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

q₁ = q₂ = Q = 14.8 pC

Step-by-step 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(kQ)/(r^2)`

Rearranging for Q

`Q= (Er^2)/(2k)\\\\Q=((50)(.005329))/(2* 9* 10^9)`

`Q=14.8* 10^(-12)\, C\\\\Q=14.8\, pC`