Answer to Question #249373 in Electricity and Magnetism for aliza

Question

A point charge is placed at each corner of a square with side length a. All charges have

magnitude q. Two of the charges are positive and two are negative. What is the direction of the

net electric field at the center of the square due to the four charges, and calculate its magnitude

in terms of q and a? (C3, CLO-2, PLO-2)

Answer 1

The magnitude of the field of a single charge in the center of the square is

$$E=\frac{kq}{r^2},$$

where r is the distance from the corner of the square to the center, which is

$$r=\frac d2=\frac{a\sqrt2}2.$$

Therefore:

$$E=\frac{2kq}{a^2}.$$

The field components of two charges of equal polarity cancel each other out at the middle, so, only vertical component is left:

$$E_v=\frac{2kq}{a^2}\cos45°=\frac{kq\sqrt2}{a^2}.$$

Two such vertical components (say from the positive charges) give

$$E_p=2E_v=\frac{2kq\sqrt2}{a^2}.$$

Added to the field vectors of negative charges, they give us the total field magnitude at the center of the square of

$$E=|2E_p|=\frac{4kq\sqrt2}{a^2}.$$