Two point charges q_1=3μ C and q_2=-4μ C are placed at points (2 i+3 j+3 k) and ( i+ j+ k) respectively. Force on charge q_2 is N. ( Take 1 4πε_0 =…

Position vector of $q_1$q1​, $\vec{r}_1 = 2\hat{i} + 3\hat{j} + 3\hat{k}$r1​=2i^+3j^​+3k^

Position vector of $q_2$q2​, $\vec{r}_2 = \hat{i} + \hat{j} + \hat{k}$r2​=i^+j^​+k^

Vector from $q_1$q1​ to $q_2$q2​ is $\vec{r}_{21} = \vec{r}_2 - \vec{r}_1 = (\hat{i} + \hat{j} + \hat{k}) - (2\hat{i} + 3\hat{j} + 3\hat{k}) = -\hat{i} - 2\hat{j} - 2\hat{k}$r21​=r2​−r1​=(i^+j^​+k^)−(2i^+3j^​+3k^)=−i^−2j^​−2k^

Magnitude $|\vec{r}_{21}| = \sqrt{(-1)^2 + (-2)^2 + (-2)^2} = \sqrt{9} = 3$∣r21​∣=(−1)2+(−2)2+(−2)2​=9​=3

Force on $q_2$q2​ due to $q_1$q1​ is given by Coulomb's law in vector form:

$$\vec{F}_{21} = \dfrac{1}{4\pi\epsilon_0} \dfrac{q_1 q_2}{|\vec{r}_{21}|^3} \vec{r}_{21}$$F21​=4πϵ0​1​∣r21​∣3q1​q2​​r21​

Substituting the given values:

$$\vec{F}_{21} = \dfrac{9 \times 10^9 \times (3 \times 10^{-6}) \times (-4 \times 10^{-6})}{3^3} (-\hat{i} - 2\hat{j} - 2\hat{k})$$F21​=339×109×(3×10−6)×(−4×10−6)​(−i^−2j^​−2k^)

$$\vec{F}_{21} = \dfrac{-108 \times 10^{-3}}{27} (-\hat{i} - 2\hat{j} - 2\hat{k})$$F21​=27−108×10−3​(−i^−2j^​−2k^)

$$\vec{F}_{21} = -4 \times 10^{-3} (-\hat{i} - 2\hat{j} - 2\hat{k})$$F21​=−4×10−3(−i^−2j^​−2k^)

$$\vec{F}_{21} = (4\hat{i} + 8\hat{j} + 8\hat{k}) \times 10^{-3} \text{ N}$$F21​=(4i^+8j^​+8k^)×10−3 N

Answer: $(4\hat{i}+8\hat{j}+8\hat{k})\times 10^{-3}$(4i^+8j^​+8k^)×10−3