- #1

sportfreak801

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## Homework Statement

A spherical distribution of charge is characterized by a constant charge density [tex] \rho [/tex] for r<= R. For radii greater than R, the charge density is zero. Find the potential [tex] \varphi [/tex] (r) by integrating Poisson's equation.

## Homework Equations

[tex]

\nabla^2(\varphi)=-\frac{\rho}{\epsilon_{0}}

[/tex]

## The Attempt at a Solution

I tried taking the triple integral of poison's equation using spherical coordinates to find the potential, u, and found

[tex]

\int_V\nabla^2(\varphi) r^2 \sin(\theta) dr d\theta d\phi

[/tex]

[tex]

\int\int\int_{0}^{\pi}\nabla^2(\varphi) r^2 \sin(\theta) d\phi dr d\theta

[/tex]

Plugging in the known value for [tex] \nabla^2(\varphi) [/tex] the equation becomes

[tex]

\int\int\int_{0}^{\pi}-\frac{\rho}{\epsilon_{0}} r^2 \sin(\theta) d\phi dr d\theta

[/tex]

The integral from 0 to [tex] \pi [/tex] of d[tex] \phi [/tex] equals [tex] \pi [/tex]. And since [tex] \rho [/tex] is constant throughout the sphere, the equation becomes:

[tex]

\frac{-\pi*\rho}{\epsilon_0}\int_{r}^{R}\int_{0}^{\pi} r^2 \sin(\theta) d\theta dr

[/tex]

The integral with respect to d[tex] \theta [/tex] is 2. The equation now becomes:

[tex]

\frac{-2*\pi*\rho}{\epsilon_0}\int_{r}^{R} r^2 dr

[/tex]

Evaluating the integral with respect to dr yields

[tex]

\frac{-2*\pi*\rho}{\epsilon_0}(\frac{R^3}{3} - \frac{r^3}{3})

[/tex]

Thus, I find the answer to be

[tex]

\varphi = \frac{-2*\pi*\rho}{3*\epsilon_0}(R^3 - r^3)

[/tex]

I don't think this is correct because the electric potential is a function of 1/r and not r^3.

Any help or clarification would be greatly appreciated. Thanks in advance!

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