Recent content by purduegirl

I knew there was an electric field here because with the orientation of the charges, when compared to a compass, would make the theoretically compass point in the east direction. But, from there, I have no idea how to solve this problem, since the method suggested by my prof. is not correct for...

2(8.99e9)(5.90e-11)/(3.525e-5) = 30094.1844

I know that the E field = 2*k*lambda/r

the answer I got was 10608.2/3.525E-5 = .030094 N/C, still incorrect

Yes, it does include the magnetic field of 3.

Yes, it is.

What is the magnitude of the electric field at the surface of the wire? I know that the E field = 2*k*lambda/r I am given lambda which I converted to m which I get 5.90E-7 C/m. I am given the diameter of the wire = 7.05E-3 cm or 0.705 m. The radius will be 0.3525 m. I found that the E...

Yes, that is the question, and the diagram is still supposed to be used.

I tried this once again, here's my math. Flux = N(BAcos 18 - BAcos 75) Flux = N [(3.0 T)(.23 m*.70 m)cos(18)-(3.0 T)(.23 m*.70 m)cos(75)] Flux = -150 (.482[(.95106) - (.25882)]) Flux = -150 ( .3343506986)/1.2 s Average Induced Current = 41.79 V Everytime, I do this problem my number...

The voltage across C1 should be 3V because the source of the charge is coming from the battery. Because C2 is in series with C1, it too should have the same charge on the battery, right?

My T.A said that I should divide the cos of angle 1 by the cos of angle 2. If I apply the subtract from 90 rule to both angles, I still get the difference is 57 degrees as I did before with the regular angles.

But, would that also apply to the 72 degrees? I talked to my T.A. about this problem yesterday and he told met that I should divide the cos theta to get the answer. Would this be the correct reasoning?

No, not really. I have tried the suggestions of getting the capitance C3 and C4 and then dividing it by the total charge from an earlier post, which didn't work. I'm still really confused.

Would it be 90-15 since the magnetic field is perpendicular to the velocity?

I tried this once again, here's my math. Flux = NBAcos\theta Flux = (3.0 T)(.23 m*.70 m)cos(72-15) Flux = 3T( 0.161 m^2)cos(57) Flux = 0.2630606539 Emf = -N( Flux/ time change) Emf = -150( 0.2630606539 V*s / 1.2 s) Emf = - 150( 0.2192172116 V) Emf = -32.88 V This and the postive...