Module 2 : Electrostatics
Lecture 8 : Electrostatic Potential
 

Hence, if the perpendicular distance of $P$from the line charge is denoted by $x$

\begin{eqnarray*} \phi(P)-\phi(P_0) &=& -\frac{\lambda}{2\pi\epsilon_0}\int_{x_... ...=& -\frac{\lambda}{2\pi\epsilon_0} \ln\left(\frac{x}{x_o}\right) \end{eqnarray*}

\includegraphics{fig25.eps} Click here for Animation

One can see that taking $x_0$to be infinite will make the integral diverge. In this case, it it is convenient to take the zero of the potential to be at unit distance from the line ( $ x=1$). With this choice the potential at a distance $x$ from the line is

\begin{displaymath}\phi(x) = -\frac{\lambda}{2\pi\epsilon_0}\ln x\end{displaymath}

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