Recent content by fornax

I have to say I like how neat your method is, without even putting numbers in you have a nice neat formula to follow. As far as that factor of 30, I suppose there is a possiblity that they are wrong? Our professor never really confirmed it was correct, so there is potential for that situation...

Homework Statement A proton is moving along the main axis of a uniformly charged thin ring. The charge density on the ring is 5.0nC/cm and the ring radius is 1.0cm. Initially the proton is 2.0cm (along the axis) from the center of the ring with the velocity towards the center of the ring. What...

Looks good, that's exactly what I got for my answer as well. I'm not sure how you go about doing your problems, but I would highly suggest drawing a simple diagram of the problem at hand. Later on it is a good practice, and it keeps things orderly and harder to mix up. Even a line with some...

Well I can tell you that you got Part a correct as far as my calculations, and here's why: At 65m the cyclist applies the brakes and starts to decelerate at 5m/s^2. The problem then asks you how far before the car he stops, so his final velocity is 0. You obviously got that and your...

Well, using the observation that you made ehild, it did turn out that was a big chunk of my error. Replacing all of the 10^-4 with 10^-12 did give me a lower result, but not the one I was hoping for, I got .022m. I'm not sure what to make of it, it seems more reasonable than 100m, but is now...

Homework Statement A very long, thin straight line of charge has a constant charge density of 2.0pC/cm. An electron is initially 1.0cm from the line and moving away with a speed of 1000km/s. How far does the electron go before it comes back? Homework Equations ΔU = ΔV*q ΔKE + ΔU = 0...

Thank you so much, I've been stuck on the rest of the problem. Thank you for clarifying that for me.

Homework Statement A +30mC point charge sits at the center of a spherical thick metal shell, of 4.0cm inner radius, and 2.0cm thickness. The net charge on the shell is -50mC. Homework Equations The Attempt at a Solution I came up with several conclusions and I'm not really sure...

Alright [i] It would have to be facing in the direction of the E field, so that it's face is perpindicular to the E field. [ii] llVll = √(x2 + y2 + z2 llVll = √-10002 + 20002 + 15002 llVll = √7250000 llVll = 2693 so then [iii] E = σ/ 2ε0 2693 = σ/ 2ε0 σ = 4.76 x 10-8...

Thanks for merging the threads, I hope I didn't confuse anybody :/ . As for another attempt, and yet again I think I have it... (20 + Q)/(4∏ε0*7.5) - 0 = -Q / (4∏ε0) * -[1/.08 - 1/.05] 20 + Q / (4∏ε0*7.5) = -Q / (4∏ε0*7.5) ---->multiply by -(4∏ε0*7.5) 20+Q = -Q 20 = -2Q Q = -10mC

that's pretty easy, a capacitor. Now to find "all possible numerical information", I just backtrack, with each individual component? ie. Ex = σ/ 2ε0 -1000 = σ/ 2ε0 σ = -5.7x10^13

I haven't worked with these much, so just put in any value in place of the Ex Ey, Ez? If I had to make an educated guess right now, I would say all of the vectors are pointingtoward the left, up and out of the screen, or page. As far as the charge distribution, it would have to be an infinite...

Ok, hopefully the last stab at this, using Vb - Va =- Q / (4∏ε0) *[1/b-1/a] 20 - 0 = -Q / (4∏ε0) * [1/.08 - 1/.05] 20 = -Q / (4∏ε0*7.5) ---->multiply by -(4∏ε0*7.5) -2.4x10^10 = Q

Ok, so I think I may have it. after using ΔV = -∫Edl Outside shell : ΔV = (20+Q)/4∏ε0r Inside shell : ΔV = Q/4∏ε0r then ΔV = (20+Q)/4∏ε0r ΔV = 20/4∏ε0r + Q/4∏ε0r -20/4∏ε0r = Q/4∏ε0r ----> multiply by 4∏ε0r -20mC = Q

Well I figure the sphere is -20mC because electric fields flow towards negative charges. Since Electric potential's strength decreases as the electric field strength increases, and the electric potential at the surface of the sphere is 0, the sphere must have a negative charge. Based off of...