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?

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:

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

A symbolic expression for the net force on a third point charge +Q located along the y axis  

Explanation:

Let the force on +Q charge y-axis due to +2Q charge be and force on +Q charge y axis due to -Q charge on x-axis be .

Distance between the +2Q charge and +Q charge = d units

Distance between the -Q charge and +Q charge = units

= Coulomb constant

Net force on +Q charge on y-axis is:

The net froce on the +Q charge on y-axis is