Electric field of a hemispheric vessel with uniform surface charge distribution

[pc2-brazil]

I'm reading a Physics (electricity and magnetism) book for self-study. I tried to do the exercise below, but I'm not getting the correct answer. This is not actually a homework (since it's not part of a course).
I will try to carefully reproduce the steps I took below, so that someone can help identify the mistake.

Thank you in advance.

Homework Statement

A hemispheric vessel of internal radius R has a total charge q uniformly distributed through its internal surface. Determine the electric field at the center of curvature.

Homework Equations

$$\vec{E} = \int d\vec{E}$$
$$E = \frac{kq}{d^2}$$, where $$k = \frac{1}{4\pi\epsilon_0}$$
Electric field produced by a charged ring at a point located at a distance z along its central axis:
$$E = \frac{kqz}{(z^2 + R^2)^{3/2}}$$

The Attempt at a Solution

First, I will call P the point at the center of curvature, and R the radius of the hemisphere.
I will divide the hemisphere into a pile of rings of charge dq and radius r at a distance z from point P. The magnitude of the electric field produced at that point by a ring is:
$$dE = \frac{kz}{(z^2 + r^2)^{3/2}} dq$$
The relation between the radius r of the ring, the radius R of the hemisphere and the distance z from the center of the ring to point P is:
$$R^2 = z^2 + r^2$$
So, I can substitute it into the expression for dE:
$$dE = \frac{kz}{R^3} dq$$
Now I have to express dq in terms of another variable. I will call σ the surface charge density of the hemisphere. So:
$$dq = \sigma dA$$
where dA is the area of the ring. Now, I can either (1) express dA in terms of z and express r in terms of z or (2) express dA in terms of r and express z in terms of r. I will choose (2):
$$dq = \sigma \times 2\pi r dr$$
Then, the expression for dE becomes:
$$dE = \frac{kz}{R^3} \times \sigma \times 2\pi r dr$$
$$dE = \frac{kz\sigma\times2\pi r}{R^3} dr$$
In terms of r, z can be written as $$\sqrt{R^2 - r^2}$$; so, the above can be rewritten as:
$$dE = \frac{k\sqrt{R^2 - z^2}\sigma\times2\pi r}{R^3} dr$$
Now, I want to obtain the electric field produced by all rings at that point. Since the electric fields produced by all rings will be at the same direction and orientation, I will just sum their magnitudes by integrating, with r going from 0 to R:
$$E = \int_0^R \frac{k\sqrt{R^2 - r^2}\sigma\times2\pi r}{R^3} dr$$
$$E =\frac{2\pi k \sigma}{R^3} \int_0^R r\sqrt{R^2 - r^2}\ dr$$
$$E = \frac{2\pi k \sigma}{R^3} \left ( -\frac{1}{3}(R^2 - R^2)^{3/2} + \frac{1}{3}(R^2 - 0^2)^{3/2} \right )$$
$$E = \frac{2\pi k \sigma}{R^3} \left (\frac{1}{3}R^3 \right )$$
$$E = \frac{2\pi k \sigma}{3}$$
Now, since the surface area of the internal surface of the hemisphere is 2πr² (half the surface area of a sphere), σ is given by:
$$\sigma = \frac{q}{2\pi R^2}$$
So, the electric field is:
$$E = \frac{kq}{3R^2}$$. Now, I rewrite k as $$\frac{1}{4\pi\epsilon_0}$$ and get:
$$E = \frac{q}{12\pi\epsilon_0 R^2}$$
This is the value I got. But the book's answer is different. It is:
$$E = \frac{q}{8\pi\epsilon_0 R^2}$$
What did I get wrong?

Thank you in advance.

 

[tiny-tim]

hi pc2-brazil!

your area A is wrong, it isn't 2πrdr, that would be the area of a flat ring

 

[eczeno]

Homework Statement

A hemispheric vessel of internal radius R has a total charge q uniformly distributed through its internal surface. Determine the electric field at the center of curvature.

Homework Equations

$$\vec{E} = \int d\vec{E}$$
$$E = \frac{kq}{d^2}$$, where $$k = \frac{1}{4\pi\epsilon_0}$$
Electric field produced by a charged ring at a point located at a distance z along its central axis:
$$E = \frac{kqz}{(z^2 + R^2)^{3/2}}$$

