Q27P - Introduction to Electrodynamics

Question

Find the potential on the axis of a uniformly charged solid cylinder,

a distance  z from the center. The length of the cylinder is L, its radius is R, and

the charge density is p. Use your result to calculate the electric field at this point.

(Assume that $\mathit{z}\mathbf{>}\mathit{L}\mathbf{/}\mathbf{2}$.)

Step-by-Step Solution

Answer

The electric field is $\frac{p}{2{\epsilon }_0}\left[L-\sqrt{R^2+{\left(\begin{array}{c}z+\frac{L}{2}\end{array}\right)}^2} +\sqrt{R^2+{\left(z-\frac{L}{2}\right)}^2}\right]\hat{z}$

Step 1: Define the uniformly charged solid cylinder and the potential at the equatorial position.

Consider the below figure, the electric filed and electric potential on the axis of solid cylinder.

                     

Here, the figure shows the uniformly charged solid cylinder and its axis is the along the axis at the center of the origin.

Here, L is the Length of the cylinder, R is the radius of the cylinder and surface charge density.

 

Write the potential at the equatorial position due to uniform surface charge of disc is given as,

$\mathit{d}\mathit{V}\mathbf{=}\frac{\mathbf{\sigma }}{\mathbf{2}{\mathbf{\in }}_{\mathbf{0}}}\mathbf{(}\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{\mathbf{Z}}^{\mathbf{2}}\mathbf{-}\mathbf{Z}\mathbf{)}}$

 Here, z distance from the center of a disc at the point P.

Step 2: Determine electric field.

Consider the thickness of each disc is dz.

Consider distance of the slice from the point P with respect to left end is .

Consider the distance of the slice from the point P with respect to right end is$z-\frac{L}{2}$.Write the formula for the potential at point due to the whole cylinder is$z-\frac{L}{2}$ obtained by integrating the equation with limits to$z-\frac{L}{2} to z+\frac{L}{2}$.

$$\begin{gathered} \mathrm{V}=\frac{\mathrm{p}}{2{\mathrm{\epsilon }}_0}{\int }_{\mathrm{z}-\frac{\mathrm{L}}{2}}^{\mathrm{z}+\frac{\mathrm{L}}{2}}\left(\sqrt{{\mathrm{R}}^2+{\mathrm{z}}^2-}\right)\mathrm{dz} \\ =\frac{\mathrm{p}}{2{\mathrm{\epsilon }}_0}\frac{1}{2}{\left|\left[\mathrm{z}\sqrt{{\mathrm{R}}^2+{\mathrm{z}}^2}+{\mathrm{R}}^2\mathrm{In}\left(\mathrm{z}+\sqrt{{\mathrm{R}}^2+{\mathrm{z}}^2}\right)-{\mathrm{z}}^2\right]\right|}_{\mathrm{z}-\frac{\mathrm{L}}{2}}^{\mathrm{z}+\frac{\mathrm{L}}{2}} \\ =\frac{\mathrm{p}}{4{\mathrm{\epsilon }}_0}\left\{\begin{array}{c}\left(\mathrm{z}+\frac{\mathrm{L}}{2}\right)\sqrt{{\mathrm{R}}^2+{\left(\mathrm{z}+\frac{\mathrm{L}}{2}\right)}^2}-\left(\mathrm{z}-\frac{\mathrm{L}}{2}\right)\sqrt{{\mathrm{R}}^2+{\left(\mathrm{z}-\frac{\mathrm{L}}{2}\right)}^2}\\ +\mathrm{RIn}\left[\frac{\mathrm{z}+{\displaystyle \frac{\mathrm{L}}{2}}+\sqrt{{\mathrm{R}}^2+{\left(\mathrm{z}+\frac{\mathrm{L}}{2}\right)}^2}}{\mathrm{z}-{\displaystyle \frac{\mathrm{L}}{2}}+\sqrt{{\mathrm{R}}^2+{\left(\mathrm{z}-\frac{\mathrm{L}}{2}\right)}^2}}\right]\end{array}\right\}-2\mathrm{zL} \\ \mathrm{Now} \mathrm{finding} \mathrm{the} \mathrm{electric} \mathrm{field} \mathrm{due} \mathrm{to} \mathrm{the} \mathrm{cylinder} \mathrm{at} \mathrm{the} \mathrm{point} \mathrm{p}, \\ \mathrm{E}=-\nabla \mathrm{V} \\ \mathrm{The} \mathrm{electric} \mathrm{field} \mathrm{along} \mathrm{the} \mathrm{z} \mathrm{axis} \mathrm{is}, \\ \mathrm{E}=-\frac{\partial \mathrm{V}}{\partial \mathrm{z}}\hat{\mathrm{z}} \end{gathered}$$

