91Ó°ÊÓ

We need to find an expression for the electric potential a distance$z$away from the center of rod on the line that bisects the rod.

We will use integration as an infinitesimal extension of the addition that comes due to the potential being a scalar quantity. That being said, letting $x$be the integration variable, $z$being the distance from the $x$$-axis$, and understanding that the infinitesimal charge- that is, the linear charge density - is $Q/L$, we will have

$V=k{}_{-L/2}{}^{L/2}\frac{Q/L}{\sqrt{z^2Â\pm x^2}}dx$,

because the expression under the denominator will be from the Pythagorean theorem, the distance of the part of charge we're considering in the integral. Shown as

$\frac{dx}{\sqrt{x^2Â\pm a^2}}=In|x+\sqrt{x^2Â\pm a^2}$

Having said this, our expression becomes

$V=\frac{kQ}{L}{}_{-L/2}{}^{L/2}\frac{dx}{\sqrt{z^2+x^2}}$

By integrating, we get

$V=kIn|L/2\_+\sqrt{x^2{(L/2)}^2}|-In|L/2+\sqrt{z^2-{(L/2)}^2}|$,

which, by the properties of the logarithm, is transform as

$V-\frac{kQ}{L}In\left|\frac{2L+\sqrt{L^2+4z^2}}{2L+\sqrt{L^2-4z^2}}\right|$