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.
  • #36
You made an algebra mistake when you changed variables from ##\theta## to ##u##.
 
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  • #37
vela said:
You made an algebra mistake when you changed variables from ##\theta## to ##u##.
Shoot, I forget to bring r'^2. Thanks now I have the solution pals!
16010995239691100979128371971110.jpg

Is it a general approach to place the point of analysis in z Axis for spherical simmetry? Thanks
 
  • #38
Yes, because the ##z## axis is a preferred axis (the polar axis) in spherical coordinates. Also note that spherical coordinates are singular along the entire ##z##-axis, because Jacobian for the volume element is ##\sqrt{g}=r^2 \sin \vartheta##, which vanishes for ##r=0## as well as for all ##r## if ##\vartheta \in \{0,\pi \}##.
 
  • #39
Well done @agnimusayoti .
Shall we, to top it off, reproduce the result with the much more effective approach proponed by @vanhees71 ?
 
  • #40
agnimusayoti said:
Thanks now I have the solution pals!
It looks like on the last line you may have accidentally moved the factor of ##R## in the first term out of the denominator. The way I like to write the final result is
$$V(r) = \frac{Q}{4\pi\epsilon_0 R}\left(\frac 32-\frac 12 \frac{r^2}{R^2}\right).$$ The factor out front is the potential at the surface, and it's multiplied by a unitless factor that varies from 3/2 at the origin to 1 at ##r=R##.

Is it a general approach to place the point of analysis in z Axis for spherical symmetry?
You might try set up the integral for an arbitrary point. You'll find you end up with an integral that's much more complicated to evaluate, and you'll see why choosing a point where ##\theta = 0## makes a lot of sense.
 
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  • #41
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!
Wait, phi is potential, isn't it?
 
  • #42
vela said:
It looks like on the last line you may have accidentally moved the factor of ##R## in the first term out of the denominator. The way I like to write the final result is
$$V(r) = \frac{Q}{4\pi\epsilon_0 R}\left(\frac 32-\frac 12 \frac{r^2}{R^2}\right).$$ The factor out front is the potential at the surface, and it's multiplied by a unitless factor that varies from 3/2 at the origin to 1 at ##r=R##.You might try set up the integral for an arbitrary point. You'll find you end up with an integral that's much more complicated to evaluate, and you'll see why choosing a point where ##\theta = 0## makes a lot of sense.
Oops. My bad. Yes there is 1/R but I write it at numerator...
 
  • #43
agnimusayoti said:
Wait, phi is potential, isn't it?
Yes!
 

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