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

#### pence

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

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

1/4∏ε∫ρ∂t/r

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

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#### szynkasz

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

#### pence

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!

#### szynkasz

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 $u=z^2+r^2-2zr\cos\theta$

#### pence

Using maple I got a ridiculous answer (attached)

What am I doing wrong?

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#### szynkasz

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

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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.

#### Bryson

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
$$\Delta V=-\frac{1}{\epsilon} \rho.$$

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