A doorframe is twice as tall as it is wide. There is a positive charge on the top left corner and an equal but negative charge in the top right corner. What is the direction of the electric force due to these charges on a negatively charged dust mite in the bottom left corner of the doorframe?

α = 141.5°  (counterclockwise)

Explanation:

If

q₁ = +q

q₂ = -q

q₃ < 0

b = 2*a

We apply Coulomb's Law as follows

F₁₃ = K*q₁*q₃ / d₁₃² = + K*q*q₃ / (2*a)² = + K*q*q₃ / (4*a²)

F₂₃ = K*q₂*q₃ / d₂₃² = - K*q*q₃ / (5*a²)

(d₂₃² = a² + (2a)² = 5*a²)

Then

∅ = tan⁻¹(2a/a) = tan⁻¹(2) = 63.435°

we apply

F₃x = - F₂₃*Cos ∅ = - (K*q*q₃ / (5*a²))* Cos 63.435°

⇒  F₃x = - 0.0894*K*q*q₃ / a²

F₃y = - F₂₃*Sin ∅ + F₁₃

⇒  F₃y = - (K*q*q₃ / (5*a²))* Sin 63.435° + (K*q*q₃ / (4*a²))

⇒  F₃y = 0.0711*K*q*q₃ / a²

Now, we use the formula

α = tan⁻¹(F₃y / F₃x)

⇒  α = tan⁻¹((0.0711*K*q*q₃ / a²) / (- 0.0894*K*q*q₃ / a²)) = - 38.5°

The real angle is

α = 180° - 38.5° = 141.5°  (counterclockwise)

wavelength divides in half and speed stays the same.