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) $(4\hat{i} + 8\hat{j} + 8\hat{k}) \times 10^{-3}$
Concept. Vector Coulomb's law: the force on $q_2$ is $\vec F_2=\dfrac{kq_1q_2}{r^2}\hat r$, where $\hat r$ points from $q_1$ to $q_2$. The signs of the charges take care of attraction.
Step 1. Separation. $\vec r=\vec r_2-\vec r_1=(1-2)\hat i+(1-3)\hat j+(1-3)\hat k=-\hat i-2\hat j-2\hat k$, so $r=3$ m and $\hat r=\dfrac{-\hat i-2\hat j-2\hat k}{3}$.
Step 2. Size factor. $\dfrac{kq_1q_2}{r^2}=\dfrac{9\times10^9\times3\times10^{-6}\times(-4\times10^{-6})}{9}=-1.2\times10^{-2}$ N.
Step 3. Force. $\vec F_2=-1.2\times10^{-2}\times\dfrac{-\hat i-2\hat j-2\hat k}{3}=(4\hat i+8\hat j+8\hat k)\times10^{-3}$ N (towards $q_1$, as unlike charges attract).
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
