First we need to determine the distance from the center of the box to each of the corners. Geometrically this distance can be defined as
[tex]d_1= \sqrt{ (\frac{a}{2})^2+(\frac{a}{2})^2+(\frac{a}{2})^2}[/tex]
[tex]d_1 = \frac{\sqrt{3}}{2}a[/tex]
Distance between center of cube and center of face is,
[tex]d_2 = \frac{a}{2}[/tex]
Then the Electric field will be given for each point as,
[tex]E_1 = \frac{kQ}{d_1^2}[/tex]
[tex]E_2 = \frac{kQ}{d_2^2}[/tex]
The relation between the two electric field then will be
[tex]\frac{E_1}{E_2} = \frac{d_2^2}{d_1^2}[/tex]
The first electric field in function of the second electric field is
[tex]E_2 = \frac{d_1^2E_1}{d_2^2}[/tex]
Replacing,
[tex]E_2 = \frac{3}{4}*4 E_1[/tex]
[tex]E_2 = 3E_1[/tex]
Replacing with the value given,
[tex]E_2 = 3*(130V/m)[/tex]
[tex]E_2 = 390V/m[/tex]
Therefore the strength of the electric field at the centers of the six square faces of the cube is 390V/m
