Recent content by Zeus

Reply to Turin Ok, I've given your post some thought: F=Magnitude of Force of Magnetic field E=Electric field theta=@ F=qvBsin(@) E=-v/d 0=F + E 0=1.60 x 10^-19 x 2000 x B x sin90 - 200/.1 0=(3.2 x 10^-16)B - 2000 B=6.25 x 10^18 Was this the idea? Thanks Turin

Please Look at Other Post We'll see Paul =) My professor is unpredictable, lol. Hey, Turin, thanks so much for your input into this problem! If you have a moment, could you look at my other post? I love these forums!

Part 5 I'm not actually sure on this part. I won't be meeting until tommorrow, so I'll ask then. I was talking to a friend of mine and he said he got 0 (he is quite the Physicist too), but he didn't have time to explain. Sorry for the lack of info, I'll give an update when I get more info.

Howdie all, I'm soo glad I found this forum, people are genuinely working together instead of ignoring their fellow physics brethren =) A pair of square parallel conducting plates, having sides of length .05 meter, are .01 meter apart and are connected to a 200 volt power supply. An electron...

Parts 4 and 5 I'm still having difficulty with Part 4. With 5, however, the work should be positive I believe because it's being brought into the system, a negative work would mean it was attracted to the other particles. A positive work would mean it was repeled slightly, but work overcame it...

Part 3 Anyone Jump in I've worked through part 1-2, works great. For part 3, U=ke(Q1Q1)/r. I've tried a number of variations on this formula. My original thought to find the answer would be: (8.99x10^9 x -2 x 8)/.2=-7.192 x 10^11 (N*m) So, the engima is twofold: 1. Is the answer...

Hello all, Physics is generally pretty hard for me, so understanding Electric forces has been a monster. I've done what I "know" how to do on the following problem, but I need some input from the professionals =) Thank you in advance. Two point charges, Q1 and Q2 are located a distance .20...