The Attempt at a Solution

First, I will call P the point at the center of curvature, and R the radius of the hemisphere.
I will divide the hemisphere into a pile of rings of charge dq and radius r at a distance z from point P. The magnitude of the electric field produced at that point by a ring is:
$$dE = \frac{kz}{(z^2 + r^2)^{3/2}} dq$$
The relation between the radius r of the ring, the radius R of the hemisphere and the distance z from the center of the ring to point P is:
$$R^2 = z^2 + r^2$$
So, I can substitute it into the expression for dE:
$$dE = \frac{kz}{R^3} dq$$
Now I have to express dq in terms of another variable. I will call σ the surface charge density of the hemisphere. So:
$$dq = \sigma dA$$
where dA is the area of the ring. Now, I can either (1) express dA in terms of z and express r in terms of z or (2) express dA in terms of r and express z in terms of r. I will choose (2):
$$dq = \sigma \times 2\pi r dr$$

hi,

i am with you up to here, but your expression for $dA$ is incorrect. you have

$$dA=2\pi r dr$$

but this implies that we should integrate over r, which is not what we want to do. we want to integrate over the inner surface of the hemisphere, so to speak. but the distance between the center and any place on the surface is always R, it is constant.

you have the right idea about using rings, this will work, but you really have to be able to picture what that looks like to get the right formula. and, we are going to have to introduce an angle, because, if you can picture it, that is what is changing as the rings sweep across the surface.

we are going to need polar coordinates for this problem (unless you want it to get really messy). Draw a right hemicircle and label the center P, the top point B and the point directly east of P, call A. This is a cross-section of the hemisphere. draw a line from P to A, this will be the line we measure $\theta$ from. now, draw a radius to a point somewhere between A and B so that it makes an angle $\theta$ with the line PA, call the point where it intersects the circle C. now, shade in a tiny dark strip on the circle at C.

ok, let me know if you followed that. i am going to continue writing as if you are with me, but i am happy to come back to this.
because of symmetry, we only need one variable, call it $\theta$.

 

[eczeno]

ok, if you have the figure drawn as i described. the small dark band you made in the last step is the cross-section of one of your rings (you might want to put one in the lower half of the circle too, so you can see that it is a complete ring). we are going to let those rings go from tiny little rings around point A to big rings at the opening of the hemisphere by integrating from $\theta=0$ to $\theta=\pi / 2$. and $dA$ should be the area of one of these rings, which will depend on $\theta$ and must include $d\theta$ because that is what we will integrate over. let me know if you need help finding $dA$.

cheers.

 

[pc2-brazil]

i am with you up to here, but your expression for $dA$ is incorrect. you have

$$dA=2\pi r dr$$

but this implies that we should integrate over r, which is not what we want to do. we want to integrate over the inner surface of the hemisphere, so to speak. but the distance between the center and any place on the surface is always R, it is constant.

What if I integrated over z (the distance between the center of each ring and point P)?
$$dA = 2\pi r dz$$
$$dq = 2 \sigma \pi r dz$$
$$dE = \frac{2kz\sigma \pi r}{R^3} dz = \frac{2kz\sigma \pi \sqrt{R^2 - z^2}}{R^3} dz$$
To integrate over z appears to make more sense, but I still get the same result.
I think I followed your explanation and made a correct picture, but I'm not very sure on how I would use $d\theta$ to find the expression for the area.
It seems that the area of the internal surface of each ring should be its circumference length multiplied by its infinitesimal height (as it were a hollow cylinder of infinitesimal height).

 

[tiny-tim]

hi pc2-brazil!

dA = 2πr dz is still wrong …

that would be the area of a slice of a cylinder (with vertical sides), not of a sphere

 

[eczeno]

that is not going to work. the width of each ring is going to be the little bit of arc length along the circle. this is going to be very hard to describe in rectangular coordinates. the best way to go is polar coordinates. are you familiar with polar coordinates?

 

[pc2-brazil]

eczeno: I've never been formally introduced to polar coordinates, but I'm more or less familiar with it, since it has already appeared many times during my self-study of some topics.
I've just managed to get the right answer. I think the method is now right.
I've used an angle, as eczeno described (refer to the attached picture).
I called ds (not dz) the height of the ring. Then, the area of each small ring is:
$$dA = 2\pi r\ ds = 2\pi r R d\theta$$
Then, dE becomes:
$$dE = \frac{kz\sigma \times 2\pi r R\ d\theta}{R^3}$$
where:
$$\cos\theta = \frac{z}{R}$$ and $$\sin\theta = \frac{r}{R}$$
So:
$$dE = \frac{kR\cos(\theta)\sigma \times 2\pi R\sin(\theta) R\ d\theta}{R^3}$$
$$dE = k\cos(\theta)\sigma \times 2\pi \sin(\theta)\ d\theta$$
$$E = 2k\sigma\pi\int_0^{\pi/2} \cos(\theta)\sin(\theta)\ d\theta$$
$$E = 2k\sigma\pi\times\frac{1}{2} = k\sigma\pi = \frac{1}{4\pi\epsilon_0}\pi\frac{q}{2\pi R^2}$$
$$E = \frac{q}{8\pi\epsilon_0R^2}$$

 

[eczeno]

yes, that's it exactly! (well, the dA anyway, i will trust you that the rest is correct). nicely done.

cheers