Question

Find out the electric force experienced by $+5\mu \text{C}$ charge at point $\text{A}(0,5,0)\text{m}$ due to a point charge $-25\mu \text{C}$ situated at point $\text{B}(2,8,\sqrt{12})\text{m}$.

• A. $-(18\hat{i}+27\hat{j}+18\sqrt{3}\hat{k})\times {10}^{-3}\text{N}$
• B. $-(2\hat{i}+3\hat{j}+\sqrt{12}\hat{k})\times {10}^{-3}\text{N}$
• C. $-(36\hat{i}+54\hat{j}+18\sqrt{3}\hat{k})\times {10}^{-3}\text{N}$
• D. $-(18\hat{i}+54\hat{j}+18\sqrt{3}\hat{k})\times {10}^{-3}\text{N}$

Answer 1

The correct option is A $-(18\hat{i}+27\hat{j}+18\sqrt{3}\hat{k})\times {10}^{-3}\text{N}$

Given:
$q_1=+5\mu \text{C};q_2=-25\mu \text{C}$

Force on charge at point $\text{A}$ due to charge at point $\text{B}$ is given by,

${\overrightarrow{F}}_{AB}=\frac{k{q}_1{q}_2}{|{\overrightarrow{r}}_{BA}{|}^3}{\overrightarrow{r}}_{BA}$ $(\because q_1{q}_2<0)$

Here, ${\overrightarrow{r}}_{BA}$ is the position vector of $\text{B}$ w.r.t $\text{A}$

$\Rightarrow {\overrightarrow{r}}_{BA}={\overrightarrow{r}}_B-{\overrightarrow{r}}_A$

$\Rightarrow {\overrightarrow{r}}_{BA}=(2\hat{i}+8\hat{j}+\sqrt{12}\hat{k})-\left(5\hat{j}\right)$

$\Rightarrow {\overrightarrow{r}}_{BA}=2\hat{i}+3\hat{j}+\sqrt{12}\hat{k}$

$\Rightarrow |{\overrightarrow{r}}_{BA}|=\sqrt{4+9+12}=5$

Hence, ${\overrightarrow{F}}_{AB}=\frac{(9\times {10}^9)(5\times {10}^{-6})(-25\times {10}^{-6})}{(5{)}^3}\times (2\hat{i}+3\hat{j}+\sqrt{12}\hat{k})$

$\therefore {\overrightarrow{F}}_{AB}=-(18\hat{i}+27\hat{j}+18\sqrt{3}\hat{k})\times {10}^{-3}\text{N}$

Hence, option (a) is the correct answer.

$\begin{array}{c}\text{Tip: In vector form, force on charge}{}^{\prime }{A}^{\prime }\text{due to}{}^{\prime }{B}^{\prime }\\ \text{can be given as}\\ {\overrightarrow{F}}_{AB}=\frac{k{q}_1{q}_2\overrightarrow{r}}{|\overrightarrow{r}{|}^3}=\frac{k{q}_1{q}_2}{|\overrightarrow{r}{|}^2}\hat{r}\\ \overrightarrow{r}\text{is position vector of A}\text{w.r.t B}\end{array}$