Question

Consider a square which is 1.0 m on a side. Charges are placed at the corners of the square as follows: +4.0 μC at (0, 0); +4.0 μC at (1, 1); +3.0 μC at (1, 0); -3.0 μC at (0, 1). What is the magnitude of the electric field at the square's center?

Answer 1

Use symmetricity, the electric fields due to the charges at (0, 0) and (1, 1) cancel each other out at the center of the square, so we need only consider the charges at (1, 0) and (0, 1).

Now, the field vectors due to those charges point in the same direction (because they are of opposite signs), so, we can just compute the field due to one of them and double it.

The formula for the magnitude of the field due to a charge q at distance r is given as -

E = (1 / *(q / r^2).

put the values -

q = 3 x 10^-6 C

r = /2 m

= 8.9 x 10^-12 F/m

So -

E = 2 x (1 / 4 pi (8.9 x 10^(-12))) (3 x 10^(-6) / (1/2))

= 1.07 x 10^5 N/C

Therefore, magnitude of the electric field at the square's center = 1.07 x 10^5 N/C (Answer)