Answer to Question #117058 in Electricity and Magnetism for Lindsay

Question

A 2D annulus (thick ring) has an inner radius Ri and outer radius Ro, and charge Q non-uniformly distributed over its surface. The 2D charge density varies with radius r by n(r)=Cr^3 for Ri<= r<=Ro and n=0 for all other r. C is a constant. Answer the following in terms of the variables given above. 1-Find an expression for C such that the total charge of the annulus is Q. Include the SI units for C next to your answer.
Draw a clear picture and use it to set-up the integral to calculate the E-field at a point on the axis of the annulus (this axis is perpendicular to the plane of the annulus) a distance z from the center of the annulus. You must complete all the steps short of actually computing the integral( your final answer must be an integral with only one variable of integration and all other variables constant.) Show that your answer has the correct SI units for the electric field.

Answer 1

1. The total charge, by definition, is the integral from $n(r)$ over all ring surface. Thus:

$$Q = \int^{2\pi}_0 \int^{Ro}_{R_i} n(r) rdrd\varphi$$

where $dS = rdrd\varphi$ is the differential surface element in the polar coordinate system.

$$Q = \int^{2\pi}_0 d\varphi\int^{Ro}_{R_i} Cr^3 rdr =2\pi \int^{Ro}_{R_i}Cr^4dr = 2\pi C \dfrac{r^5}{5} |^{Ro}_{Ri}=\\ = \dfrac{2\pi}{5} C (R_o^5 - R_i^5)$$

Express $C$ from this equation and obtain final result:

$$C = \dfrac{5Q}{2\pi (R_o^5 - R_i^5)} [C/m^5]$$

2. In order to find the field at the point on the z axis one should integrate the fields created by the charge of each differential surface element $dq$ of the ring (see figure below).

This small part of toltal field will be equal to $dE = \dfrac{1}{4\pi\varepsilon_0}\dfrac{dq}{r'^2}$ and directed in the direction of $r'$. But the total field, due to the symmetry of the problem, will be directed alon the z axis.

By definition, the charge of the differential element is $dq = n(r)dS = Cr^3\cdot rdrd\varphi = Cr^4\cdot drd\varphi$.

According to the Pythagorean theorem: $r' = \sqrt{r^2 + z^2}$ , where $z$ is the coorditate of the point on the axis.

Putting it all together:

$$dE = \dfrac{1}{4\pi\varepsilon_0}\dfrac{dq}{r'^2} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Cr^4\cdot drd\varphi }{r^2 + z^2}$$

The module of the total field:

$$E_{tot} = \int dE = \dfrac{1}{4\pi\varepsilon_0} \int^{2\pi}_0 d\varphi\int^{Ro}_{R_i} \dfrac{Cr^4}{r^2 + z^2}dr = \dfrac{C}{2\varepsilon_0} \int^{Ro}_{R_i} \dfrac{Cr^4}{r^2 + z^2}dr$$

Finally, subsitute the value of $C$ from the previous question and obtain the answer:

$$E_{tot} = \dfrac{5Q}{4\pi \varepsilon_0 (R_o^5 - R_i^5)} \int^{Ro}_{R_i} \dfrac{Cr^4}{r^2 + z^2}dr$$