Force on Charge Placed at Position Vector

To find the force acting on charge $q_2$ due to charge $q_1$, we use Coulomb's Law in vector form:

$\vec{F}_{21} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{|\vec{r}_2 - \vec{r}_1|^3} (\vec{r}_2 - \vec{r}_1)$

1. Identify the given values:

• Charge $q_1 = 3\text{ }\mu\text{C} = 3 \times 10^{-6}\text{ C}$
• Charge $q_2 = -4\text{ }\mu\text{C} = -4 \times 10^{-6}\text{ C}$
• Position vector of $q_1$: $\vec{r}_1 = 2\hat{i} + 3\hat{j} + 3\hat{k}$
• Position vector of $q_2$: $\vec{r}_2 = \hat{i} + \hat{j} + \hat{k}$
• Electrostatic constant: $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$

2. Calculate the displacement vector $(\vec{r}_2 - \vec{r}_1)$: $\vec{r}_2 - \vec{r}_1 = (\hat{i} + \hat{j} + \hat{k}) - (2\hat{i} + 3\hat{j} + 3\hat{k}) = -\hat{i} - 2\hat{j} - 2\hat{k}$

3. Calculate the distance between the two charges: $|\vec{r}_2 - \vec{r}_1| = \sqrt{(-1)^2 + (-2)^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3\text{ m}$

4. Calculate the force $\vec{F}_{21}$: $\vec{F}_{21} = (9 \times 10^9) \frac{(3 \times 10^{-6}) \times (-4 \times 10^{-6})}{(3)^3} (-\hat{i} - 2\hat{j} - 2\hat{k})$

$\vec{F}_{21} = (9 \times 10^9) \frac{-12 \times 10^{-12}}{27} (-\hat{i} - 2\hat{j} - 2\hat{k})$

$\vec{F}_{21} = \frac{9 \times 10^9}{27} (-12 \times 10^{-12}) (-\hat{i} - 2\hat{j} - 2\hat{k})$

$\vec{F}_{21} = \frac{1}{3} \times 10^9 \times (12 \times 10^{-12}) (\hat{i} + 2\hat{j} + 2\hat{k})$

$\vec{F}_{21} = 4 \times 10^{-3} (\hat{i} + 2\hat{j} + 2\hat{k})\text{ N}$

$\vec{F}_{21} = (4\hat{i} + 8\hat{j} + 8\hat{k}) \times 10^{-3}\text{ N}$

Thus, the force on charge $q_2$ is $(4\hat{i} + 8\hat{j} + 8\hat{k}) \times 10^{-3}\text{ N}$.

Correct Option: B