JEE Main 2026 Apr 8 Shift 2, Physics Q16: Coulomb's Law And Electric Field

Displacement from $q_1$ to $q_2$: $\vec{r}={\vec{r}}_2-{\vec{r}}_1=-\hat{i}-2\hat{j}-2\hat{k}$, with $|\vec{r}|=\sqrt{1+4+4}=3$.

Coulomb force on $q_2$ due to $q_1$: $\vec{F}=\frac{k{q}_1{q}_2}{|\vec{r}{|}^3}\vec{r}$.

Since $q_1{q}_2=(3)(-4)\times 10^{-12}=-12\times 10^{-12}$ C${}^2$, the force is opposite to $\vec{r}$ (attractive, pulling $q_2$ toward $q_1$).

$\vec{F}=\frac{(9\times 10^9)(-12\times 10^{-12})}{27}(-\hat{i}-2\hat{j}-2\hat{k})$

$=\frac{-108\times 10^{-3}}{27}(-\hat{i}-2\hat{j}-2\hat{k})=-4\times 10^{-3}(-\hat{i}-2\hat{j}-2\hat{k})$

$=(4\hat{i}+8\hat{j}+8\hat{k})\times 10^{-3}$ N.