Recent content by carloscariell

Yes. You are right. Guess that what I first meant to say is that lim r→−∞ V(r, α/2 ± ε) = 0 , ε ≥ 0 In some area around the two planes where r→∞, V→0; The other option would be V→∞ when r→∞. I don't think so, but I can't strongly argue the reason.

Or maybe, lim _{r→ -∞ }V(r,θ) = 0

You're right. I'll will impose that lim_{r→∞} V(r,\theta) = 0

Yes. I have just find out that I should have ϕ(r) = r^{−λ}. Thanks again!PS: I need to finish this and other questions in 1 hour at least. Back to work!

Lol.. it was very easy to solve the ODE... I think I can re write the last equations to: * r² ϕ''(r) + rϕ'(r) − λ²ϕ(r) = 0 (*) * ψ''(θ) + λ²ψ(θ) = 0 (**) So, I'll have ψ(θ) = A cos λθ + B sin λθ ϕ(r) = r^{λ} My solution, so, would be V(r,θ) = r^{λ} ( A cos λθ + B...

I've forgot to write this: λ = µ² I''ll try to solve using your suggestion. I'll post it here as soon as I can... Thanks!

After some brain spanking, got to this: Starting with u(r, θ) = ϕ(r) ψ(θ) and putting it in the Laplace Equation(polar) ϕ''(r) ψ(θ) + (1/r) ϕ'(r) ψ(θ) + (1/r²)ϕ(r) ψ''(θ) = 0 Dividing by (1/r²) ϕ(r)ψ(θ) , I have: r² ϕ''(r) ϕ(r) + rϕ'(r)ϕ(r) = ψ''(θ)ψ(θ) = λ Then: * r² ϕ''(r) +...

Homework Statement Two semi-infinite conductor planes (like this ∠ ) have an angle β at a constant potential. Whats the potencial close to the origin? Homework Equations ∇²V = (1/r)(∂/∂r){ r ( ∂V/∂r ) } + (1/r²)(∂²V/∂\theta²) The Attempt at a Solution Trying for laplacian equation on...