# Electrodynamics Potential from charged sphere. I am lost :/

~Electrodynamics~ Potential from charged sphere. I am lost :/

## Homework Statement

A sphere of radius R, centered at the origin, carries a charge density ρ(r,θ)=κ/r^2(R-2r)sin^2(θ).
κ is constant. Find exact potential.

1/4∏ε∫ρ∂t/r

## The Attempt at a Solution

Question and attempt via attachments :)
I do not think my solution is correct please help!
Thanks all for looking!

## Answers and Replies

Shouldn't it be ##\frac{...}{\sqrt{z^2+r^2-2zr\cos\theta}}## instead of ##\frac{...}{\sqrt{R^2+r^2-2Rr\cos\theta}}## ?

as ##\sqrt{(z-r\cos\theta)^2+r^2\sin^2\theta}=\sqrt{z^2+r^2-2rz\cos\theta}## is a distance from the point in the sphere to the point in z-axis

Okay, yeah that makes sense. So the numerator looked good? If possible could you please work it out? I MUST have the integration done correctly! Thank you for responding!

Yes, the numerator is correct, you get:

##V(z)=\frac{K}{4\pi\varepsilon}\int_0^{2\pi}\int_0^R\int_0^\pi\frac{(R-2r)\sin^3\theta}{\sqrt{z^2+r^2-2zr\cos\theta}}\,d\theta drd\varphi##

so try a substitution ##u=z^2+r^2-2zr\cos\theta##

I don't know what is wrong. I calculated it in such a way:

##V(z)=\frac{K}{4\pi\varepsilon}\int_0^{2\pi}\int_0^R\int_0^\pi\frac{(R-2r)\sin^3\theta}{\sqrt{z^2+r^2-2zr\cos\theta}}\,d\theta drd\varphi=\frac{K}{2\varepsilon}\int_0^R\int_0^\pi\frac{(R-2r)(1-\cos^2\theta)\sin\theta}{\sqrt{z^2+r^2-2zr\cos\theta}}\,d\theta dr=(*)\\
z^2+r^2-2zr\cos\theta=u\\
\cos\theta=\frac{r^2+z^2-u}{2zr}\\
\sin\theta\,d\theta=\frac{du}{2zr}\\
(*)=\frac{K}{2\varepsilon}\int_0^R\int_{(z-r)^2}^{(z+r)^2}\frac{(R-2r)\left(1-\left(\frac{r^2+z^2-u}{2zr}\right)^2\right)}{2zr\sqrt{u}}\,dudr=...##

vela
Staff Emeritus
Science Advisor
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Okay, yeah that makes sense. So the numerator looked good? If possible could you please work it out? I MUST have the integration done correctly! Thank you for responding!
That's not how it works here. We don't do the problem for you. That's your job.

Using maple I got a ridiculous answer (attached)

What am I doing wrong?
Using Maple for one thing. The point of this problem really is to learn how to do the math.

You should recognize the expression
$$\frac{1}{\sqrt{z^2+r^2-2zr\cos\theta}}.$$ Expand it in terms of Legendre polynomials.

If I had a nickle every time someone says "I plugged it into maple". . .

Also, a table of integrals comes in very handy for EM problems.

vanhees71
Science Advisor
Gold Member
One should also say that the correct formula for the potential everywhere is (in SI units)
$$V(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\rho(\vec{x}')}{4 \pi \epsilon_0 |\vec{x}-\vec{x}'|}.$$
Here, however, I'd rather use the multipole expansion of the potential and solve Poisson's Equation
$$\Delta V=-\frac{1}{\epsilon} \rho.$$