Question

A point charge $q_A=+100\mu \text{C}$ is placed at point $\text{A}(1,0,2)\text{m}$ and another point charge $q_B=+200\mu \text{C}$ is placed at point $\text{B}(4,4,2)\text{m}$. The magnitude of electrostatic force acting between them is:
(Take $k=9\times {10}^9\frac{{\text{N-m}}^2}{{\text{C}}^2}$ for calculation)

• A. $10\text{N}$
• B. $8.5\text{N}$
• C. $7.2\text{N}$
• D. $15\text{N}$

Answer 1

The correct option is C $7.2\text{N}$

Given:
$q_A=+100\mu \text{C}=100\times {10}^{-6}\text{C}$
$q_B=+200\mu \text{C}=200\times {10}^{-6}\text{C}$

Position vector of $\text{B}$ w.r.t. $\text{A}$ is,

${\overrightarrow{r}}_{BA}=(4-1)\hat{i}+(4-0)\hat{j}+(2-2)\hat{k}$

$\Rightarrow {\overrightarrow{r}}_{BA}=3\hat{i}+4\hat{j}+0\hat{k}$

Thus, the distance between charged particles is,

$r=|{\overrightarrow{r}}_{BA}|=\sqrt{3^2+4^2}$

$\Rightarrow r=5\text{m}$

Applying Coulomb's law,

$F_e=\frac{k{q}_1{q}_2}{r^2}$

$\Rightarrow F_e=\frac{9\times {10}^9\times (100\times {10}^{-6})\times (200\times {10}^{-6})}{25}$

$\Rightarrow F_e=\frac{180}{25}=7.2\text{N}$

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

$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: Whenever the coordinate of position of charges is given,}\\ \text{find out the distance (r) between them by taking magnitude of}\\ \text{position vector of one charge w.r.t. to another one.}\end{array}$