Question

A point charge Q is placed on the x axis at x = 2.0 m . A second point charge , -Q , is placed at x = 3.0 m . If Q = 40 uC , what is the magnitude of the electrostatic force on a 30 - uC charge placed at the origin ?

Answer 1

Answer:

Electrostatic force on a point charge due to another,

Fe = k q1 q2 / d^2

due to x = 2m charge,

same sign charge so replusive force.

F1 = - (9 x 10^9) (40 x 10^-6) (30 x 10^-6) / (2^2)

F1 = - 2.7 N

due to x = 3m charge,

F2 = + (9 x 10^9) (40 x 10^-6) 30 x 10^-6 / 3^2

F2 = 1.2 N

Fnet = F1 + F2 = - 1.5 N

magnitude = 1.5 N (toward -ve x axis )

Explanation: