- #36

#### vela

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You made an algebra mistake when you changed variables from ##\theta## to ##u##.

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- Thread starter agnimusayoti
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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

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You made an algebra mistake when you changed variables from ##\theta## to ##u##.

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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##.

Is it a general approach to place the point of analysis in z Axis for spherical simmetry? Thanks

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Shall we, to top it off, reproduce the result with the much more effective approach proponed by @vanhees71 ?

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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!

$$V(r) = \frac{Q}{4\pi\epsilon_0 R}\left(\frac 32-\frac 12 \frac{r^2}{R^2}\right).$$ 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.Is it a general approach to place the point of analysis in z Axis for spherical symmetry?

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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!

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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.

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Yes!agnimusayoti said:Wait, phi is potential, isn't it?

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