Question

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

• A. $\frac{q_1{q}_2}{56\pi {ϵ}_0}(2\hat{i}-\hat{j}+3\hat{k})$
• B. $\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{(\sqrt{14}{)}^3}(-2\hat{i}+\hat{j}-3\hat{k})$
• C. $\frac{q_1{q}_2}{56\pi {ϵ}_0}(\hat{j}-2\hat{i}-3\hat{k})$
• D. $\frac{q_1{q}_2}{56\sqrt{14}\pi {ϵ}_0}(2\hat{i}-\hat{j}+3\hat{k})$

Answer 1

Answer: (D)

Position vector, $\overrightarrow{d}$
$<br/>\begin{array}{l}<br/>\vec{d}=(2-0)\hat{i}-(1-0)\hat{j}+(3-0)\hat{k}=2\hat{i}-\hat{j}+3\hat{k}\\ <br/>d^3=[(2\hat{1}-\hat{j}+3\hat{k})\cdot (2\hat{ı}-\hat{j}+3\hat{k}){]}^{\frac{3}{2}}\\ <br/>d^3=14\sqrt{14}<br/>\end{array}<br/>$
Electrostatic force F
$<br/>\begin{array}{l}<br/>F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{d^3}\overrightarrow{d}=\frac{q_1{q}_2}{4\pi {\epsilon }_0}\frac{(2\hat{i}-\hat{j}+3\hat{k})}{14\sqrt{14}}\\ <br/>F=\frac{q_1{q}_2}{56\sqrt{14}\pi {\epsilon }_0}(2\hat{i}-\hat{j}+3\hat{k})<br/>\end{array}<br/>$