Question

A thin cylindrical shell of radius $R$and length $L$, like a soda straw, is uniformly charged with surface charge density $\eta$. What is the electric field strength at the center of one end of the cylinder?

Short answer

The electric field strength at the center of one end of the cylinder is$\frac{\mathbf{\eta }}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{[}\mathbf{1}\mathbf{-}\frac{\mathbf{R}}{\sqrt{{\mathbf{L}}^{\mathbf{2}}\mathbf{+}{\mathbf{R}}^{\mathbf{2}}}}\mathbf{]}$

Step-by-step solution

Given information and formula used

Given :

Cylindrical shell has radius and length : $R\mathrm{and}L$respectively.

It is uniformly charged with surface charge density : $\eta$

Theory used :

Charge on the cylinder is by:

$$\begin{gathered} \eta =\frac{Q}{A} \\ =\frac{Q}{2\pi RL} \\ \Rightarrow Q=2\pi RL\eta (WhereisA=2\mathrm{\pi }RL.) \end{gathered}$$
For all rings, use this formula:

$$\begin{gathered} E_r=dE \\ =\frac{1}{4\pi {\epsilon }_0}\frac{3dq}{{(z^2+R)}^{3/2}} \end{gathered}$$

Finding the electric field strength at the center of one end of the cylinder.

Now we integrate $E_r=dE$:

role="math" localid="1649488487347" $$\begin{gathered} E_c=\int dE \\ =\frac{1}{4\pi {\epsilon }_0}{\int }_0^L\frac{zdq}{{(z^2+R)}^{3/2}} \\ =\frac{1}{4\pi {\epsilon }_0}{\int }_0^{L}2\pi R\eta \frac{zdz}{{(z^2+R)}^{3/2}} \\ =\frac{R\eta }{2{\epsilon }_0}{\frac{-1}{\sqrt{z^2+R^2}}}_0^L \\ =\frac{R\eta }{2{\epsilon }_0}[\frac{-1}{\sqrt{L^2+R^2}}-\frac{-1}{R}] \\ =\boxed{\frac{\mathbf{\eta }}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{[}\mathbf{1}\mathbf{-}\frac{\mathbf{R}}{\sqrt{{\mathbf{L}}^{\mathbf{2}}\mathbf{+}{\mathbf{R}}^{\mathbf{2}}}}\mathbf{]}} \end{gathered}$$