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vela
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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!vela said:You made an algebra mistake when you changed variables from ##\theta## to ##u##.
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 isagnimusayoti said:Thanks now I have the solution pals!
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.Is it a general approach to place the point of analysis in z Axis for spherical symmetry?
Wait, phi is potential, isn't it?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!
Oops. My bad. Yes there is 1/R but I write it at numerator...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.
Yes!agnimusayoti said:Wait, phi is potential, isn't it?