Q66 / 75 · JEE Main 2026 (08 April Shift 2) · Physics
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. $\left(\text{Take } \dfrac{1}{4\pi\epsilon_0} = 9\times 10^9 \text{ SI Units}\right)$
- A$(12\hat{i}+24\hat{j}+24\hat{k})\times 10^{-3}$
- Correct answerB$(4\hat{i}+8\hat{j}+8\hat{k})\times 10^{-3}$
- C$(3\hat{i}+6\hat{j}+6\hat{k})\times 10^{-3}$
- D$(-4\hat{i}-8\hat{j}-8\hat{k})\times 10^{-3}$
Solution
Position vector of $q_1$, $\vec{r}_1 = 2\hat{i} + 3\hat{j} + 3\hat{k}$
Position vector of $q_2$, $\vec{r}_2 = \hat{i} + \hat{j} + \hat{k}$
Vector from $q_1$ to $q_2$ is $\vec{r}_{21} = \vec{r}_2 - \vec{r}_1 = (\hat{i} + \hat{j} + \hat{k}) - (2\hat{i} + 3\hat{j} + 3\hat{k}) = -\hat{i} - 2\hat{j} - 2\hat{k}$
Magnitude $|\vec{r}_{21}| = \sqrt{(-1)^2 + (-2)^2 + (-2)^2} = \sqrt{9} = 3$
Force on $q_2$ due to $q_1$ is given by Coulomb's law in vector form:
$\vec{F}_{21} = \dfrac{1}{4\pi\epsilon_0} \dfrac{q_1 q_2}{|\vec{r}_{21}|^3} \vec{r}_{21}$
Substituting the given values:
$\vec{F}_{21} = \dfrac{9 \times 10^9 \times (3 \times 10^{-6}) \times (-4 \times 10^{-6})}{3^3} (-\hat{i} - 2\hat{j} - 2\hat{k})$
$\vec{F}_{21} = \dfrac{-108 \times 10^{-3}}{27} (-\hat{i} - 2\hat{j} - 2\hat{k})$
$\vec{F}_{21} = -4 \times 10^{-3} (-\hat{i} - 2\hat{j} - 2\hat{k})$
$\vec{F}_{21} = (4\hat{i} + 8\hat{j} + 8\hat{k}) \times 10^{-3} \text{ N}$
Answer: $(4\hat{i}+8\hat{j}+8\hat{k})\times 10^{-3}$
Position vector of $q_2$, $\vec{r}_2 = \hat{i} + \hat{j} + \hat{k}$
Vector from $q_1$ to $q_2$ is $\vec{r}_{21} = \vec{r}_2 - \vec{r}_1 = (\hat{i} + \hat{j} + \hat{k}) - (2\hat{i} + 3\hat{j} + 3\hat{k}) = -\hat{i} - 2\hat{j} - 2\hat{k}$
Magnitude $|\vec{r}_{21}| = \sqrt{(-1)^2 + (-2)^2 + (-2)^2} = \sqrt{9} = 3$
Force on $q_2$ due to $q_1$ is given by Coulomb's law in vector form:
$\vec{F}_{21} = \dfrac{1}{4\pi\epsilon_0} \dfrac{q_1 q_2}{|\vec{r}_{21}|^3} \vec{r}_{21}$
Substituting the given values:
$\vec{F}_{21} = \dfrac{9 \times 10^9 \times (3 \times 10^{-6}) \times (-4 \times 10^{-6})}{3^3} (-\hat{i} - 2\hat{j} - 2\hat{k})$
$\vec{F}_{21} = \dfrac{-108 \times 10^{-3}}{27} (-\hat{i} - 2\hat{j} - 2\hat{k})$
$\vec{F}_{21} = -4 \times 10^{-3} (-\hat{i} - 2\hat{j} - 2\hat{k})$
$\vec{F}_{21} = (4\hat{i} + 8\hat{j} + 8\hat{k}) \times 10^{-3} \text{ N}$
Answer: $(4\hat{i}+8\hat{j}+8\hat{k})\times 10^{-3}$
Hence, the correct answer is option: 'B'.