrightarrow{r_{\mathrm{A}}}}\left(\overrightarrow{r_{\mathrm{B}}}-\overrightarrow{r_{\mathrm{A}}}\right)$
$r_{\mathrm{A}}=2\hat{\mathrm{i}}-\hat{\mathrm{j}}+3\hat{\mathrm{k}};{\vec{r}}_{\mathrm{B}}=0$
Putting all the values in the formula,
We get ${\vec{F}}_{\mathrm{BA}}=\frac{q_1{q}_2}{56\sqrt{14}\pi {\epsilon }_0}(-2\hat{\mathrm{i}}+\hat{\mathrm{j}}-3\hat{\mathrm{k}})$