Question

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.

Answer 1

10250 N/C leftwards

Step-by-step 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

`E_(A)=(KQ_(A))/(AP^(2))=(9*10^(9)*4*10^(-6))/(6^(2))=1000 N/C (rightwards)`

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

`E_(B)=(KQ_(B))/(BP^(2))=(9*10^(9)*5*10^(-6))/(2^(2))=11250 N/C (leftwards)`

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

E = EB - EA

E = 11250 - 1000 = 10250 N/C leftwards