Question

Charge is distributed uniformly throughout the volume of an infinitely long solid cylinder of radius R.

(a) Show that, at a distance r < R from the cylinder axis,$\mathit{E}\mathbf{=}\frac{\mathit{p}\mathit{r}}{\mathbf{2}{\mathit{\epsilon }}_{\mathbf{0}}}$where is the volume charge density.

(b) Write an expression for E when r > R.

Short answer

1. The electric field at a distance r < R from the cylinder axis is $\frac{pr}{2{\epsilon }_0}$.
2. The expression for the electric field when r > R is $E_{ext}=\frac{p{R}^2}{2{\epsilon }_{0}r}$.

Step-by-step solution

The given data

A charge is distributed uniformly through the volume V of an infinitely long cylinder of the radius R with the volume density p.

Understanding the concept of the electric flux

Using the concept of Volume charge density, the value of the net charge q is calculated for both cases. Then using the flux concept from the Gauss theorem, we can get the required value of the electric field.

Formula:

The electric flux distribution within an enclosed surface,$\varphi =EA=q/{\epsilon }_0$ (i)

The volume charge density of a material, p = q / V (ii)

a) Calculation of the electric field at r < R

The diagram shows a cross-section (or, perhaps more appropriately, “end view”) of the charged cylinder (solid circle).

Consider a Gaussian surface in the form of a cylinder with radius and length l, coaxial with the charged cylinder. An “end view” of the Gaussian surface is shown as a dashed circle.

Thus, the charge enclosed by it is given using equation (ii) as:

$$\begin{gathered} q=pV \\ ={\mathrm{\pi r}}^2/p \end{gathered}$$ (a)

Where $V={\mathrm{\pi r}}^2/$is the volume of the cylinder.

If ρ is positive, the electric field lines are radially outward, normal to the Gaussian surface, and distributed uniformly along with it.

Thus, the total flux through the Gaussian cylinder is given as:

$$\begin{gathered} \varphi =E{A}_{cylinder} \\ =E\left(2\pi rl\right) \end{gathered}$$ (b)

Now, comparing equation (a) with equation (b) and substituting in equation (i), we get the electric field at r < R as follows:

$$\begin{gathered} 2\pi {\epsilon }_{0}rlE=\pi r^2/p \\ E=\frac{pr}{2{\epsilon }_0} \end{gathered}$$

Hence, it can be seen that the value of the electric field in this case is $\frac{pr}{2{\epsilon }_0}$.

b) Calculation of the electric field at r > R

Next, we consider a cylindrical Gaussian surface of the radius r > R.

If the external field is $E_{ext}$then the flux is using equation (i) and (a) can be given as:

$\varphi =2\pi rl{E}_{ext}$ (c)

The charge enclosed is the total charge in a section of the charged cylinder with length. That is given using equation (ii) as:

$q=\pi R^2/p$ (d)

Now, comparing equation (c) with equation (d) and substituting in equation (i), we get the electric field at r > R as follows:

$$\begin{gathered} 2\pi {\epsilon }_{0}rl{E}_{ext}=\pi R^2/p \\ {E}_{ext}=\frac{p{R}^2}{2{\epsilon }_{0}r} \end{gathered}$$

Hence, the expression of the electric field is given as $E_{ext}=\frac{p{R}^2}{2{\epsilon }_{0}r}$.