Question

Find the electric field a distance z above the midpoint between equal and opposite charges,a distance d apart

Answer 1

Answer:

Ex = $2 k q \frac{x}{(x^2 + z^2) ^{3/2}}$ ,  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 = $k \frac{q}{x^2 + z^2}$

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 = $2 k \frac{q}{x^2 + z^2} \ \frac{x}{\sqrt{x^2 + z^2} }$

     Ex = $2 k q \frac{x}{(x^2 + z^2) ^{3/2}}$

     E_z = 0