Question

Point charge A has a charge of –1.0 nC, and point charge B has a charge of 4.0 nC. They are separated by 1.0 cm. What is the electric field strength along the line connecting the charges at a point 1.0 cm to the right of the positive charge?

---------(-)------------(+)-----------(•)--------

1 cm 1cm

Answer 1

Magnitude of Electric field due to point charge q, at distance r from itself,
= kq/r²
direction of field is away from , if charge is +ve and towards charge if it is -ve.

Electric field due to charge A, at given point, 2 cm on right
E_a = k (10^-9) / 4x10^-4 , Direction of the field is towards left
Electric field due to charge B, at given point, 1 cm on right
E_a = k (4x10^-9) /10^-4 , Direction of the field is towards right.

Net field E = E_b - E_a ( toward right)
                  = k ( 3.75x10^-5)
                  = 9x10⁹ * 3.75x10^-5
                   = 3.4x10⁵ N/C