Question

Two point charges of equal magnitude are 7.3 cm apart. At the midpoint of the line connecting them, their combined electric field has a magnitude of 50 N/C .

Answer 1

Answer:

q₁ = q₂ = Q = 14.8 pC

Explanation:

Given that

q₁ = q₂ = Q = ?

Distance between charges = r =7.3 cm = 0.073 m

Combined electric field = E₁ + E₂ = E = 50 N/C

Using formula

$E=2\frac{kQ}{r^2}$

Rearranging for Q

$Q= \frac{Er^2}{2k}\\\\Q=\frac{(50)(.005329)}{2\times 9\times 10^9}$

$Q=14.8\times 10^{-12}\, C\\\\Q=14.8\, pC$