Wo fixed charges, a and b are located at x axis. a is at x = 0 m, b is at x = 4 m. qa = +4.0 μc and qb = -5.0 μc. calculate the electric field at point x = 6 m.

10250 N/C leftwards

Explanation:

QA = 4 micro Coulomb

QB = - 5 micro Coulomb

AP = 6 m

BP = 2 m

A is origin, B is at 4 m and P is at 6 m .

The electric field due to charge QA at P is EA rightwards

The electric field due to charge QB at P is EB leftwards

The resultant electric field at P due the charges is given by

E = EB - EA

E = 11250 - 1000 = 10250 N/C leftwards

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