In order to solve this problem, we will first need to find the electric field at the origin without the 3rd charge
E1 = (9x10^9)(13.4x10^-9)/(9.4x10^-2)^2 = 13648.7 V/m towards the negative y-axis
E2 = (9x10^9)(4.23x10^-9)/(4.99x10^-2)^2 = 15289.1 V/m towards the positive x-axis
The red arrow shows the direction of which the electric field points.
To make the electric field at the origin 0, we must find a location where q3 = the magnitude of q1 and q2
Etotal = sqrt(E1+E2) = 20494.97 V/m
E3 = 20494.97 = (9x10^9)(14.23x10^-9)/(d)^2
d = 0.079 m = 7.9 cm
The magnitude of the source charge is 3 μC which generates 4286 N/C of the electric field. Option B is correct.
What does Gauss Law state?
It states that the electric flux across any closed surface is directly proportional to the net electric charge enclosed by the surface.
Where,
= electric force = 4286 N/C
= Coulomb constant =
= charges = ?
= distance of separation = 2.5 m
Put the values in the formula,
Therefore, the magnitude of the source charge is 3 μC.
Learn more about Gauss's law:
brainly.com/question/1249602