Recent content by Doc Z

Thank you all for your help! Seriously, I cannot tell you how grateful I am. This forum is really appreciated, especially with members like you! Thanks again.

So anything up to 7cm would be 0, anything above would be the equation I used earlier?

That charge is located 20cm away from the axis. Is it because 4cm is within the 7cm of the real cylinder, thus making it 0 since the inside charge of a closed surface is always 0?

Apparently 0 lol. But I don't understand why...it's the same concept as part a isn't it? Why is the net charge 0 in this case?

I don't understand why the answer to part b would be 0 though. It's asking to find the electric field. That means E= q/(epsilon0*2*pi*R*l) The problem states that it is 4.0 cm from the axis which means that r=.04m. Using the net charge found from part a (1.05 * 10^-6)/(epsilon0 * 2 * pi *...

Ah yes you are absolutely correct. Your description is perfect! So am I understanding that the net charge I am solving for in this problem is that of the imaginary cylinder which is the SAME as the real cylinder? But...even with the mistakes I made, I'm still getting the wrong answer for some...

Ok, so q = (E*2*pi*r*l) * (epsilon0) That's q= 36000*2*pi*.07*.260 * (8.88542 x 10^-12) = 3.66 x 10^-8. It's still the wrong answer.

That would be E*4pieR^2 right? and my answer to that was 2216.708 using a radius of .07m. Then multiplying that number by epsilon0 should give me the total charge right?

So the equation I just used helps me find the net electric flux which equals the net charge/e0. So to find the net electric charge, I should multiply epsilon0 by the net electric flux...but I get a wrong answer.

Ugh still a little confused. So I use E * 4pie R^2 --> 36000 * 4pie (.07^2) How do I find the net charge though?

Thank you very much, I really appreciate it.

Thank you so much! I am so confused with all this electric stuff...it's like a foreign language to me. It's the first time I'm completely lost in my Physics class.

Homework Statement A cylindrical shell of radius 7.00 cm and length 260 cm has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 20.0 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C. Find values for...

Doc Al - PLEASE help me. So I understand that the electric F equals the gravitational force. Then I divide that by the charge given in the problem to get the electric field. Now, to get the charge per unit area we use the equation E=sigma/2e0. We know the electric field so we rearrange the...

Thanks for the help!