## JenniferSmart1 2 years ago Capacitance question

1. JenniferSmart1

|dw:1362261509427:dw| I'm gonna write out what I know, let me know if I'm doing anything wrong...

2. JenniferSmart1

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

3. JenniferSmart1

I think I should be able to sub for Q

4. JenniferSmart1

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

5. JenniferSmart1

Yes???

6. JenniferSmart1

7. JenniferSmart1

|dw:1362263230248:dw| circumference =$$2\pi r$$ $E 2\pi rL=\frac{Q}{\epsilon_0}$ |dw:1362263350589:dw| $E=\frac{Q}{\epsilon_0 2\pi r L}$ $dV=Edr$ $V=-\int_a^b Edr$ $V=\int_b^a \frac{Q}{\epsilon_0 2\pi r L} dr$ $V=\frac{Q}{2\pi\epsilon_0L}\int_b^a \frac 1 r dr$ $V=\frac{Q}{2\pi\epsilon_0L}\ln(b/a)=\frac{\lambda}{2\pi\epsilon_0 }\ln(b/a)$ $C=\frac{Q}{V}=\frac{Q}{\frac{\lambda \ln(b/a)}{2\pi \epsilon_0}}$

8. JenniferSmart1

$C=\frac{\lambda \;L\;2\pi\epsilon_0}{\lambda \ln(b/a)}=\frac{L2\pi \epsilon_0}{\ln(b/a)}$