Potential inside a uniformly charged solid sphere

In summary: For question setters, it is not their intent to confuse you. If they denote a distance by a symbol normally used for a coordinate axis, then they are giving you a hint as to how you should set up your problem.
  • #1
agnimusayoti
240
23
Homework Statement
Use Eq 2.29 to calculate potential inside a uniformly charged solid sphere of radius R and total charge q. Compare tour answer to Prob 2.21
Relevant Equations
Eq. 2.29:
$$V(\vec r)=\frac{1}{4 \pi \epsilon_0} \int \frac{\rho (\vec r')}{\mu} d\tau' $$
where ##\mu## is distance from ##d\tau'##
Well, in this problem, I try to use
$$d \tau '= \mu ^2 \sin {\theta} {d\mu} {d\theta} {d\phi}$$
With these domain integration:
$$0<\mu<r$$
$$0<\theta<\pi$$
$$0<\phi<2\pi$$
, I get $$V=\frac{1}{4\pi \epsilon_0} \frac{3Qr^2}{2R^3}$$

This result is wrong because doesn't match with Prob 2.21, which potential is determined with line integral.
I suspect that I made a mistake when define the ##\mu##, which is distance from volume element to point of analysis. Could you please what is wrong and how to fix it? Thanks
 
Physics news on Phys.org
  • #2
agnimusayoti said:
Could you please what is wrong and how to fix it?
Using telepathy ? show us what you do to get that answer !
 
  • #3
BvU said:
Using telepathy ? show us what you do to get that answer !
As I've explained before, i think the distance ##\mu## is the same with integration variable in ##d \tau##.
For details, I attach my work
PicsArt_09-16-04.40.19.jpg
 
  • #4
In my work, I use variable V for volume and potential. So, i changed after take a look again in grifth. He used tau to define volume. So in thia forum I use tau
 
  • #5
Using ##\tau## or ##\tau'## is just fine (it is only a name).
But I don't see the distance ##\mu## from a point ##\vec r## to ##\vec r'## of your volume element ##d\tau'## anywhere.
Make a sketch to convince yourself that ##\mu\ne|\vec r'|##

[edited formula]
 
  • #6
BvU said:
Using ##\tau## or ##\tau'## is just fine (it is only a name).
But I don't see the distance ##\mu## from a point ##\vec r## to ##\vec r'## of your volume element ##d\tau'## anywhere.
Make a sketch to convince yourself that ##\mu\ne|\vec r'|##

[edited formula]
IMG_20200916_184628.jpg

Well, then what is the relation between ##\mu## and ##r##?
 
  • #7
It might go easier if you choose the axis orientation in such a way that P is on the ##z## axis ...
1600272454098.png
 
  • Like
  • Love
Likes PhDeezNutz and pasmith
  • #8
BvU said:
It might go easier if you choose the axis orientation in such a way that P is on the ##z## axis ...

This is strongly hinted at by the question defining [itex]z[/itex] as the distance from [itex]O[/itex] to [itex]P[/itex].
 
  • #9
Having seen multiple threads from you, I've noticed that you always seem to get stuck on what ##\vec r## is, what ##\vec r'## is, and what Griffith's denotes as script r is. Can you describe what each represents in words, both in general and in the context of this problem?
 
  • #10
BvU said:
It might go easier if you choose the axis orientation in such a way that P is on the ##z## axis ...
View attachment 269522
Well, this sketch really helpful to answer Vela's question. From this sketch, (also in general):
##\vec r## is position of the infinitesimal element from origin.
##\vec r'## is position of point of analysis (in this case is P)
##\vec \mu## is position of point of analysis wrt infinitesimal element.
Is that right?
 
  • #11
pasmith said:
This is strongly hinted at by the question defining [itex]z[/itex] as the distance from [itex]O[/itex] to [itex]P[/itex].
Which part of the question suggest the P is at z Axis?
 
  • #12
agnimusayoti said:
Well, this sketch really helpful to answer Vela's question. From this sketch, (also in general):
##\vec r## is position of the infinitesimal element from origin.
##\vec r'## is position of point of analysis (in this case is P)
##\vec \mu## is position of point of analysis wrt infinitesimal element.
Is that right?
Based on the BvU's figure, that's right. Griffiths, however, uses ##\vec r## for the position of P and ##\vec r'## for the position of the charge element, and he defines ##\vec \mu = \vec r - \vec r'##. In my copy of the book (2nd edition), Figure 2.3 illustrates this convention for discrete charges, but it's the same idea.

