Answer:
The electric field at a point midway between the two charges is 2 x 10⁹ N/C.
Explanation:
Given;
first point charge, Q₁ = 3.0 C
second point charge, Q₂ = 5.0 C
distance between the two point charges, R = 6.0 m
The mid-point between the two charges, r = R/2 = 3.0 m
3.0 C <-------------------->3.0m<----------------------->5.0 C
The forces are acting in opposite direction, the electric field strength of each charge is calculated as;
[tex]E_1 = \frac{kQ_1}{r^2} (-\bar x)\\\\E_1 = \frac{9\times 10^9 \times 3}{3^2}(-\bar x)[/tex]
[tex]E_2 = \frac{kQ_1}{r^2} (+\bar x)\\\\E_2 = \frac{9\times 10^9 \times 5}{3^2} (+\bar x)\\\\[/tex]
The net electric field is calculated as;
[tex]E_{net} = \frac{9\times 10^9 \times 5}{3^2} \ - \ \frac{9\times 10^9 \times 3}{3^2}\\\\E_{net} = \frac{9\times 10^9 }{3^2} (5-3)\\\\E_{net} = 2 \times 10^9 \ N/C[/tex]
Therefore, the electric field at a point midway between the two charges is 2 x 10⁹ N/C.
