four point charges lie on the vertices of a square with side length . two adjacent vertices have charge while, the other two have charge . what is the magnitude of the electric field at the center of the square?

The magnitude of the electric field at the center is

E=22 k qa2E=\frac{2\sqrt{2},k,q}{a^{2}}E=a222​kq​

if two adjacent vertices carry charge +q+q+q and the other two adjacent vertices carry charge −q-q−q, where aaa is the side length of the square and kkk is Coulomb’s constant.

Step-by-step idea (short)

1. Distance from each corner to the center of a square of side aaa is

r=a2.r=\frac{a}{\sqrt{2}}.r=2​a​.

2. Magnitude of the field at the center due to any one charge of magnitude ∣q∣|q|∣q∣ is

E0=k∣q∣r2=k∣q∣(a2/2)=2k∣q∣a2.E_0=\frac{k|q|}{r^2}=\frac{k|q|}{(a^2/2)}=\frac{2k|q|}{a^2}.E0​=r2k∣q∣​=(a2/2)k∣q∣​=a22k∣q∣​.

3. Because the charges are at the four corners and the center is along the diagonals, each field vector makes a 45∘45^\circ 45∘ angle with the sides. The components along one diagonal add; the components along the other diagonal cancel pairwise.
4. Adding the contributing components gives a net field of magnitude

E=22 E0=22 2k∣q∣a2=22 k qa2,E=2\sqrt{2},E_0=2\sqrt{2},\frac{2k|q|}{a^2}=\frac{2\sqrt{2},k,q}{a^2},E=22​E0​=22​a22k∣q∣​=a222​kq​,

directed from the negative charges toward the positive charges along the diagonal joining the like charges.

TL;DR

For a square of side aaa with two adjacent corners at +q+q+q and the other two at −q-q−q, the electric field at the center has magnitude

E=22 k qa2.E=\frac{2\sqrt{2},k,q}{a^{2}}.E=a222​kq​.