Recent content by sliperyfrog

Thanks for the help I got the answer!

So in the problem when I get the dY(s)/ds after using the product rule what am I supposed to do? Is it the same as having the laplace of y'(t)?

Homework Statement ty'' + y' = 2t2, y(0) = 0Homework Equations laplace(f''(t)) = s2laplace(f(t)) -sf(0) - f'(0) (-1) (d/ds) (F(s))The Attempt at a Solution I know how to solve the problem except for the ty'' part. I tried using the equation and I got -d/ds(s2Y(s) - 0 - f'(0)) which becomes...

Homework Statement An electron is launched between two parallel, neutral, conducting plates that are each L long and separated by a distance d with a uniform magnetic field of magnitude B permeated between them. (a) What is the minimum speed the particle must have to traverse the region...

So since they are in parallel that means the R in the problem is (1/90 + 1/90)-1 = 45Ω so for the answer I would get t = (45)(200E-6)ln(2)/2 = .0031s

So then the circuit would look like this And now I am back to the problem i was struggling with before is the two 90Ω resistors in parallel or is the capacitor in the the middle stopping it?

I was thinking that since that loop that connects the battery was in the original circuit that the junction that connect to the battery would still be there.

Okay so i redrew it and i found out it was parallel so I did (1/60 + 1/30)-1 = 20Ω but now i am confused on whether the R in the equation is 20Ω or 40Ω since the current goes through the first 20Ω then the capacitor then another 20Ω.

Homework Statement In the circuit below (drew it the best I can) the switch is opened, how much time elapses before the capacitor loses half of its initial stored energy.Homework Equations [/B] q(t) = Qe-t/RC U = q2/2C The Attempt at a Solution [/B] So first thing I did was I set (1/2)Q2/2C...

Thanks for the help!

I just realized I keeping trying to turn ∫dV into V like i am taking the integral. So it would just be ρ = (6Q0b5)/a6

So would I do (Q0b6)/a6 = ∫ρdV then i take the derivative of both sides so (6Q0b5)/a6 = (ρ4πb3)/3 so ρ = (9Q0b3)/2a6

Why when i take the surface integral of dA would it come out to 4πb2 instead of 4πa2 isn't the area supposed to be base of the radius of the sphere?

So what you are sayings is I take my answer for a E0 = (a2Q0)/4πb4ε0 since b=a when E0 = E I replace b for a so E0 = Q0/4πa2ε0 Then i take that answer for a and plug it into E so I get E = (Q0b4)/4πa6ε0 Than i plug that into guass's law so ε0∫E⋅dA = (ε04πa2Q0b4)/4πa6ε0 = (Q0b4)/a4 so then...

That is typo I meant that when you integrate dQ is comes out to the total charge of Q which is Q0 I am confused on why i would need to find dQ/db or what dQ/db is supposed to be. That when the sphere is at the maximum of E0 the electrostatic sphere is a distance a from the center of the...