Recent content by darkpsi

To clarify... To be more specific, is the division going to give a value of 1 or is there ever a case where this doesn't happen? If it helps, the equation E is both twice differentiable and convex.

Hello all, I am reading a research paper and have found the equation below: http://latex.codecogs.com/gif.latex?\mathbf{z}%20=%20\mathbf{a}%20-%20%28\nabla^2E%28\mathbf{t}%29%29^{-1}\Delta%20E%28\mathbf{t}%29%29 in which E is some function with the variable t being the vector input, and a...

Homework Statement A square loop of mass m resistance R and side l] is halfway inside a magnetic field pointing into the page. It is then released from rest and gravity pulls it out of the magnetic field. a) Calculate the induced emf and the current as functions of the velocity. b) Calculate...

Oh now I see. I wasn't even thinking that because r_hat wasn't pointing vertically that the z component wasn't all of the electric field. I was thinking the x-fields canceled so the field in the z direction was exactly equal to the electric field, but its actually only cos(theta) of it. Thanks a...

Homework Statement Find the electric field a distance z above the midpoint of a straight line segment of length 2L, which carries a uniform line charge λ. Homework Equations 1) my textbook says : E(r) = 1/4πεo ∫V λ(r')/r2 r dl' 2) and this also works? : E(r) = 1/4πεo ∫V λ(r')/r3...

Homework Statement A particla of charge q enters a region of uniform magnetic field B (pointing into the page). The field deflects the particle a distance d above the original line of flight. In terms of a, d, B, and q, find the momentum of the particle. Homework Equations F=dp/dt...

since it's a constant of integration, it wouldn't matter either way.

either you went from Q to P instead of the other way, or you forgot to take the negative of the integral

Oh I see. I'm actually learning this right now, so that's nice to know. I tried it the way you suggested and it does seem less tedious.

Because I was distributing the sin2 to just that term and then trying to solve that integral. But I see that you're way is much easier. Thanks!

Well just for future reference, when couldn't I change everything to one variable to solve?

Here goes: V(r,θ) = 1/4πεo * 1/r3 ∫∫∫0R (3/2 cos2θ' - 1/2) kRsin2θ' (Rr'2 - 2r'3) dr'dθ'dφ' = 1/4πεo * 1/r3 ∫∫0π (3/2 cos2θ' - 1/2) kRsin2θ' (-1/6 kR4) dθ'dφ' | Using 3/2 cos2θ'(cos2θ' - 1) | = 3/2 (cos4θ' - cos2θ') | = 3/2 [(1/2 + 1/2cos2θ')2 - cos2θ'] = 1/4πεo * 1/r3 ∫∫0π [3/2...

Yes but then when you dot E with dL you'll get VPQ = ∫∫∫ 40xy dx + 20x2 dy + 2 dz Am I missing something or does changing of variables need to take place?

But then the answer I'm supposed to be aiming for is wrong? At least I know maybe my qudrapole term wasn't wrong then, and maybe it's right? 3πkR5/64εoz3 ? I got the anwser that it is suppose to be of this website http://einstein1.byu.edu/~masong/emsite/S2Q50/S2Q50.html

Well it tells you the component in the x direction, it's 40xy. Maybe your confusing ux as having an actual magnitude? The magnitude is the 40xy. ux is simply saying the electric field (40xy) in the x direction, because the electric field is the same direction as vector u. To be accurate they...