Recent content by treynolds147

Ahhh, I see that now. Thank you so much for your help!

Ah, I see, the sine already incorporates the periodicity. Let's see, calculating the coefficients then, we should have ##\int_{-a}^{a}V_{0}\sin\frac{n\pi x}{a}\,\mathrm{d}x=\int_{-a}^{a}\sum_{n}A_{n}\sin\frac{n\pi x}{a}\sinh\frac{n\pi b}{a}\sin\frac{n'\pi x}{a}\,\mathrm{d}x##, which, if I did my...

I think I see why it'd work for the full problem. The sine term is going to give me something that falls on both the upper and the lower boundary, i.e. something like a square wave, right? As far as fixing the constants is concerned, I know the condition that has to be met (i.e., that...

Homework Statement A potential satisfies ##\nabla^{2}\Phi=0## in the 2d slab ##-\infty<x<\infty##, ##-b<y<b##, with boundary conditions ##\Phi(x,b)=V_{s}(x)## on the top and ##\Phi(x,-b)=-V_{s}(x)## on the bottom, where ##V_{s}(x)=-V_{0}## for ##-a<x<0##, and ##V_{s}(x)=V_{0}## for ##0<x<a##...

For the indefinite integral I get ##\frac{1}{15r^{2}}(3a^{5}+ae^{-r^{2}/a^{2}}(3a^{4}+8a^{2}r^{2}-r^{4})+e^{-r^{2}/2a^{2}}(-6a^{5}-8a^{3}r^{2}+8ar^{4})-\sqrt{\pi}r^{5}\text{erf}\frac{r}{a}+4\sqrt{2\pi}r^{5}\text{erf}(\frac{r}{\sqrt{2}a}))##. Wolframalpha isn't really able to compute between the...

Ah, you're totally right about ##e^{-r^{2}/a^{2}}##, I forgot to square that for some reason. ##|\mathbf{E}|^{2}## should be ##e^{-r^{2}/a^{2}}(a^{4}+r^{4}+2a^{2}r^{2})-2e^{-r^{2}/2a^{2}}(a^{4}+r^{2}a^{2})+a^{4}##. At any rate, is there any particular rule for when it's safe to ignore a...

Well upon squaring the field, I get ##\frac{\epsilon_{0}}{2}\int|\mathbf{E}|^{2}\,\mathrm{d}^{3}x'=\frac{q^{2}}{8\pi\epsilon_{0}a^{4}}\int\frac{1}{r^{4}}(a^{4}(1-e^{-r^{2}/2a^{2}})+r^{4}e^{-r^{2}/2a^{2}})r^{2}\,\mathrm{d}r## (where I already integrated over the solid angle). The term...

Ah, for some reason I had blindly plugged in the total charge into the RHS of Gauss's law. Being a bit more careful I get ##\oint_{S}\mathbf{E}\cdot\mathrm{d}\mathbf{A}=\frac{1}{\epsilon_{0}}\int\rho(\mathbf{x}')\,\mathrm{d}^{3}x'##, and upon using the charge distribution the RHS becomes...

Well I worked out from Gauss's law that the electric field would just be ##-\frac{1}{4\pi\epsilon_{0}}\frac{q}{r^{2}}##. What's striking to me about this result though is that I get different energies depending on which equation I use. If I use...

Is it at all possible to obtain the electric field from ##\nabla\cdot\mathbf{E}=\rho/\epsilon_{0}##? I'd rather avoid having to work out the field from the proper integral definition if I can help it.

Homework Statement A charge distribution has the form ##\rho=-\frac{q}{4\pi ra^{2}}(1-\frac{r^{2}}{a^{2}})\exp(-\frac{r^{2}}{2a^{2}})##. Compute the total charge Q, the electric field E, the potential ##\Phi##, and the electrostatic energy W for this charge distribution. Homework Equations...

Ah, that was my hold up, I was treating the the sine of C as being a vanishing quantity. Yes, this makes much more sense, thank you so much. I didn't think this problem would trip me up as much as it did, I guess I'm really out of practice!

Alright, so we'd have a wavefunction that looks like ψtotal=aψAB+aψCD, where c is the expansion coefficient, right? They'd have the same coefficient since the probability of obtaining the particle in either well is supposed to be equal. And since the two wells are approximately independent we...

If then the integrals between AB and CD are equal, would that not mean that the distance between C and D is ΔL as well? I can't imagine that the ground state wavefunction would be altered by CD's non-zero bottom, just the energy levels would be shifted. Since both regions AB and CD are...

Homework Statement Consider a one-dimensional, nonrelativistic particle of mass m which can move in the three regions defined by points A, B, C, and D. The potential from A to B is zero; the potential from B to C is (10/m)(h/ΔL)2; and the potential from C to D is (1/10m)(h/ΔL)2. The distance...