Two point charges q_1 = 3 mu text{C} and q_2 | vedclass

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(B) The force on charge $q_2$ due to $q_1$ is given by Coulomb's Law in vector form: $\vec{F} = k \frac{q_1 q_2}{r^3} \vec{r}_{21}$, where $\vec{r}_{2

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

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(B) The force on charge $q_2$ due to $q_1$ is given by Coulomb's Law in vector form: $\vec{F} = k \frac{q_1 q_2}{r^3} \vec{r}_{21}$, where $\vec{r}_{21} = \vec{r}_2 - \vec{r}_1$.Position vectors are $\vec{r}_1 = 2\hat{i} + 3\hat{j} + 3\hat{k}$ and $\vec{r}_2 = \hat{i} + \hat{j} + \hat{k}$.$\vec{r}_{21} = (1-2)\hat{i} + (1-3)\hat{j} + (1-3)\hat{k} = -\hat{i} - 2\hat{j} - 2\hat{k}$.The magnitude $r = |\vec{r}_{21}| = \sqrt{(-1)^2 + (-2)^2 + (-2)^2} = \sqrt{1+4+4} = \sqrt{9} = 3$.Substituting the values: $\vec{F}_2 = \frac{(9 	imes 10^9)(3 	imes 10^{-6})(-4 	imes 10^{-6})}{3^3} (-\hat{i} - 2\hat{j} - 2\hat{k})$.$\vec{F}_2 = \frac{-108 	imes 10^{-3}}{27} (-\hat{i} - 2\hat{j} - 2\hat{k}) = -4 	imes 10^{-3} (-\hat{i} - 2\hat{j} - 2\hat{k}) = (4\hat{i} + 8\hat{j} + 8\hat{k}) 	imes 10^{-3} 	ext{ N}$.

Two point charges $q_1 = 3 \mu \text{C}$ and $q_2 = -4 \mu \text{C}$ 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 $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ SI Units}$)

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