Question

Three charges q1=q2=+6 and q3=4 are located at - 0.6i, 0.6i and 0.8j respectively. Find the net force on q3

Answer 1

The net force on the third charge is obtained from the forces between each couple of charges:

$$F_1=kq_1q_3/r^2=kq_1q_3/((-0.6)^2+0.8^2),\\ F_1=2.2·10^{11}.\\ F_2=kq_2q_3/r^2=kq_1q_3/(0.6^2+0.8^2),\\ F_2=2.2·10^{11}.\\$$

The angle between the forces is

$$\theta=180°-2\tan^{-1}\frac{0.6}{0.8}=106°.$$

The magnitude of the resultant is

$$F_R=\sqrt{F_1^2+F_2^2-2F_1F_2\cos\theta}=3.1·10^{10}.$$

It is oriented in +j direction.