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JenniferSmart1

  • 3 years ago

Capacitance question

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  1. JenniferSmart1
    • 3 years ago
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    |dw:1362261509427:dw| I'm gonna write out what I know, let me know if I'm doing anything wrong...

  2. JenniferSmart1
    • 3 years ago
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    \[\oint E\cdot dA=\frac Q{\epsilon_0} \] \[E 4\pi rL=\frac{Q}{\epsilon_0}\] 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
    • 3 years ago
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    I think I should be able to sub for Q

  4. JenniferSmart1
    • 3 years ago
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    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
    • 3 years ago
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    Yes???

  6. JenniferSmart1
    • 3 years ago
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    I made several mistakes already....

  7. JenniferSmart1
    • 3 years ago
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    |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
    • 3 years ago
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    \[C=\frac{\lambda \;L\;2\pi\epsilon_0}{\lambda \ln(b/a)}=\frac{L2\pi \epsilon_0}{\ln(b/a)}\]

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