Question

A hollow, thin-walled insulating cylinder of radius R and length L (like the cardboard tube in a roll of toilet paper) has charge Q uniformly distributed over its surface. (a) Calculate the electric potential at all points along the axis of the tube. Take the origin to be at the center of the tube, and take the potential to be zero at infinity. (b) Show that if L << R, the result of part (a) reduces to the potential on the axis of a ring of charge of radius R. (See Example 23.11 in Section 23.3.) (c) Use the result of part (a) to find the electric field at all points along the axis of the tube.

Short answer

(a) The electric potential at all points along the axis of the tube is $\frac{\mathrm{kQ}}{\mathrm{L}}\ln \left(\frac{\sqrt{{\left(\frac{\mathrm{L}}{2}-\mathrm{x}\right)}^2+{\mathrm{R}}^2}+{\displaystyle \frac{\mathrm{L}}{2}}-\mathrm{x}}{{\left(\frac{\mathrm{L}}{2}+\mathrm{x}\right)}^2+{\mathrm{R}}^2-\frac{\mathrm{L}}{2}-\mathrm{x}}\right)$.

(b) If L<<R,${\mathrm{V}}_{\mathrm{cylinder}}={\mathrm{V}}_{\mathrm{ring}}=\frac{\mathrm{kQ}}{({\mathrm{R}}^2+{\mathrm{x}}^2)}$

(c) The electric field at all points along the axis of the tube is$2\mathrm{kQ}\frac{\left(\sqrt{(\mathrm{L}-2\mathrm{x}{)}^2+4{\mathrm{R}}^2}-\sqrt{(\mathrm{L}+2\mathrm{x}{)}^2+4{\mathrm{R}}^2)}\right)}{{\sqrt{\mathrm{L}(\mathrm{L}-2\mathrm{x}{)}^2+4\mathrm{R}}}^2\sqrt{(\mathrm{L}+2\mathrm{x}{)}^2+4{\mathrm{R}}^2}}$.

Step-by-step solution

Step 1:

For the thin ring, the voltage across the ring is given by

$\mathbf{V}\mathbf{=}\frac{\mathbf{kQ}}{\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{R}}^{\mathbf{2}}}}$

Where x= distance from the center, R=radius and k=$\frac{1}{4\pi {\epsilon }_{\circ }}=9.0\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2$

Here the slice be at coordinate z along x axis. So the potential of infinitesimal slice of finite cylinder

$$\begin{gathered} dV=\frac{kdQ}{\sqrt{{\left(x-z\right)}^2}+R^2} \\ =\frac{kQ}{L}\frac{dz}{\sqrt{{\left(x-z\right)}^2}+R^2} \end{gathered}$$

role="math" localid="1664880157914" $$\begin{gathered} dV=\frac{kdQ}{\sqrt{{\left(x-z\right)}^2}+R^2} \\ =\frac{kQ}{L}\frac{dz}{\sqrt{{\left(x-z\right)}^2}+R^2}.............\left(1\right) \\ \mathrm{Integration}\left(1\right)\mathrm{gives} \\ \mathrm{V}=\frac{\mathrm{kQ}}{\mathrm{L}}{\int }_{-\frac{\mathrm{L}}{2}-\mathrm{x}}^{\frac{\mathrm{L}}{2}-\mathrm{x}}\frac{\mathrm{dz}}{\sqrt{{\left(\mathrm{x}-\mathrm{z}\right)}^2+{\mathrm{R}}^2}} \\ \mathrm{Let}\mathrm{x}-\mathrm{z}=\mathrm{u} \\ \mathrm{V}=\frac{\mathrm{kQ}}{\mathrm{L}}{\int }_{-\frac{\mathrm{L}}{2}-\mathrm{x}}^{\frac{\mathrm{L}}{2}-\mathrm{x}}\frac{\mathrm{du}}{\sqrt{{\left(\mathrm{u}\right)}^2+{\mathrm{R}}^2}} \\ \frac{kQ}{L}ln\frac{\sqrt{{\left(\frac{L}{2}-x\right)}^2}+R^2=\frac{L}{2}-x}{\sqrt{{\left(\frac{L}{2}+x\right)}^2}+R^2-\frac{L}{2}-x} \end{gathered}$$

Therefore, the electric potential at all points along the axis of the tube is

role="math" localid="1664880243784" $\frac{kQ}{L}ln\left(\frac{\sqrt{{\left(\frac{L}{2}-x\right)}^2}+R^2=\frac{L}{2}-x}{\sqrt{{\left(\frac{L}{2}+x\right)}^2}+R^2-\frac{L}{2}-x}\right)$.

