- #1

zenterix

- 466

- 69

- Homework Statement
- I am self-studying MIT OCW's 8.02 Electromagnetism. In one of the chapters, there is the following problem (with no solution provided):

A uniformly charged circular cylindrical shell of radius ##R## and height ##h## has a total charge ##Q##. What is the electric field at a point ##P## a distance ##z## from the bottoms side of the cylinder as shown in figure 2.16.6. (Hint: Treat the cylinder as a set of charged rings).

- Relevant Equations
- I assume that the cylinder has no bottom or top.

The charge density per area is ##\rho=\frac{Q}{2\pi Rh}##.

$$dq = dl\cdot dz_{dq}\cdot \rho=\frac{Q}{2\pi h}dz_{dq}d\theta$$

where I used ##dl=Rd\theta##.

The vector from ##dq## to ##P## is

$$\vec{r}=R\cos{\theta}\hat{i}+R\sin{\theta}\hat{j}+z_{dq}\hat{k}$$

and this vector has length ##\sqrt{R^2+(z-z_{dq})^2}##.

Then, if ##\vec{u}## is the unit vector in the direction of ##\vec{r}##, we have

$$d\vec{E}=k_e\frac{dq}{d^2}\vec{u}$$

$$=k_e\frac{Q}{2\pi h}\frac{1}{(R^2+(z-z_{dq}))^{3/2}}\vec{r} dz_{dq}d\theta$$

We want to compute

$$\int_0^{2\pi}\int_0^h d\vec{E}$$

$$=\frac{k_eQ}{2\pi h}\left [-\int_0^{2\pi}\int_0^h \frac{R\cos{\theta}}{(R^2+(z-z_{dq}))^{3/2}} dz_{dq}d\theta \hat{i} \right .$$ $$- \int_0^{2\pi}\int_0^h \frac{R\sin{\theta}}{(R^2+(z-z_{dq}))^{3/2}} dz_{dq}d\theta \hat{j}$$ $$\left . + \int_0^{2\pi}\int_0^h \frac{z-z_{dq}}{(R^2+(z-z_{dq}))^{3/2}} dz_{dq}d\theta\hat{k} \right ]$$

Here is figure 2.16.6

Here is the picture I drew to set up the problem

My first question is if the reasoning and integrals are correct. I used Maple to compute the three integrals. The first two result in 0, which makes sense by symmetry.

Maple can't seem to solve the last integral.

Here is the picture I drew to set up the problem

Maple can't seem to solve the last integral.