Question

Given the electric potential V(θ)=V_0(1+cos(θ)) on a sphere of radius R, find V(r,θ) inside the sphere where the is no charge.

Answer 1

Q

Answer 2

im jk

Answer 3

I know this is the equation I need to start, but I don't know much more than that. \[V(r,\theta)=\sum_{n=o}^{\infty}[A_nr^n+Br ^{-(n+1)}]P_n(\cos(\theta))\]

Answer 4

I have a solution, but it's from an example problem in a book and not mah brainz. Electrodynamics wasn't my best class.... :P so I can't explain a bit of it very well - not sure how much help it'll be :/

\[ V(\theta)=V_0(1+cos \theta)\]

First off note that at theta = 0 V is max at 2V_0, and at theta = pi V = 0. Just good to know.

So we get that equation you have there as the general solution for Poisson's equation with no charge inside ( rho = 0), aka the Laplacian, of V in spherical coords with azimuthal symmetry
\[ \nabla ^2V{(r,\theta)} = 0\]
\[V(r,\theta) = (R_{(r)})( \Theta_{(\theta)}) \]
\[V(r,\theta) = \sum_{\ell=0}^{\infty} \left( A_{\ell}r^{\ell}+B_{\ell}r^{-(\ell+1)} \right) \Big( P_{\ell} (\cos \theta) \Big) \]

The l's are just a separation constant, with P corresponding to Legendre polynomials P(x) with x = cos theta.

We can rewrite the given potential using the Legendre polynomials

\[V(\theta)=V_0(1+cos \theta)=V_0 \Big(P_0(\cos\theta)+P_1( \cos\theta)\Big) \]

So, first off, the constant B has to be set to zero because that term blows up at r=0 otherwise (since r is in the denominator), leaving us with

\[ V(r,\theta) = \sum_{\ell=0}^{\infty} \left( A_{\ell}r^{\ell} \right) \Big( P_{\ell} (\cos \theta) \Big) \]

At r=R the potential has to match the given potential (because that's the basis for the problem)
\[ V(R,\theta) = \sum_{\ell=0}^{\infty} \left( A_{\ell}R^{\ell} \right) \Big( P_{\ell} (\cos \theta) \Big) = V( \theta)\]

This is the solution, now we just have to find the constants.....

Answer 5

For that we can use "Fourier's trick" knowing that the Legendre polynomials are orthogonal functions, so when you integrate them

\[\int_{-1}^{1}P_{n}(x)P_m(x) \mathrm{d}x = \frac{2}{2n+1}\delta_{nm}
\\ \
\\ \ \hspace{20px} \textrm{where} \ \ \delta_{nm} \ \ \textrm{is the Kronecker delta; equals 1 for m=n, equals for 0 n≠m}\]
So for us we can use

\[\int_{0}^{\pi}P_{\ell}(\cos \theta)P_{\ell \textbf '}(\cos \theta) \sin \theta \ \mathrm{d} \theta= \frac{2}{2n+1}\delta_{\ell \ell \textbf '}\]

So by multiplying both sides of the solution by

\[ P_{\ell \textbf '}(\cos \theta) \sin \theta \ \mathrm{d} \theta\]

and adding an integrand we get

\[ \int_0^{\pi}A_{\ell}R^{\ell} P_{\ell} (\cos \theta) P_{\ell \textbf '}(\cos \theta) \sin \theta \ \mathrm{d} \theta =\int_0^{\pi} V(\theta) P_{\ell \textbf '}(\cos \theta) \sin \theta \ \mathrm{d} \theta\]

\[ A_{\ell}R^{\ell} \frac{2}{2\ell+1}= \int_0^{\pi} V(\theta) P_{\ell \textbf '}(\cos \theta) \sin \theta \ \mathrm{d} \theta \]

giving

\[ A_{\ell \textbf '} = \frac{2{\ell \textbf '}+1}{2R^{\ell \textbf '}} \int_0^{\pi} V_0 \Big(P_0(\cos\theta)+P_1( \cos\theta)\Big) P_{\ell \textbf '}(\cos \theta) \sin \theta \ \mathrm{d} \theta \]

or

\[ A_{\ell} = \frac{2\ell+1}{2R^{\ell}} \int_0^{\pi} V_0 \Big(P_0(\cos\theta)+P_1( \cos\theta)\Big) P_{\ell \textbf '}(\cos \theta) \sin \theta \ \mathrm{d} \theta \]


Examining the left hand side shows that, evidently, the only two values for l are 0 and 1 because all others vanish!, so you can solve for your two constants A0 and A1

\[ A_{0} = \frac{V_0}{2} \int_0^{\pi} P_0(\cos\theta) P_{0}(\cos \theta) \sin \theta \ \mathrm{d} \theta \]
or
\[ A_{1} = V_0\frac{2+1}{2R} \int_0^{\pi}P_1( \cos\theta) P_{1}(\cos \theta) \sin \theta \ \mathrm{d} \theta \]

evaluate then plug everything back into the solution.

Luckily they gave you a really, really nice potential to work with! ^_^

Answer 6

Hope this helps ^_^

Answer 7

Oh my god you went all out. Thank you! I definitely need to go through this slowly to make sure I understand it. Thanks!

Answer 8

^^