Two point charges $q_1 = 3\,\mu C$ and $q_2 = -4\,\mu C$ are placed at points $(2\hat{i} + 3\hat{j} + 3\hat{k})$ and $(\hat{i} + \hat{j} + \hat{k})$ respectively. Force on charge $q_2$ is ______ N. (Take $\dfrac{1}{4\pi\epsilon_0} = 9 \times 10^9$ SI Units)
Answer: (B) $(4\hat{i} + 8\hat{j} + 8\hat{k}) \times 10^{-3}$
Vector from $q_1$ to $q_2$: $\vec{r} = (\hat{i} + \hat{j} + \hat{k}) - (2\hat{i} + 3\hat{j} + 3\hat{k}) = -\hat{i} - 2\hat{j} - 2\hat{k}$, with $r = 3$ m.
$$\vec{F}_2 = \frac{kq_1q_2}{r^3}\vec{r} = \frac{9 \times 10^9 \times 3 \times 10^{-6} \times (-4 \times 10^{-6})}{27}(-\hat{i} - 2\hat{j} - 2\hat{k})$$
$= -4 \times 10^{-3}(-\hat{i} - 2\hat{j} - 2\hat{k}) = (4\hat{i} + 8\hat{j} + 8\hat{k}) \times 10^{-3}$ N.
(The force on $q_2$ points towards $q_1$, as expected for attraction.)
Solution by Sreeraj P, M.Sc Physics