Superposition of electric fields

Question: Charges q and -q are at a distance b from each other, as shown in the figure. What is the electric field at the point P, which lies at a third corner of the square?

Solution: The field at P is the vector sum of the fields that would have been created by the two charges independently. Let positive x be to the right and let positive y be up.

Negative charges have fields that point at them, so the charge -q makes a field that points to the right, i.e. has a positive x component. Using the answer to the self-check, we have

E _-q,x = kq/b²

E _-q,y = 0 .

Note that if we had blindly ignored the absolute value signs and plugged in -q to the equation, we would have incorrectly concluded that the field went to the left.

By the Pythagorean theorem, the positive charge is at a distance √2b from P, so the magnitude of its contribution to the field is kq/(√2b) ² = kq/2b². Positive charges have fields that point away from them, so the field vector is at an angle of 135° counterclockwise from the x axis.

The total field is