The electric field at this point is given by the formula:
$$E=\frac{1}{4\pi\epsilon_{0}}\times\frac{4\lambda az}{z^{2}+\frac{a^{2}}{4}}\sqrt{z^{2}+\frac{a^{2}}{2}}$$
where $\lambda$ is the linear charge density and is equal to $\sigma\times\frac{dA}{2}$.
Show more…