What is the electric field at a point midway between a -8.50μC and a +6.38μC charge 7.79 cm apart? Take the direction towards the positive charge to be positive. Assume no other charges are nearby.
E(net) = E(1) + E(2)
E = (k*q)/r^2
The Attempt at a Solution
E(1) = (8.99 x10^-6 * 6.38x10^-6)/ (0.03895)^2
E(2) = (8.99x10^-6 * -8.5x10^-6)/ (0.03895)^2
E(net) = 3.78x10^7 - 5.037x10^7
This is not one of the answer choices, and I am not sure what I'm doing wrong. I thought I understood how to do the problem, so it might just be a complete misunderstanding of what the question is asking? I'm not sure...