Electric field from a cylindrical shell

[phil ess]

Homework Statement

Consider a uniformly charged thin-walled right circular cylindrical shell having total charge Q, radius R, and height h. Determine the electric field at a point a distance d from the right side of the cylinder as shown in the figure below.

Homework Equations

E = kQ/r²

The Attempt at a Solution

Ok so I started by finding the electric field at a distance d from a ring of uniform charge:

E = $$\stackrel{kdQ}{\sqrt{(d\stackrel{2}{}+R\stackrel{2}{})\stackrel{3}{}}}$$

Now I want to treat this shell as a collection of infinitely thin rings, so I need to sum the electric field due to the rings and integrate right? This is where I am having trouble:

E = $$\stackrel{k(d+dx)Q}{\sqrt{((d+dx)\stackrel{2}{}+R\stackrel{2}{})\stackrel{3}{}}}$$?

Then integrate from 0 to h? What about the charge Q? Should it be Q/h?

I am confused as to how I set this integral up. Any help is greatly appreciated! thanks!

 

[ideasrule]

Ok so I started by finding the electric field at a distance d from a ring of uniform charge:

E = $$\stackrel{kdQ}{\sqrt{(d\stackrel{2}{}+R\stackrel{2}{})\stackrel{3}{}}}$$

Almost. However, a simpler way to think about this problem is to use a single variable, x, to denote the distance to a specific ring making up the cylinder. Instead of dQ, you'd use Qdx/h; instead of d^2, you'd use x. Then you just have to integrate.

 

[phil ess]

Almost. However, a simpler way to think about this problem is to use a single variable, x, to denote the distance to a specific ring making up the cylinder. Instead of dQ, you'd use Qdx/h; instead of d^2, you'd use x. Then you just have to integrate.

So then I would have something like

E = $$\frac{kQdx}{h\sqrt{(x\stackrel{2}{}+R\stackrel{2}{})}\stackrel{3}{}}$$?

and integrate from x=d to x=d+h?

 

[ideasrule]

Oops, there's one more error I didn't catch. The electric field due to a ring is kdQ/r^2 *cos(theta)=kdQ/r^2 * x/r, where x is the distance to the ring's center. I think you left out the x.

 

[phil ess]

oh, I had d as the distance to the ring's center, now I am confused? what is the d in your equation for a ring?

 

[ideasrule]

Sorry for being so confusing.


Originally you had this:

E = $$\stackrel{kdQ}{\sqrt{(d\stackrel{2}{}+R\stackrel{2}{})\stackrel{3}{}}}$$

You might want to check this equation. It should be kdQ/(d^2+R^2) * cos(theta), and cos(theta)=d/sqrt(d^2+R^2). You just missed the "d". Now, apply the changes I suggested: x=d, dQ=Qdx/h, etc.

 

[phil ess]

But my textbook has a different formula than that, one without cos theta. here:

So which one do I use?

In my work I replaced z with d to match the diagram, and that 1/4piepsilon = k

 

[phil ess]

Oh I see, it should be d² in the numerator?

 

[phil ess]

Sorry for being so confusing.


Originally you had this:



You might want to check this equation. It should be kdQ/(d^2+R^2) * cos(theta), and cos(theta)=d/sqrt(d^2+R^2). You just missed the "d". Now, apply the changes I suggested: x=d, dQ=Qdx/h, etc.

So youre saying d=x, but then you say that d=dx/h. I think youre getting confused because of the dQ. I didnt write that to mean Delta Q or differential of Q or anything like that. the "d" in "dQ" is the same as the one in the denominator.

EDIT: Ok so this is what I have afer your changes:

$$\int\stackrel{d+h}{d}\frac{kxQ}{h(x\frac{2}{}+R\frac{2}{})\frac{3/2}{}}dx$$

It came out sloppy, should be integral from d to d+h

Please tell me if this is right!