MIT OCW, 8.02, Electromagnetism: Charged Cylindrical Shell

zenterix
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.

Homework Helper
Gold Member
I assume that the cylinder has no bottom or top.
You cannot assume that. The cylinder has a flat bottom in the ##xy##-plane and a flat top in the plane ##z=h##. Look at the picture. It's not a thin-walled tube.

Your ##dq## doesn't look right. If the cylinder has total charge ##Q##, the volume charge density is ##\rho=\frac{Q}{\pi R^2h}.## Then ##dq=\rho dV=\rho r'dr'~d\theta'~ dz'## is a charge element at location ##\mathbf{r}'=r'\cos\theta'~\mathbf{\hat x}+r'\sin\theta'~\mathbf{\hat y}+z'~\mathbf{\hat z}##. You need to put that in your expression for ##d\mathbf{E}## and integrate over all three primed coordinates. Yes, by symmetry only the ##z##-component is non-zero. However, your integrals ignore the contributions from charge elements inside the cylinder at points ##r'<R.##

Homework Helper
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2022 Award
You cannot assume that. The cylinder has a flat bottom in the xy-plane and a flat top in the plane z=h. Look at the picture. It's not a thin-walled tube.
I disagree. From the shading on the 'top', what you are seeing is the inside of the cylinder.
It is described as a cylindrical shell.

zenterix and nasu
Staff Emeritus
Homework Helper
Gold Member
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).

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

Maple can't seem to solve the last integral.
You have a typo. The ##(z-z_{dq})## in the denominator should be squared.

Steve4Physics
Homework Helper
Gold Member
I disagree. From the shading on the 'top', what you are seeing is the inside of the cylinder.
It is described as a cylindrical shell.
You are, of course, correct in disagreeing. I cannot imagine how I read "ring" and interpreted it as "disk". I guess my reading skills are deteriorating . . .

zenterix
You have a typo. The ##(z-z_{dq})## in the denominator should be squared.

Indeed, thanks.

So the integral itself is correct? It's just a question of solving it.