Question

12) Two-point charges each have a value of 3.0 C and 5.0 C and are separated by a distance of 6.0 m. What is the electric field at a point midway between the two charges?

Answer 1

The electric field at a point midway between the two charges is 2 x 10⁹ N/C.

Step-by-step 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;

`E_1 = (kQ_1)/(r^2) (-\bar x)\\\\E_1 = (9* 10^9 * 3)/(3^2)(-\bar x)`

`E_2 = (kQ_1)/(r^2) (+\bar x)\\\\E_2 = (9* 10^9 * 5)/(3^2) (+\bar x)\\\\`

The net electric field is calculated as;

`E_(net) = (9* 10^9 * 5)/(3^2) \ - \ (9* 10^9 * 3)/(3^2)\\\\E_(net) = (9* 10^9 )/(3^2) (5-3)\\\\E_(net) = 2 * 10^9 \ N/C`

Therefore, the electric field at a point midway between the two charges is 2 x 10⁹ N/C.