Answer:
The electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C
Step-by-step explanation:
Let the midpoint of the two charges be considered as the origin, and charge A = 30.0 * 10⁻⁹ C be moving in the +x- axis and the charge B = 60.0 * 10⁻⁹ C be moving in the -x-axis.
Electric field, E = kQ/r² where k is a constant = 9.0 * 10⁹ N.m²/C², Q = quantity of charge, r = distance of separation
In the given question,r = 30.0 cm = 0.03 m; the midway point between A and B = 0.03/2 = 0.015 m
Electric field due to charge A
Ea = +(9.0 * 10⁹ N.m²/C² * 30.0 * 10⁻⁹ ) / ( 0.015 m)²
Ea = +1.8 * 10⁴ N/C
Electric field due to charge B
Eb = -(9.0 * 10⁹ N.m²/C² * 60.0 * 10⁻⁹ ) / ( 0.015 m)²
Eb = -3.6 * 10⁴ N/C
The resultant electric field E = Ea + Eb
E = (+1.8 * 10⁴ + -3.6 * 10⁴) N/C
E = -1.8 * 10⁴ N/C
Therefore, the electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C