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. \(\left(\text{Take } \dfrac{1}{4\pi\epsilon_0} = 9\times 10^9 \text{ SI Units}\right)\)

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: \((4\hat{i}+8\hat{j}+8\hat{k})\times 10^{-3}\)