Question

Find the electrostatic force on a charged particle of $1C$ charge located at a point $\text{P(1, 2, 3)}$ due to another charge $2C$ located at point $\text{Q(2, 3, 4)}$ in the cartesian coordinate system. $(K=\frac{1}{4\pi {ϵ}_0})$

• A. $\frac{-2K}{3\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$
• B. $\frac{-K}{3\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$
• C. $\frac{-K}{3\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$
• D. $\frac{K}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$

Answer 1

The correct option is A $\frac{-2K}{3\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$


From coulomb's law in vectors form, the electrostatic force between two charges is given by,

$\overrightarrow{F}=\frac{K{q}_1{q}_2\overrightarrow{r}}{|\overrightarrow{r}{|}^3}$

Here,

position vector, $\overrightarrow{r_{PQ}}=\overrightarrow{r}=(1-2)\hat{i}+(2-3)\hat{j}+(3-4)\hat{k}$

$\Rightarrow \overrightarrow{r}=-\hat{i}-\hat{j}-\hat{k}$

$|\overrightarrow{r}|=\sqrt{(1{)}^2+(1{)}^2+(1{)}^2}$

$|\overrightarrow{r}|=\sqrt{3}$

Substituting the values in the above equation, we get,

$\overrightarrow{F}=\frac{K\left(2\right)\left(1\right)}{3\sqrt{3}}(-\hat{i}-\hat{j}-\hat{k})$

$\therefore \overrightarrow{F}=\frac{-2K}{3\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$

Hence, option (a) is the correct answer.