$$\begin{gathered} \mathrm{Substitute}\frac{\mathrm{p}}{4{\mathrm{\epsilon }}_0}\left\{\begin{array}{c}\left(\mathrm{z}+\frac{\mathrm{L}}{2}\right)\sqrt{{\mathrm{R}}^2+{\left(\mathrm{z}+\frac{\mathrm{L}}{2}\right)}^2}-\left(\mathrm{z}-\frac{\mathrm{L}}{2}\right)\sqrt{{\mathrm{R}}^2+{\left(\mathrm{z}-\frac{\mathrm{L}}{2}\right)}^2}\\ +\mathrm{RIn}\left[\frac{\mathrm{z}+{\displaystyle \frac{\mathrm{L}}{2}}+\sqrt{{\mathrm{R}}^2+{\left(\mathrm{z}+\frac{\mathrm{L}}{2}\right)}^2}}{\mathrm{z}-{\displaystyle \frac{\mathrm{L}}{2}}+\sqrt{{\mathrm{R}}^2+{\left(\mathrm{z}-\frac{\mathrm{L}}{2}\right)}^2}}\right]-2\mathrm{zL}\end{array}\right\}\mathrm{for} \mathrm{V} \mathrm{in} \mathrm{above} \\ \mathrm{equation}. \\ \mathrm{E}=-\frac{\hat{\mathrm{z}}\mathrm{p}}{4{\mathrm{\epsilon }}_0\partial \mathrm{z}}\left\{\begin{array}{c}\left(\mathrm{z}+\frac{\mathrm{L}}{2}\right)\sqrt{{\mathrm{R}}^2+{\left(\mathrm{z}+\frac{\mathrm{L}}{2}\right)}^2}-\left(\mathrm{z}-\frac{\mathrm{L}}{2}\right)\sqrt{{\mathrm{R}}^2+{\left(\mathrm{z}-\frac{\mathrm{L}}{2}\right)}^2}\\ +\mathrm{RIn}\left[\frac{\mathrm{z}+{\displaystyle \frac{\mathrm{L}}{2}}+\sqrt{{\mathrm{R}}^2+{\left(\mathrm{z}+\frac{\mathrm{L}}{2}\right)}^2}}{\mathrm{z}-{\displaystyle \frac{\mathrm{L}}{2}}+\sqrt{{\mathrm{R}}^2+{\left(\mathrm{z}-\frac{\mathrm{L}}{2}\right)}^2}}\right]-2\mathrm{zL}\end{array}\right\} \\ \mathrm{Now} \mathrm{partially} \mathrm{differentiating} \mathrm{the} \mathrm{above} \mathrm{equation} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{z}. \end{gathered}$$

$E=-\frac{\hat{z}p}{4{\epsilon }_0}\left\{\begin{array}{c}\left\{\sqrt{R^2+{\left(z+\frac{L}{2}\right)}^2}+\frac{{\left(z+\frac{L}{2}\right)}^2}{\sqrt{R^2+{\left(z+\frac{L}{2}\right)}^2}}-\sqrt{R^2+{\left(z-\frac{L}{2}\right)}^2}-\frac{{\left(z-\frac{L}{2}\right)}^2}{\sqrt{R^2+{\left(z-\frac{L}{2}\right)}^2}}\right\}\\ +R^2\left[\frac{1+{\displaystyle \frac{z+{\displaystyle \frac{L}{2}}}{\sqrt{R^2+{\left(z+\frac{L}{2}\right)}^2}}}}{\left(z+\frac{L}{2}\right)+\sqrt{R^2+{\left(z+\frac{L}{2}\right)}^2}}-\frac{1+{\displaystyle \frac{z-{\displaystyle \frac{L}{2}}}{\sqrt{R^2+{\left(z-\frac{L}{2}\right)}^2}}}}{\left(z-\frac{L}{2}\right)+\sqrt{R^2+{\left(z-\frac{L}{2}\right)}^2}}\right]-2L\\ \end{array}\right\}$

Simplify the above equation,

$$\begin{gathered} \mathrm{E}=\frac{-\hat{\mathrm{z}}\mathrm{p}}{4{\mathrm{\epsilon }}_0}\left\{2\sqrt{{\mathrm{R}}^2+{\left(\mathrm{z}+\frac{\mathrm{L}}{2}\right)}^2}-2\sqrt{{\mathrm{R}}^2+{\left(\mathrm{z}-\frac{\mathrm{L}}{2}\right)}^2}-2\mathrm{L}\right\} \\ =\frac{\mathrm{p}}{2{\mathrm{\epsilon }}_0}\left[\mathrm{L}-\sqrt{{\mathrm{R}}^2+{\left(\mathrm{z}+\frac{\mathrm{L}}{2}\right)}^2}+\sqrt{{\mathrm{R}}^2+{\left(\mathrm{z}-\frac{\mathrm{L}}{2}\right)}^2}\right]\hat{\mathrm{z}} \\ \mathrm{Therefore}, \mathrm{the} \mathrm{electric} \mathrm{field} \mathrm{is} \frac{\mathrm{p}}{2{\mathrm{\epsilon }}_0}\left[\mathrm{L}-\sqrt{{\mathrm{R}}^2+{\left(\mathrm{z}+\frac{\mathrm{L}}{2}\right)}^2}+\sqrt{{\mathrm{R}}^2+{\left(\mathrm{z}-\frac{\mathrm{L}}{2}\right)}^2}\right]\hat{\mathrm{z}} \end{gathered}$$