# Electric field strength between 2 positive charges [closed]

I'm having a lot of trouble trying to solve this problem:

Three identical point charges of magnitude $+1.0$ $mC$ are located at the corners of a square that is $1.0$ $m$ on a side. A point charge of magnitude $-1.5$ $mC$ is placed at the remaining corner. What is the electric field strength at a point halfway between 2 adjacent positive charges?

My thinking was that you can just find the strength of one charge to the center and just double it because they're identical charges. Why is this method wrong and how do I solve this problem?

The correct answer is $8.63\times 10^6$ $N/C$, but I have no idea how to get this.

## closed as too localized by dmckee♦Mar 5 '13 at 15:56

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

The electric field is a vector. You just added the magnitudes of the two vector, without taking the direction into account.

• Then wouldn't the answer be 0 because the two charges are pushing each other apart in opposite directions? – Andrew Mar 5 '13 at 6:46
• For two charges, yes. But you have four charges. – Bernhard Mar 5 '13 at 6:51
• oh so the top two charges cancel each other out, but I have to add the bottom two charges using vectors. – Andrew Mar 5 '13 at 6:53