A cube of side $$a$$ has point charges +Q located at each of its vertices except at the origin where the charge is -Q. The electric field at the centre of cube is: 

Using x, y, z components only:

Take the corner with charge −Q as origin

(0,0,0)

Center of cube is at

First imagine all 8 corners have charge +Q.
By symmetry, electric field at cube center is zero.

Now replacing the +Q at origin by −Q is equivalent to adding an extra charge

−2Q

at origin.

So net field at center is simply field due to charge −2Q at origin.

Distance of center from origin:

$$r=\frac{\sqrt{3}a}{2}$$

Magnitude of field:

Direction is from center toward origin, i.e. along

(−1,−1,−1)

Unit vector in this direction:

So components are

$$E_y=-\frac{8kQ}{3\sqrt{3}a^2}$$

$$E_z=-\frac{8kQ}{3\sqrt{3}a^2}$$

Hence field is