Question

A total charge Q is distributed uniformly along a straight rod of Length L. The potential at a point P at a distance h from the midpoint of the rod is $(Hint:\int \frac{1}{\sqrt{x^2+a^2}}dx=ln(x+\sqrt{x^2+a^2}))$

• A. $\frac{Q}{4\pi {\epsilon }_{0}L}ln(L+\sqrt{L^2+3h^2})$
• B. $\frac{Q}{4\pi {\epsilon }_{0}L}ln\left(\frac{L+\sqrt{L^2+4h^2}}{2}\right)$
• C. $\frac{Q}{4\pi {\epsilon }_{0}L}ln\left(\frac{L+\sqrt{L^2+4h^2}}{L-\sqrt{L^2+4h^2}}\right)$
• D. $\frac{Q}{4\pi {\epsilon }_{0}L}ln\left(\frac{\sqrt{L^2+4h^2}+L}{\sqrt{L^2+4h^2}-L}\right)$

Answer 1

The correct option is A $\frac{Q}{4\pi {\epsilon }_{0}L}ln(L+\sqrt{L^2+3h^2})$

Total charge $=Q$

Straight rod of length $=L$

distance from $P$ is $h$ midpoint of the rod

$\int \frac{dx}{\sqrt{x^2+a^2}}=ln(x+\sqrt{x^2+a^2})$

$V=\frac{1}{4\pi {ϵ}_0}\frac{Q}{L}\int \frac{dx}{\sqrt{x^2+a^2}}$ (due to the perpendicular to the value of $x$)

when $x=L$ and $a=\sqrt{3}h$

$=\frac{1}{4\pi {ϵ}_0}\frac{Q}{L}ln(x+\sqrt{x^2+a^2})$

$=\frac{1}{4\pi {ϵ}_0}\frac{Q}{L}ln(x+\sqrt{L^2+{\left(\sqrt{3}h\right)}^2})$

$=\frac{1}{4\pi {ϵ}_0}\times \frac{Q}{L}ln(L+\sqrt{L^2+{3h}^2})$