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

Solution :

Position vectors of charges :

$$\vec r_1 = 2\hat i + 3\hat j + 3\hat k$$

$$\vec r_2 = \hat i + \hat j + \hat k$$

Vector from $$q_1$$ to $$q_2$$ :

$$\vec r = \vec r_2 - \vec r_1$$

$$= (\hat i + \hat j + \hat k) - (2\hat i + 3\hat j + 3\hat k)$$

$$= -\hat i -2\hat j -2\hat k$$

Magnitude of separation :

$$r = \sqrt{(-1)^2+(-2)^2+(-2)^2}$$

$$= \sqrt{1+4+4}$$

$$= 3$$

Therefore,

$$r^2 = 9$$

Magnitude of electrostatic force :

$$F = \frac{1}{4\pi\epsilon_0}\frac{|q_1q_2|}{r^2}$$

Given :

$$q_1 = 3 \times 10^{-6}\text{ C}$$

$$q_2 = -4 \times 10^{-6}\text{ C}$$

Substituting values :

$$F = 9 \times 10^9 \times \frac{(3 \times 10^{-6})(4 \times 10^{-6})}{9}$$

$$= 9 \times 10^9 \times \frac{12 \times 10^{-12}}{9}$$

$$= 12 \times 10^{-3}$$

$$= 1.2 \times 10^{-2}\text{ N}$$

Since charges are unlike, force on $$q_2$$ is towards $$q_1$$.

Unit vector along direction :

$$\hat r = \frac{\hat i +2\hat j +2\hat k}{3}$$

Therefore,

$$\vec F = 1.2 \times 10^{-2}\left(\frac{\hat i +2\hat j +2\hat k}{3}\right)$$

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

Final Answer :

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