Question

Two particles, each of charge Q, are fixed at opposite corners of a square that lies in the plane of the page. A positive test charge q is placed at a third corner. If F is the magnitude of the force on the q test charge due to only one of the other charges, what is the magnitude of the net force acting on the test charge due to both of these charges?

Answer 1

Answer:

The magnitude of the net force is √2F.

Explanation:

Since the two particles have the same charge Q, they exert the same force on the test charge; both attractive or repulsive. So, the angle between the two forces is 90° in any case. Now, as we know the magnitude of these forces and that they form a 90° angle, we can use the Pythagorean Theorem to calculate the magnitude of the resultant net force:

$F_N=\sqrt{F^{2}+F^{2}}\\\\F_N=\sqrt{2F^{2}}\\\\F_N=\sqrt{2}F$

Then, it means that the net force acting on the test charge has a magnitude of √2F.