Question

A very long insulating cylinder has radius R and carries positive charge distributed throughout its volume. The charge distribution has cylindrical symmetry but varies with perpendicular distance from the axis of the cylinder. The volume charge density is p(r) = a(l – r/R), where a is a constant with units C/m' and r is the perpendicular distance from the center line of the cylinder. Part A Derive an expression for E(r), the electric field as a function of r for r < R. Express your answer in terms of electric constant en and some or all of the variables a, Rr. Part B Derive an expression for E(r), the electric field as a function of r for r > R. Express your answer in terms of electric constant en and some or all of the variables a, Rr. Part C Do your results agree for r = R?

Answer 1

Answer:

(a) $E(r) = \frac{a}{4\pi \epsilon_0}(2-\frac{r}{R})$

(b) $E(r) = \frac{1}{4\pi\epsilon_0}\frac{aR}{r}$

(c) The results from part (a) and (b) agree at r = R.

Explanation:

(a) r < R:

The electric field can be found by Gauss' Law, since the insulating cylinder is very long.

We will choose an cylinder shaped-Gaussian surface which is imaginary, only to use the Gauss' Law. The radius of the imaginary cylinder will be 'r'. Then by Gauss' Law, the electric flux flowing through this imaginary surface is equal to the enclosed charge trapped inside the imaginary cylinder.

$\int {\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

Let us denote the length of the imaginary cylinder as 'h'.

Since we chose a symmetric geometry (cylinder) as Gaussian surface, we do not need to take the integral in the left-hand side of the Gauss' Law. It is simply

$\int \, da = 2\pi rh$

The enclosed charge can be calculated by integrating the volume charge density over the imaginary cylinder.

$Q_{enc} = \int\limits^r_0 {\rho(r)h} \, dr = ah\int\limits^r_0 {(1 - \frac{r}{R})} \, dr  = ah(r - \frac{r^2}{2R})\left \{ {{r=r} \atop {r=0}} \right. = ah(\frac{2rR - r^2}{2R})$

The Gauss' Law can now be solved.

$E(r)2\pi rh = ah\frac{2rR - r^2}{2\epsilon_0 R}\\E(r) = \frac{a}{4\pi \epsilon_0}\frac{2R - r}{R}\\E(r) = \frac{a}{4\pi \epsilon_0}(2-\frac{r}{R})$

It can be seen that 'h' is canceled out, because the length of the imaginary surface has no effect on the Electric field at r.

(b) r > R:

A similar approach will be followed, although this time the enclosed charge will be the cylinder itself.

$Q_{enc} = \int\limits^R_0 {\rho(r)h} \, dr = ah\int\limits^R_0 {(1-\frac{r}{R})} \, dr = ah(r - \frac{r^2}{2R})\left \{ {{r=R} \atop {r=0}} \right. = ah(R-\frac{R^2}{2R}) = ah\frac{R}{2}$

Now, Gauss' Law can be solved.

$E(r)2\pi rh = \frac{ahR}{2\epsilon_0}\\E(r) = \frac{1}{4\pi\epsilon_0}\frac{aR}{r}$

(c) r = R:

$E(r) = \frac{a}{4\pi \epsilon_0}(2-\frac{r}{R})\\E(r) = \frac{1}{4\pi\epsilon_0}\frac{aR}{r}\\\\E(r) = \frac{a}{4\pi\epsilon_0}$

For both parts E(R) is equal to each other.