A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
Capacitance question
anonymous
 3 years ago
Capacitance question

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362261509427:dw I'm gonna write out what I know, let me know if I'm doing anything wrong...

anonymous
 3 years ago
Best 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 \[Q=\frac {\lambda}{L} \] ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think I should be able to sub for Q

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I made several mistakes already....

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[C=\frac{\lambda \;L\;2\pi\epsilon_0}{\lambda \ln(b/a)}=\frac{L2\pi \epsilon_0}{\ln(b/a)}\]
Ask your own question
Sign UpFind more explanations on OpenStudy
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.