Question

Capacitance question

Answer 1

I'm gonna write out what I know, let me know if I'm doing anything wrong...

Answer 2

\[\oint E\cdot dA=\frac Q{\epsilon_0} \]
\[E 4\pi rL=\frac{Q}{\epsilon_0}\] 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
\[Q=\frac {\lambda}{L} \]
?

Answer 3

I think I should be able to sub for Q

Answer 4

Ok i'll try it
\[C=\frac Q V=\frac {\lambda }{L\frac{\lambda ln(b/a)}{\epsilon_0\pi}}=\frac{\epsilon_0 \pi}{Lln(b/a)}\]

Answer 5

Yes???

Answer 6

I made several mistakes already....

Answer 7

circumference =\(2\pi r\)
\[E 2\pi rL=\frac{Q}{\epsilon_0}\]

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