Electric field a distance z from the center of a spherical surface

  • #1
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Homework Statement:

Find the electric field a distance z from the center of a spherical surface of radius R which carries a uniform charge density ##\sigma##. Hint: use the law of cosines to write r in terms of R and ##\theta## where ##\theta## is inclination/elevation angle of sphere.

Relevant Equations:

Coulomb's Law: $$dE=\frac{1}{4 \pi \epsilon_0} \frac{dq}{r^2}$$
Well, I really don't understand what is the use of the hint.
I try to solve this problem with Coulomb's Law and try to do in spherical coordinates and got very messy infinitesimal field due to the charge of infinitesimal surface element of the sphere.
Here what I got:
$$\vec{r}=\vec{r_P} + \vec{R}$$ where ##\vec{r}## is position of the point of field (P in question) relative to infinitesimal surface element; ##\vec{r_P}## is position of the point of field relative to the origin, and ##\vec{R}## is position of vector radius of the sphere; as shown in figure.

I get, $$\vec{r} = (z cos \theta - R)\hat{r} - z sin\theta \hat{\theta}$$.
With the domain of integration ##0<=\theta<=\pi## ##0<=\\phi<=2\pi##; how I can solve the integration and find the electric field?

Second, is there another method which is easier? Thankss
 

Answers and Replies

  • #2
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This is Prob 2.7 at Griffith Introduction to Electrodynamics
 
  • #3
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1) The hint was to use Geometry of problem to find ##r^2## rather then vectors. Since, you used Vectors, that is just fine.

2) The integral is easy to do in in this case.
 
  • #4
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Yes I Made a mistake. The ##\vec{r_P} = z \hat {r} ## so that the integral can be done easily in spherical coordinate. I get $$E = \frac{1}{4\pi \epsilon_0} \frac {Q}{(z-R)^2}$$. Is that true?
 
  • #5
vanhees71
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Check it again! If you want to do it calculating the integral (which I don't recommend) here are some hints (I've no clue what to make out of the hint in the problem. What's "the law of cosines"?).

(a) the electric field is a vector
(b) you have to carefully distinguish the cases ##z<R## and ##z>R##

However, to take the integral is overcomplicating things since the problem doesn't force you to do so. It's much simpler to solve the potential problem
$$\Delta \Phi=-\frac{1}{\epsilon_0} \rho$$
with
$$\rho(\vec{r})=\sigma \delta(r-R).$$
Note that the charge distribution is spherically symmetric and thus the problem is very simple in spherical coordinates using directly the differential equation.
 
  • #7
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Unfortunately I havent learnt potential yet. :(
I expect zero electric field inside the Shell, which I don't get on my answer
 
  • #8
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Check your solution again.

It is better to write the position vectors and unit vectors using Cartesian unit vectors since they are fixed during integration whereas spherical units vary.
 
  • #9
vela
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Yes I Made a mistake. The ##\vec{r_P} = z \hat {r} ## so that the integral can be done easily in spherical coordinate. I get $$E = \frac{1}{4\pi \epsilon_0} \frac {Q}{(z-R)^2}$$. Is that true?
No. It's a spherically symmetric charge distribution, so the field outside of the sphere will be identical to that of a point charge at the origin.
 
  • #10
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$$\rho(\vec{r})=\sigma \delta(r-R).$$

Shouldn't this be:

$$\rho(\vec{r})=\frac {\sigma}{4\pi R^2}\delta(r-R)$$
 
  • #11
vanhees71
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No, ##\sigma## is the surface-charge density. What you have in mind is ##\sigma=Q/(4 \pi R^2)## I guess.
 
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  • #12
vela
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I'm going to stick with Griffiths' notation because I find yours hard to keep straight. The location of point ##P## is ##\vec r = z\,\hat k##, and the location of the infinitesimal area on the sphere is ##\vec r' = R\,\hat r'##. Their difference is what Griffiths denotes by script r, which I'll write as ##\vec r_s = \vec r - \vec r'##.

One mistake you made was assuming ##\hat r## is the same for ##\vec r## and ##\vec r'##. It's not.

The hint given is to use the law of cosines to find ##r_s = \| \vec r_s \|##, but you can find the same result using the vectors:
$$\| \vec r_s \|^2 = (\vec r - \vec r')^2 = \vec r^2 + \vec r'^2 - 2 \vec r \cdot \vec r' = z^2 + R^2 - 2 z R \cos\theta$$ where ##\theta## is the angle between the ##z##-axis and ##\vec r'##. Note that this is different than the expression you obtained. Because you effectively assumed ##\vec r## and ##\vec r'## point in the same direction (same ##\hat r##), you had ##\theta = 0##, so you ended up with ##r_s = z^2 + R^2 - 2zR = (z-R)^2##.
 
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  • #13
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I'm going to stick with Griffiths' notation because I find yours hard to keep straight. The location of point ##P## is ##\vec r = z\,\hat k##, and the location of the infinitesimal area on the sphere is ##\vec r' = R\,\hat r'##. Their difference is what Griffiths denotes by script r, which I'll write as ##\vec r_s = \vec r - \vec r'##.

