Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
dw:1362261509427:dw I'm gonna write out what I know, let me know if I'm doing anything wrong...
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
\[\oint E\cdot dA=\frac Q{\epsilon_0} \] \[E 4\pi rL=\frac{Q}{\epsilon_0}\] where \(a\le r \le b\) The charge density =\(\lambda=\frac Q L\) is this the charge density of the inner cylinder? \[E=\frac{Q}{\epsilon_0 r \pi L}=\frac{\lambda}{\epsilon_0r\pi}\] \[dV= Edr\] \[V=E\int_a^b \frac 1 r dr\] \[V=\frac{\lambda}{\epsilon_0\pi} [ln(b)ln(a)]\] \[C=\frac Q V\] I'm trying to sub for Q \[Q=\frac {\lambda}{L} \] ?
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
I think I should be able to sub for Q
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
Ok i'll try it \[C=\frac Q V=\frac {\lambda }{L\frac{\lambda ln(b/a)}{\epsilon_0\pi}}=\frac{\epsilon_0 \pi}{Lln(b/a)}\]
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
Yes???
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
I made several mistakes already....
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
dw:1362263230248:dw circumference =\(2\pi r\) \[E 2\pi rL=\frac{Q}{\epsilon_0}\] dw:1362263350589:dw \[E=\frac{Q}{\epsilon_0 2\pi r L}\] \[dV=Edr\] \[V=\int_a^b Edr\] \[V=\int_b^a \frac{Q}{\epsilon_0 2\pi r L} dr \] \[V=\frac{Q}{2\pi\epsilon_0L}\int_b^a \frac 1 r dr\] \[V=\frac{Q}{2\pi\epsilon_0L}\ln(b/a)=\frac{\lambda}{2\pi\epsilon_0 }\ln(b/a)\] \[C=\frac{Q}{V}=\frac{Q}{\frac{\lambda \ln(b/a)}{2\pi \epsilon_0}}\]
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
\[C=\frac{\lambda \;L\;2\pi\epsilon_0}{\lambda \ln(b/a)}=\frac{L2\pi \epsilon_0}{\ln(b/a)}\]
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.