A positive point charge 50 $μ$ C is located in the plane x-y at the point with radius vector ${\vec{r}}_{0} = 2 \hat{i} + 3 \hat{j}$ where $\hat{i} and \hat{j}$ are the unit vectors of the x and y axes. The magnitude of electrostatic force $\vec{F}$ on a charge of $2 μC$ placed at the point with radius vector $\vec{r} = 8 \hat{i} - 5 \hat{j}$ is (Here ${\vec{r}}_{0} and \vec{r}$ are expressed in meter)

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9 mN

9 $μ$ N

3 mN

answer is B.

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Detailed Solution

$\vec{F} = {\vec{F}}_{21} = \frac{q_{1} q_{2}}{4 {πε}_{0} {|\vec{r} - {\vec{r}}_{0}|}^{3}} (\vec{r} - {\vec{r}}_{0})$
where $\vec{r} - {\vec{r}}_{0} = 6 \hat{i} - 8 \hat{j}$
$\Rightarrow |\vec{r} - {\vec{r}}_{0}| = 10 m$
So, $\vec{F} = \frac{50 \times 10^{- 6} \times 2 \times 10^{- 6} \times 9 \times 10^{9}}{(10)^{3}} (6 \hat{i} - 8 \hat{j})$
$\begin{array}{l} \Rightarrow \vec{F} = 9 \times 10^{- 4} (6 \hat{i} - 8 \hat{j}) N \\ \Rightarrow | \vec{F} | = 9 \times 10^{- 4} \sqrt{6^{2} + (- 8)^{2}} \\ \Rightarrow | \vec{F} | = 9 \times 10^{- 3} N = 9 mN \end{array}$