Potential inside a uniformly charged solid sphere

[agnimusayoti]

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

 

[BvU]

Could you please what is wrong and how to fix it?

Using telepathy ? show us what you do to get that answer !

 

[agnimusayoti]

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

 

[agnimusayoti]

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

 

[BvU]

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]

 

[agnimusayoti]

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]

Well, then what is the relation between $\mu$ and $r$?

 

[BvU]

It might go easier if you choose the axis orientation in such a way that P is on the $z$ axis ...

 

[pasmith]

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 z as the distance from O to P.

 

[vela]

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?

 

[agnimusayoti]

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?

 

[agnimusayoti]

This is strongly hinted at by the question defining z as the distance from O to P.

Which part of the question suggest the P is at z Axis?

 

[vela]

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.

 

[vela]

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.

 

[pasmith]

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 P. It i therefore open to you to orient the axes so that P is at (0,0,z).

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.

 

[agnimusayoti]

Can I assume that z is $r' cos \theta$?
So far: (with Griffith Notation)
$\mu = \sqrt(z^2 +r'^2 - 2zr' cos \theta)$
$$dV = \frac {\rho}{4 \pi \epsilon_0} \frac{r'^2 sin \theta' dr' d\phi' d\theta'}{\sqrt{z^2 +r'^2 - 2zr' cos \theta}} $$

 

[agnimusayoti]

If now I assumme $zz=r'cos\theta$ then:
$$dV=\frac{\rho}{4 \pi \epsilon_0} \frac{r'^2 \sin \theta' dr' d\phi' d\theta'}{\sqrt{r'^2 \cos^2 \theta' +r'^2 - 2r'^2 \cos^2 \theta'}}$$
$$dV=\frac{\rho}{4 \pi \epsilon_0} \frac{r'^2 \sin \theta' dr' d\phi' d\theta'}{\sqrt{r'^2 - r'^2 \cos^2 \theta'}}$$
$$dV=\frac{\rho}{4 \pi \epsilon_0} \frac{r'^2 \sin \theta' dr' d\phi' d\theta'}{\sqrt{r'^2 \sin^2 \theta'}}$$
$$dV=\frac{\rho}{4 \pi \epsilon_0} \frac{r'^2 \sin \theta' dr' d\phi' d\theta'}{r'\sin \theta'}$$
$$dV=\frac{\rho}{4 \pi \epsilon_0} r' dr' d\phi' d\theta'$$

 

[agnimusayoti]

I think this is Wrong. ...

 

[agnimusayoti]

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?

 

[BvU]

I think this is Wrong. ...

Yes, $z$ is fixed !

 

[vela]

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.

 

[agnimusayoti]

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?

 

[agnimusayoti]

Or my domain of integration was wrong?

 

[BvU]

Domain seems OK, but

Because z inside the sphere, so $\ \sqrt{z^2+r'^2 - 2zr'} = r' - z$

is wrong for $\ r' - z<0$

How to express rho in terms of Q ?

How about $Q = \rho V$ ?

 

[vanhees71]

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!

 

[BvU]

I agree, but Griffith clearly intended the current approach ...

 

[agnimusayoti]

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.

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...

 

[vanhees71]

First get the Laplacian in spherical coordinates. Then, due to symmetry, the task reduces to a very simple ordinary differential equation.

 

[BvU]

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 $R$.

 

[agnimusayoti]

Well, until now, I haven't solve the problem yet. I don't knw where is the mistake in my integration.. TT

 

[BvU]

See #23: you need to split the interval in two