Question

Find out the electric force experienced by $+1\mu \text{C}$ charge at point $\text{A}(0,1,2)\text{m}$ due to a point charge $-20\mu \text{C}$ situated at point $\text{B}(\sqrt{2},0,1)\text{m}$.

• A. $-22.5\times {10}^3(-\sqrt{2}\hat{i}+\hat{j}+\hat{k})\text{N}$
• B. $22.5\times {10}^{-3}(-\sqrt{2}\hat{i}+\hat{j}+\hat{k})\text{N}$
• C. $44.5\times {10}^{-3}(\hat{i}+\sqrt{2}\hat{j}-\hat{k})\text{N}$
• D. $25\times {10}^{-3}(\hat{i}-\hat{j}+\hat{k})\text{N}$

Answer 1

The correct option is B $22.5\times {10}^{-3}(-\sqrt{2}\hat{i}+\hat{j}+\hat{k})\text{N}$

Given:
$q_1=+1\mu \text{C};q_2=-20\mu \text{C}$

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

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

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

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

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

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

$\Rightarrow |{\overrightarrow{r}}_{BA}|=\sqrt{2+1+1}=2$

Hence, ${\overrightarrow{F}}_{AB}=\frac{(9\times {10}^9)(1\times {10}^{-6})(-20\times {10}^{-6})}{(2{)}^3}\times (\sqrt{2}\hat{i}-\hat{j}-\hat{k})$

$\therefore {\overrightarrow{F}}_{AB}=22.5\times {10}^{-3}(-\sqrt{2}\hat{i}+\hat{j}+\hat{k})\text{N}$

Hence, option $\left(b\right)$ is the correct answer.

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