Recent content by Slusho

Alright, thanks for the advice everyone. I guess I always assumed it would be easier to get a startup position since it's mostly young people at those companies so I'm not sure how they'd have a lot of experience, and I thought since they're less traditional they'd care less about GPA.

What boards would you recommend? I saw someone say Monster and Indeed are a waste of time and that they recommend using Glassdoor exclusively. And my most successful NYC friend told me to just use AngelList. There, however, I think it's mostly applying for individual jobs rather than putting...

I've been trying to figure out what jobs I can use my BS in physics for and I see data analyst being recommended a lot. I have limited python and SQL experience from college. I was thinking of trying for startups in NYC, but now I see articles saying startups want people that need no training so...

Okay, so that gives me A=sech(nπ/2)/2 which means V(x,y)=∑sech(nπ/2)cosh(nπx/L)cos(nπy/L) (summed over n=1,3). And E=∑(nπ/L)sech(nπ/2)[sinh(nπx/L)cos(nπy/L[PLAIN]http://mathworld.wolfram.com/images/equations/Hat/Inline1.gif-cosh(nπx/L)sin(nπy/L)ŷ]

Thanks for catching that. I accidentally dropped the A. It makes sense to me that C=0, as we only have cos terms on the left of cos(πy/L)+cos(3πy/L)=2Acosh(kL/2)(Csin(ky)+Dcos(ky)) but what about the 2Acosh(kL/2) term? If k=nπ/L then that would become 2Acosh(π/2) or 2Acosh(3π/2). Also, I don't...

Okay, I used both right and left boundaries, dividing one by the other, to see that A=B. This allowed me to simplify (AekL/2+Be-kL/2) to be 2cosh(kx). Now I tried going to the other boundary conditions: 0=2cosh(kx)(Csin(kL/2)+Dcos(kL/2)) which lead to tan(kL/2)=-D/C which is where I'm stuck now.

Okay, so the example problem in the book has k=nπ/a, with n=1,2,3,... For the equation to work here, (AekL/2+Be-kL/2) would have to equal 1 (not sure how that would work), C=0, D=1, and k=nπ/L, where n=1,3 But wouldn't the right side of the equation need a summation ∑ over n? And doesn't n...

For the right boundary condition, I get cos(πy/L)+cos(3πy/L)=(AekL/2+Be-kL/2)(Csin(ky)+Dcos(ky)) And the left is the same except the signs of the exponents are switched.

I just edited my above reply, which should answer your question.

The textbook does not have a general solution. It gives an example and works through it. Therefore, because it has different boundary conditions, it is not applicable to this exact problem. They plug in the values from a single boundary condition at a time into the equation I gave to determine...

Sorry, it was late when I wrote this. I've bolded the change: 1. Homework Statement A square is made up of four plates with a potential of zero on the top and bottom plates at (x,L/2) and (x,-L/2), and a potential of cos(πy/L)+cos(3πy/L) on the left and right plates at (-L/2,y) and (L/2,y)...

Homework Statement A square is made up of four plates with a potential of zero on the top and bottom plates at (x,L/2) and (x,-L/2), and a potential of cos(πy/L)+cos(3πy/L). Find the potential and electric fields inside the square. The Attempt at a Solution I start with...

Hmm. Maybe it's a way of simplifying the problem to be one-dimensional from three-dimensional due to inherent symmetry? I doubt it's a typo as this comes from a solutions manual with plenty of errata documents available.

Homework Statement Find the total charge Q given the charge density ρ(r)=ε0A(4πδ3(r)-π2e-λr/r The Attempt at a Solution I know the solution's steps start with: Q=∫ρdr=ε0A(4π∫δ3(r)dr-λ2∫e-λr(4πr2)/rdr) What I don't understand is where that 4πr at the end comes from. That last step is only...

Homework Statement A particle of mass m is subjected to a net force F(t) given by F(t)=F0(1-t/T)i; that is F(t) equals F0 at t=0 and decreases linearly to zero in time T. The particle passes the origin x=0 with velocity v0i. Show that at the instant t=T and F(t) vanishes, the speed v and...