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Introductory Physics Homework Help
Determine the total electrostatic potential energy
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[QUOTE="hitemup, post: 5049332, member: 525312"] [h2]Homework Statement [/h2] Determine the total electrostatic potential energy of a nonconducting sphere of radius [itex]r_0[/itex] carrying a total charge [itex]Q[/itex] distributed uniformly thorughout its volume. [h2]Homework Equations[/h2] U = qV [h2]The Attempt at a Solution[/h2] [tex]\rho = \frac{Q}{4/3\pi r_0^3} = \frac{dq}{4\pi r^2dr}[/tex] Electric field inside a nonconductor sphere [tex]V = \frac{kQ}{2r_0}(3-\frac{r^2}{r_0^2})[/tex] [tex]dU = Vdq[/tex] [tex]= \frac{kQ}{2r_0} (3 - \frac{r^2}{r_0^2}) \rho4\pi r^2 dr[/tex] [tex]= \frac{2kQ\pi\rho}{r_0}(3r^2 -\frac{r^4}{r_0^2} )dr[/tex] After integrating it over [from zero to [itex]r_0[/itex]], I end up with the following result [tex] \frac{3Q^2}{10\pi r_0 \epsilon_0}[/tex] But the correct answer according to the textbook is this. [tex]\frac{3Q^2}{20\pi r_0 \epsilon_0}[/tex] It's almost the same result but missing a factor of two in the denominator. Is it because of the potential equation? Solutions manual uses potential at the surface, but in my answer I use potential inside a nonconductor. That may be the reason of 1/2. (kq/r_0 vs kq/[B]2[/B]r_0* (3 - r^2/r_0^2))[ATTACH=full]176935[/ATTACH] [/QUOTE]
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Introductory Physics Homework Help
Determine the total electrostatic potential energy
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