One mistake you made was assuming ##\hat r## is the same for ##\vec r## and ##\vec r'##. It's not.

The hint given is to use the law of cosines to find ##r_s = \| \vec r_s \|##, but you can find the same result using the vectors:
$$\| \vec r_s \|^2 = (\vec r - \vec r')^2 = \vec r^2 + \vec r'^2 - 2 \vec r \cdot \vec r' = z^2 + R^2 - 2 z R \cos\theta$$ where ##\theta## is the angle between the ##z##-axis and ##\vec r'##. Note that this is different than the expression you obtained. Because you effectively assumed ##\vec r## and ##\vec r'## point in the same direction (same ##\hat r##), you had ##\theta = 0##, so you ended up with ##r_s = z^2 + R^2 - 2zR = (z-R)^2##.
Nah. At the first, I think ##\vec {r}## is different with ##\vec{r'}## but in spherical coordinate the integral was not good, because I need to change ##\hat{k}## in to unit vector in spherical coordinate.

Then, If I want to use Cartesian, then I must define ##\hat{r'}## in to unit vector Cartesian,
which is the same for unit vector radial in spherical coordinate. Is this true?
So, $$\hat{r'}=\sin \theta \cos\phi \hat{i} +\sin \theta \sin\phi \hat{j} + \cos \theta \hat{k}$$

Or, $$\hat{r'}=\sin \theta \hat{j} + \cos \theta \hat {k}$$?

I think the second is the true one, becouse it give the same result as you have shown.
But, how can the problem become two dimensional problem? At first, I think ##\hat {r'}## is a 3D vector..
 
  • #14
vela
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Nah. At the first, I think r→ is different with r′→ but in spherical coordinate the integral was not good, because I need to change k^ in to unit vector in spherical coordinate.
You wrote ##\vec r = z\,\hat r## (your ##\vec r_P##) and ##\vec r' = R\,\hat r## (your ##\vec R##). Then you said ##\vec r_s = z\,\hat r - R \,\hat r = (z-R)\,\hat r##. By treating the ##\hat r## in ##\vec r## and the ##\hat r## in ##\vec r'## as equal (because that's what you did when you factored it out), you're making ##\vec r## and ##\vec r'## point in the same direction. That may not be what you meant to do, but that's what you did.
 
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  • #15
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Yup, I wrote and assume the unit vector is radially outward because I hope the integration is easy to do.
IMG_20200907_120453.jpg


At this stage, am I correct?
 
  • #16
vela
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Yup, I wrote and assume the unit vector is radially outward because I hope the integration is easy to do.
I'm not sure you're understanding my point.

At this stage, am I correct?
No. It's wrong for the reason you suspected. This isn't a two-dimensional problem. You can, however, use symmetry to simplify the problem.
 
  • #17
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Hmm. Then How can I solve this problem? What symmetry do you mean?
 
  • #18
vanhees71
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Well, let's write down the correct "relevant equation",
$$\vec{E}(\vec{r})=\frac{\sigma}{4 \pi \epsilon_0} \int_{S_R} \mathrm{d}^2 f' \frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^3}.$$
Now parametrize the spherical shell with radius ##R## with spherical coordinates. Think about first how to express ##\mathrm{d}^2 f'## in terms of the spherical coordinates.

Further you can assume from symmetry that ##\vec{E}=E_r \vec{e}_r## and then choose ##\vec{r}=r \vec{e}_z##, which implifies the integral considerably.
 
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  • #19
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Now parametrize the spherical shell with radius ##R## with spherical coordinates. Think about first how to express ##\mathrm{d}^2 f'## in terms of the spherical coordinates.
Is ##d^2f## same with ##dA## surface infinitesimal element? Then, ##d^2f=R^2 \sin \theta d\theta d\phi## in spherical coordinates. Is this right?
 
  • #20
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I look Youtube video about this problem, but I don't understand how one can use spherical coordinate, but at the same moment, he got the electric field is at the z direction, which is Cartesian.
 
  • #21
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Further you can assume from symmetry that ##\vec{E}=E_r \vec{e}_r## and then choose ##\vec{r}=r \vec{e}_z##, which implifies the integral considerably.
If I choose ##\vec{r}=r \vec{e}_z##; then what kind of coordinate that I am going to use?
 
  • #22
vanhees71
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Is ##d^2f## same with ##dA## surface infinitesimal element? Then, ##d^2f=R^2 \sin \theta d\theta d\phi## in spherical coordinates. Is this right?
right!
 
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  • #23
vanhees71
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If I choose ##\vec{r}=r \vec{e}_z##; then what kind of coordinate that I am going to use?
Just calculate ##\vec{r}-\vec{r}'## and ##|\vec{r}-\vec{r}'|## to get the integral and then solve it!
 
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  • #25
vanhees71
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Looks good. Now express ##\hat{e}_r## in terms of ##\theta## and ##\varphi## and Cartesian basis vectors. Then you can go on and evaluate the surface integral.
 
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