Find the potential a distancesfrom an infinitely long straight wire

that carries a uniform line charge$\lambda$. Compute the gradient of your potential, and

check that it yields the correct field.

The electric field and potential is expressed as,

Here, E is the electric field and V is the potential.

Consider the diagram for the linear cross section.

The above diagram shows the infinite long with linear charge density$\lambda$.And is the radius of circular cross-section.

Consider the Gaussian cylinder of length l,

The radius of the Gaussian cylinder is s.

Now, apply Gauss Law,

$$\begin{gathered} \int E.da=\frac{q}{{\epsilon }_0} \\ E\left(2\mathrm{\pi sl}\right)=\frac{\lambda l}{{\epsilon }_0} \\ E=\frac{\lambda }{2{\mathrm{\pi S\epsilon }}_0}S \end{gathered}$$

The expression for electric field is,

$E=\frac{\lambda }{2{\mathrm{\pi S\epsilon }}_0}S$

Consider the reference point$\infty$, to charge itself extend infinite.

Integrate the electric file with the limits s = bto s = sdetermine the potential of the wire.

role="math" localid="1655958331580" $$\begin{gathered} V\left(s\right)=-{\int }_b^{s}E.ds \\ =-{\int }_b^s\frac{\lambda }{2{\mathrm{\pi \epsilon }}_0\mathrm{s}}ds \\ =-\frac{\lambda }{2{\mathrm{\pi \epsilon }}_0\mathrm{s}}In\left[\frac{S}{b}\right] \\ \mathrm{Therefore},\mathrm{the}\mathrm{potential}\mathrm{due}\mathrm{to}\mathrm{infinitely}\mathrm{long}\mathrm{wire}\mathrm{is}V\left(s\right)=-\frac{\lambda }{2{\mathrm{\pi \epsilon }}_0\mathrm{s}}In\left[\frac{S}{b}\right]. \end{gathered}$$

The electric field in potential term is described as,

$E=-\nabla V$

Now, compute the gradient potential,

$$\begin{gathered} \nabla V=\nabla \left[-\frac{\lambda }{2{\mathrm{\pi \epsilon }}_0}n\left[\frac{S}{b}\right]\right] \\ =-\frac{\lambda }{2{\mathrm{\pi \epsilon }}_0\mathrm{s}}\frac{d}{ds}\left[in\left[\frac{s}{b}\right]\right] \\ =-\frac{\lambda }{2{\mathrm{\pi \epsilon }}_0\mathrm{s}}\frac{b}{s}\left[\frac{1}{b}\right] \\ =-\frac{\lambda }{2{\mathrm{\pi \epsilon }}_0\mathrm{s}} \\ \mathrm{Therefore},\mathrm{the}\mathrm{expression}\mathrm{of}\mathrm{the}\mathrm{electric}\mathrm{field}\mathrm{is}\mathrm{E}=\frac{\lambda }{2{\mathrm{\pi \epsilon }}_0\mathrm{s}}\hat{s}. \\ \mathrm{The}\mathrm{expression}\mathrm{for}\mathrm{the}\mathrm{electric}\mathrm{field}\mathrm{and}\mathrm{the}\mathrm{expression}\mathrm{for}\mathrm{the}\mathrm{gauss}\mathrm{law}\mathrm{are}\mathrm{same}. \\ \mathrm{Thus},\mathrm{the}\mathrm{gradient}\mathrm{of}\mathrm{potential}\nabla V=-E\mathrm{is}\mathrm{proved}. \end{gathered}$$