Q22P - Introduction to Electrodynamics

Question

Find the potential a distance s from 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.

Step-by-Step Solution

Answer

The gradient of potential$\nabla V=-E$is proved.

Step 1: Determine the electric field

The electric field and potential is expressed as,

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

Step 2: Determine the diagram for the condition.

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.

Step 3: Determine expression for infinite wire

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 = b to s = sdetermine the potential of the wire.

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

Step 4: Determine gradient potential.

The electric field in potential term is described as,

$E=-\nabla V$

 

Now, compute the gradient potential,

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