Answer:
Ex = [tex]2 k q \frac{x}{(x^2 + z^2) ^{3/2}}[/tex] , E_z =0
Explanation:
To find the electric field of each charge, let's use
E = k q / r²
where the field for a positive charge is outgoing and for a negative charge it is incoming.
Since the electric field is a vector magnitude, let's add its components at the point z
X axis
Eₓ = E₁ₓ + E₂ₓ
Eₓ = 2 E₁ₓ
Z axis
E_z = E_{1z} -E_{2z}
E_z= 0
Let's find the expression for the electric field
the distance from the charge to the test point, using the Pythagorean theorem is
r² = (x-0) ² + (0-z) ²
E = [tex]k \frac{q}{x^2 + z^2}[/tex]
with trigonometry we can find its components
cos θ = E₁ₓ / E
E₁ₓ = E cos θ
we substitute
Eₓ = 2 E cos θ
where the cosine
cos θ = x / r
Ex = [tex]2 k \frac{q}{x^2 + z^2} \ \frac{x}{\sqrt{x^2 + z^2} }[/tex]
Ex = [tex]2 k q \frac{x}{(x^2 + z^2) ^{3/2}}[/tex]
E_z = 0
