Question

A +3.4 x 10-6 C test charge experiences forces from two other nearby charges: a 3 N force due east and a 15 N force due west. What are the magnitude and direction of the electric field st the location of the test charge?

Answer 1

Answer:

3.53×10⁶ N/c due west

Explanation:

From the question

E = F'/q........................ Equation 1

Where E =  Electric Field, F = Net Force, q = Charge.

But,

F' = F₂-F₁...................... Equation 2

Substitute equation 2 into equation 1

E = (F₂-F₁)/q................ Equation 3

Given: F₁ = 3 N due east, F₂ = 15 N due west,  q = 3.4×10⁻⁶ C

Substitute these values into equation 1

E = (15-3)/(3.4×10⁻⁶)

E = 12/(3.4×10⁻⁶)

E = 3.53×10⁶ N/c due west