In the integral, the volume element ##d\tau## is always ##d\tau = dx'\,dy'\,dz'## in cartesian coordinates and ##d\tau = r'^2\sin\theta'\,dr'\,d\theta'\,d\phi'## in spherical coordinates. Fix that in your attempt and see where you get to.
 
  • #13
agnimusayoti said:
Which part of the question suggest the P is at z Axis?
The question doesn't, but if you look at the various figures in Griffiths, you might notice the distance from O to P is typically labeled ##z##.

When you have spherical symmetry, as you do in this problem, you should know that ##V## can only depend on ##r## and not on ##\theta## or ##\phi##. You have the freedom to choose any point that's a distance ##r## from the origin, so you might as well choose the point on the ##z## axis to simplify the math.
 
  • #14
agnimusayoti said:
Which part of the question suggest the P is at z Axis?

The charged sphere is spherically symmetric. This symmetry is broken only by the designation of a particular point as [itex]P[/itex]. It i therefore open to you to orient the axes so that [itex]P[/itex] is at [itex](0,0,z)[/itex].

Question setters are not trying to confuse you. If they denote a distance by a symbol normally used for a coordinate axis, then they are giving you a hint as to how you should set up your axes.
 
  • #15
Can I assume that z is ##r' cos \theta##?
So far: (with Griffith Notation)
##\mu = \sqrt(z^2 +r'^2 - 2zr' cos \theta)##
$$dV = \frac {\rho}{4 \pi \epsilon_0} \frac{r'^2 sin \theta' dr' d\phi' d\theta'}{\sqrt{z^2 +r'^2 - 2zr' cos \theta}} $$
 
  • #16
If now I assumme ##zz=r'cos\theta## then:
$$dV=\frac{\rho}{4 \pi \epsilon_0} \frac{r'^2 \sin \theta' dr' d\phi' d\theta'}{\sqrt{r'^2 \cos^2 \theta' +r'^2 - 2r'^2 \cos^2 \theta'}}$$
$$dV=\frac{\rho}{4 \pi \epsilon_0} \frac{r'^2 \sin \theta' dr' d\phi' d\theta'}{\sqrt{r'^2 - r'^2 \cos^2 \theta'}}$$
$$dV=\frac{\rho}{4 \pi \epsilon_0} \frac{r'^2 \sin \theta' dr' d\phi' d\theta'}{\sqrt{r'^2 \sin^2 \theta'}}$$
$$dV=\frac{\rho}{4 \pi \epsilon_0} \frac{r'^2 \sin \theta' dr' d\phi' d\theta'}{r'\sin \theta'}$$
$$dV=\frac{\rho}{4 \pi \epsilon_0} r' dr' d\phi' d\theta'$$
 
  • #17
I think this is Wrong. ...
 
  • #18
vela said:
The question doesn't, but if you look at the various figures in Griffiths, you might notice the distance from O to P is typically labeled ##z##.

When you have spherical symmetry, as you do in this problem, you should know that ##V## can only depend on ##r## and not on ##\theta## or ##\phi##. You have the freedom to choose any point that's a distance ##r## from the origin, so you might as well choose the point on the ##z## axis to simplify the math.
Thanks for the information. Now, what I should to do is the integration right? Does the integration easy to be solved?
 
  • #19
agnimusayoti said:
I think this is Wrong. ...
Yes, ##z## is fixed !
 
  • #20
agnimusayoti said:
Can I assume that z is ##r' cos \theta##?
You mean ##r' \cos\theta'##, right? ##\theta## and ##\theta'## aren't interchangeable.

You don't need to assume anything. What do ##r' \cos\theta'## and ##z## represent? If they correspond to the same thing, you can set them equal to each other. If not, you can't.

##r' \cos\theta'## is the ##z##-coordinate of ##\vec r'##, the position of the infinitesimal element of charge. ##z## is the ##z##-coordinate of ##\vec r##, the point ##P## of interest. They're clearly not the same.
 
