Recent content by NotCarlSagan

Great, thanks.

If I have a +5 nC charge on the inside of the shell, the inside surface would be -5nC, the outside would be +5 nC and between those surfaces there would a 0 charge, right? So just to make sure I have it all straight, the INSIDE of the shell would actually be 0 because the INNER SURFACE is -5...

Homework Statement Homework Equations 1/Ceq= 1/C1 + 1/C2 The Attempt at a Solution 1/C13eq= 1/3C1 + 1/5C3 = 8C/3 = 2.67C 1/C12eq= (3C x C)/(3C + C) = 0.75C Ceqtotal = 2.67C + 0.75C = 3.42C I think the series circuit is throwing me off?

Well I couldn't figure it out in time. This was the only HW problem I missed luckily. Apparently the answer is 2.34 x 10 5 I wasn't too far off with my first answer...if anyone could help me with the solution I'd be really thankful.

Crap. I think I already made a mistake...I was supposed to ADD ECX... Doing that gave me a final answer of 1.92 x 105 N/C

Ok, I gave it a try but I think I'm still wrong. EA = 7.91 x 104 N/C EB = 2.70 x 105 N/C EC = 1.40 x 105 N/C EAX = (2.70 x 105) + (7.91 x 104)cos 60 = 3.10 x 105 EAY = (2.70 x 105) + (7.91 x 104)sin 60 = 3.39 x 105 ECX = (2.70 x 105) - (1.40 x 105)cos 60 = 2.00 x 105 ECY =...

So would the E-field simply be: E=ke(2.2x10-6)/(0.52) = 7.91 x 104 N/C?

I thought it was the same formula? Shoot. Idk. I'll review the problem and post here again if I have trouble. I guess I was just looking for the wrong thing. Thanks for pointing that out.

Homework Statement Homework Equations F=Keq1q2/r2 The Attempt at a Solution Using Coulomb's Law, here's what I've worked out: FAB=0.593N FAC=0.309N Fx=0.309 - 0.593cos60°= 0.0125 Fy=0.593sin60° =0.514 √(0.01252 + 0.5142) = 0.514 N/C and θ = tan-1(0.514/0.0125) = 88.6° I'm not sure why...