In vacuum, four charges each equal to q are placed at each of the four vertices of a square. Find the intensity and potential of the electric field at the point of intersection of two diagonals.

Explanation

Let the length of each side of the square be a. According to the fig.
`AC = sqrt(AB^(2)+BC^(2)) = sqrt(a^(2)+a^(2))`
` = sqrt(2)a`
Now, `AC = BD`
and `AO = BO = CO = Do = 1/2 AC`
`=1/2 cdot sqrt(2)a = a/(sqrt(2))`
Therefore, intensity at O due to the change q at A is given by
`E_(1) = 1/(4 pi epsilon_(0)) cdot (q)/((a/(sqrt(2)))^(2)), along vec(OC)`
Intensity at O due to the charge at C is given by
`E_(2) = 1/(4 pi epsilon_(0)) cdot (q)/((a/(sqrt(2)))^(2)), along vec(OA)`
These two intensities being equal and opposite balance each other.
Similarly, intensities at O due to the charges at B and D being equal and opposite balance each other. So the intensity at the point of intersection of the two diagonals is zero.
Potential at O = `1/(4 pi epsilon_(0)) cdot [q/((a)/(sqrt(2)))+q/((a)/(sqrt(2)))+q/((a)/(sqrt(2)))+q/((a)/(sqrt(2)))] = (4sqrt(2)q)/(4 pi epsilon_(0)a)`