Question

Two point charges of equal magnitude are

7.5 cm apart. At the midpoint of the line
connecting them, their combined electric field
has magnitude of 45 N/C. Find the magnitude
of the charges.
Pls show me the complete calculation,

Answer 1

We know that the electric field generated by to charges at a certain point between them is given by:
E = 1/(4πε₀) · (q₂ / d₂² - q₁ / d₁²)
Please note that distances must be in [m] in order to have the correct units of measurements.

In our case, we are at the midpoint between the two charges, therefore d₁ = d₂ = d and:
E = 1/(4πε₀d²) · (q₂ - q₁)

In order for the term (q₂ - q₁) to be ≠0, since the two charges have equal magnitude, they must be of opposite sign. For convenience, we decide
q₂ = q
q₁ = -q

Therefore we get:
E = 1/(4πε₀d²) · (q - (-q)) = 1/(4πε₀d²) · (2q) = q / (2πε₀d²)

We now solve for charge q:
q = 2π · ε₀ · d² · E
= 2π · (8.854×10⁻¹²) · (3.75×10⁻²)² · (45)
= 3.52×10⁻¹²C

Therefore:
q₂ = +3.52×10⁻¹²C
q₁ = -3.52×10⁻¹²C