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JenniferSmart1
 2 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...

JenniferSmart1
 2 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 \[Q=\frac {\lambda}{L} \] ?

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0I think I should be able to sub for Q

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

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0I made several mistakes already....

JenniferSmart1
 2 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 \[E=\frac{Q}{\epsilon_0 2\pi r L}\] \[dV=Edr\] \[V=\int_a^b Edr\] \[V=\int_b^a \frac{Q}{\epsilon_0 2\pi r L} dr \] \[V=\frac{Q}{2\pi\epsilon_0L}\int_b^a \frac 1 r dr\] \[V=\frac{Q}{2\pi\epsilon_0L}\ln(b/a)=\frac{\lambda}{2\pi\epsilon_0 }\ln(b/a)\] \[C=\frac{Q}{V}=\frac{Q}{\frac{\lambda \ln(b/a)}{2\pi \epsilon_0}}\]

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0\[C=\frac{\lambda \;L\;2\pi\epsilon_0}{\lambda \ln(b/a)}=\frac{L2\pi \epsilon_0}{\ln(b/a)}\]
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