Step 2:

$$\begin{gathered} v\left(x\right)\frac{\mathrm{kQ}}{\mathrm{L}}\ln \left(\frac{\sqrt{{\left(\frac{\mathrm{L}}{2}-\mathrm{x}\right)}^2}+{\mathrm{R}}^2+\frac{\mathrm{L}}{2}-\mathrm{x}}{\sqrt{{\left(\frac{\mathrm{L}}{2}+\mathrm{x}\right)}^2}+{\mathrm{R}}^2-\frac{\mathrm{L}}{2}-\mathrm{x}}\right) \\ =\frac{kQ}{L}ln\left(\frac{\sqrt{x^2-xL+R^2}+\frac{L}{2}-x}{\sqrt{x^2+xL+R^2}-\frac{L}{2}-x}\right) \\ =\frac{kQ}{L}ln\left(\frac{\sqrt{{\displaystyle \frac{1-xL}{\left(R^2+x^2\right)}}}+{\displaystyle \frac{\frac{L}{2}-x}{\sqrt{R^2+x^2}}}}{\sqrt{\frac{1-xL}{\left(R^2+x^2\right)}}+\frac{\left(-\frac{L}{2}-x\right)}{\sqrt{R^2+x^2}}}\right) \end{gathered}$$

$$\begin{gathered} simplifing \\ V=\frac{kQ}{L}In\frac{1+{\displaystyle \frac{L}{2\sqrt{R^2+x^2}}}}{1-\frac{L}{2\sqrt{R^2+x^2}}} \\ =\frac{kQ}{L}\frac{2L}{2\sqrt{R^2+x^2}} \\ =\frac{kQ}{\sqrt{R^2+x^2}} \\ {V}_{cylinder}=V_{ring}=\frac{kQ}{\sqrt{R^2+x^2}} \end{gathered}$$

Thus, if L << R, the result of Step 1 reduces to the potential on the axis of a ring.

Step 3:

The electric field is given by:

$$\begin{gathered} E=\frac{-\partial V}{\partial X} \\ E\left(x\right)=-\frac{\partial }{\partial X}\left(\frac{\mathrm{kQ}}{\mathrm{L}}\ln \left(\frac{\left(\sqrt{{\left(\frac{\mathrm{L}}{2}-\mathrm{x}\right)}^2}+{\mathrm{R}}^2+\frac{\mathrm{L}}{2}-\mathrm{x}\right)}{\left(\sqrt{{\left(\frac{\mathrm{L}}{2}+\mathrm{x}\right)}^2}+{\mathrm{R}}^2-\frac{\mathrm{L}}{2}-\mathrm{x}\right)}\right)\right) \end{gathered}$$

$E\left(x\right)=\frac{2\mathrm{kQ}\left(\sqrt{(\mathrm{L}-2\mathrm{x}{)}^2+4{\mathrm{R}}^2}-\sqrt{(\mathrm{L}+2\mathrm{x}{)}^2+4{\mathrm{R}}^2)}\right)}{\mathrm{L}\sqrt{(\mathrm{L}-2\mathrm{x}{)}^2+4{\mathrm{R}}^2}\sqrt{(\mathrm{L}+2\mathrm{x}{)}^2+4{\mathrm{R}}^2}}$

Therefore, the electric field at all points along the axis of the tube is

$\mathrm{E}\left(\mathrm{x}\right)=\frac{2\mathrm{kQ}\left(\sqrt{(\mathrm{L}-2\mathrm{x}{)}^2+4{\mathrm{R}}^2}-\sqrt{(\mathrm{L}+2\mathrm{x}{)}^2+4{\mathrm{R}}^2)}\right)}{\mathrm{L}\sqrt{(\mathrm{L}-2\mathrm{x}{)}^2+4{\mathrm{R}}^2}\sqrt{(\mathrm{L}+2\mathrm{x}{)}^2+4{\mathrm{R}}^2}}$