Question

Calculate the electric potential at the point $P$ due to the given charged rod of length $L$ having charge per unit length $\lambda$.

• A. $$V=k\lambda ln\left\{\frac{a+L}{a}\right\}$$
• B. $$V=2k\lambda ln\left\{\frac{a+L}{a}\right\}$$
• C. $V=2k\lambda ln\left\{\frac{a+L+\sqrt{b^2+{(a+L)}^2}}{a+\sqrt{b^2+a^2}}\right\}$
• D. $V=k\lambda ln\left\{\frac{a+L+\sqrt{b^2+{(a+L)}^2}}{a+\sqrt{b^2+a^2}}\right\}$
  

Answer 1

The correct option is D $V=k\lambda ln\left\{\frac{a+L+\sqrt{b^2+{(a+L)}^2}}{a+\sqrt{b^2+a^2}}\right\}$


Let the total charge on the rod is $Q$.
The charge per unit length, $\lambda =\frac{Q}{L}$
Let's take a small length of charged rod $dx$ at a distance $x$ from the vertical $y-$axis as shown below.


Total charge in $dx$ element is $dq$
$\Rightarrow dq=\frac{Q}{L}dx=\lambda dx$
Let the distance between point $P$ and $dx$ is $r$.
$\Rightarrow r=\sqrt{x^2+b^2}$.

Now the electric potential at point $P$ due to charge $dq$ is given by

$dV=\frac{k\lambda dx}{r}$
The electric potential at point $P$ due to entire rod is obtained by integration.

$\Rightarrow \int dV=k\lambda {\int }_a^{a+L}\frac{dx}{\sqrt{x^2+b^2}}$

$\Rightarrow V=k\lambda ln{\left(\frac{\sqrt{x^2+b^2}+x}{b}\right)}_a^{a+L}$

$\Rightarrow V=k\lambda ln\left\{\frac{\sqrt{{(a+L)}^2+b^2}+a+L}{b}\right\}-k\lambda ln\left\{\frac{\sqrt{a^2+b^2}+a}{b}\right\}$

$\Rightarrow V=k\lambda ln\left\{\frac{a+L+\sqrt{b^2+{(a+L)}^2}}{a+\sqrt{a^2+b^2}}\right\}$