Answer to Question #150049 in Physics for red

Question

Three charges q1 = 8 uC, q2 =3 uC , and q3 = -2 uC. Particle 1 from particle 2 has a distance of 0.06 m and particle 2 to particle 3 is separated by a distance of 0.04 m. a) Calculate the resultant force on q2. b) Solve for its direction.

Answer 1

1)

$$F=k\frac{q_1q_2}{d}+k\frac{q_2q_3}{x}\\ F=9\cdot10^9\frac{(8\cdot10^{-6})(3\cdot10^{-6})}{0.06}+9\cdot10^9\frac{(3\cdot10^{-6})(2\cdot10^{-6})}{0.04}\\F=4.95\ N$$

2) Direction: towards to the particle 3.