  • #21
Ok, now this is my correction:
$$V= \frac{\rho}{4 \pi \epsilon_0} (2\pi) \int_{r'_1 = 0}^{r'_2=R} \int_{\theta'_1 = 0}^{\theta'_2 = \pi} \frac{r'^2 \sin \theta'}{\sqrt{z^2+r'^2 - 2zr' cos \theta'}} d\theta' dr'$$
$$V= \frac{(2\pi)\rho}{4 \pi \epsilon_0} \int_{r'_1 = 0}^{r'_2=R} \frac{r'}{z} \left[\sqrt{z^2+r'^2 + 2zr'} - \sqrt{z^2+r'^2 - 2zr'}\right] dr'$$
Because z inside the sphere, so ##\sqrt{z^2+r'^2 - 2zr'} = r' - z##
And the other root is ##\sqrt{z^2+r'^2 +2zr'} =r' + z##
Therefore:
$$V= \frac{(2\pi)\rho}{4 \pi \epsilon_0} \int_{r'_1 = 0}^{r'_2=R} \frac{r'}{z} \left[(r' + z) - (r' - z)\right] dr'$$
$$V= \frac{(4\pi)\rho}{4 \pi \epsilon_0} \int_{r'_1 = 0}^{r'_2=R} r'dr'$$
$$V= \frac{(2\pi)\rho}{4 \pi \epsilon_0} \left[r'^2\right]_0^R dr'$$
$$V= \frac{(2\pi)\rho}{4 \pi \epsilon_0} R^2$$
$$V= \frac{(2\pi)\rho}{4 \pi \epsilon_0} R^2$$
How to express rho interms of Q?
 
Last edited:
  • #22
Or my domain of integration was wrong?
 
  • #23
Domain seems OK, but
agnimusayoti said:
Because z inside the sphere, so ##\ \sqrt{z^2+r'^2 - 2zr'} = r' - z##
is wrong for ##\ r' - z<0##

agnimusayoti said:
How to express rho in terms of Q ?
How about ##Q = \rho V## 😁 ?
 
  • Like
Likes pasmith
  • #24
This thread shows, why differential equations are so much preferable to using general solutions. After the confusion is hopefully solved in this thread, I strongly recommend to solve the problem, using the local form of electrostatics,i .e.,
$$\Delta \Phi=-\frac{1}{\epsilon_0} \rho,$$
making use of the simplifying fact that in this problem ##\rho=\rho(r)## only and thus also ##\Phi## should be a function of ##r## only. Expressing then the Laplacian in spherical coordinates and applying it to this highly symmetric case, makes the solution a no-brainer!
 
  • Like
Likes BvU and etotheipi
  • #25
I agree, but Griffith clearly intended the current approach ...
 
  • #26
BvU said:
Domain seems OK, but
is wrong for ##\ r' - z<0##

How about ##Q = \rho V## 😁 ?
Hmm, so the ##\rho## should evaluated in the integration? I excluded it from integration because I think this charge density is constant.

vanhees71 said:
This thread shows, why differential equations are so much preferable to using general solutions. After the confusion is hopefully solved in this thread, I strongly recommend to solve the problem, using the local form of electrostatics,i .e.,
$$\Delta \Phi=-\frac{1}{\epsilon_0} \rho,$$
making use of the simplifying fact that in this problem ##\rho=\rho(r)## only and thus also ##\Phi## should be a function of ##r## only. Expressing then the Laplacian in spherical coordinates and applying it to this highly symmetric case, makes the solution a no-brainer!

Well, I just started to begin the study of special technique using Laplacian. But, I still haven't get the idea what the special techniques told to me...
 
  • #27
First get the Laplacian in spherical coordinates. Then, due to symmetry, the task reduces to a very simple ordinary differential equation.
 
  • #28
agnimusayoti said:
Hmm, so the ρ should evaluated in the integration?
No, you asked how ##\rho##, a constant, could be expressed in terms of ##q##, the total charge of a uniformly charged sphere with radius ##R##.
 
  • #29
Well, until now, I haven't solve the problem yet. I don't knw where is the mistake in my integration.. TT
 
  • #30
See #23: you need to split the interval in two
 
  • #31
Isn't it now at the time to give the solution?
 
  • #32
I expect @agnimusayoti already has the solution (from the Gauss approach, problem 2.21) and wants to see the integral yields the same result.
 
  • #33
That's what I meant: Shouldn't we just show, how to get the integral now? I think it's good to finally have the solution in the thread, even if the OP couldn't completely solve it him/herself.
 
  • #34
Ok, should I integrate from r' = 0 to z and from r' = z to R ?
Yeah, I can calculate the potential using Gauss Law and want to determine the same potential using another method. ..
I really appreciate if I can find the answer in this thread...
 
  • Like
Likes vanhees71
  • #35
1601036171459..jpg
 

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
354
  • Advanced Physics Homework Help
Replies
7
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
2K
  • Advanced Physics Homework Help
Replies
6
Views
2K
  • Advanced Physics Homework Help
Replies
0
Views
536
  • Advanced Physics Homework Help
Replies
7
Views
1K
  • Advanced Physics Homework Help
Replies
16
Views
2K
  • Advanced Physics Homework Help
Replies
3
Views
1K
Replies
16
Views
1K
  • Advanced Physics Homework Help
Replies
10
Views
2K
Back
Top