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The capacitance of a parallel plate capacitor \[C=\frac{Q}{V}\] \[dV=-\int Edr\] \[E=\]
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|dw:1362265815092:dw|
|dw:1362265934285:dw| should I use gauss's law to find the electric field?
yes .. use the gauss law. consider a single plate first.
\[\oint E dA=\frac{Q}{\epsilon_0}\] |dw:1362266179361:dw| \[Q=\sigma A\] \[EL^2=\frac{Q}{\epsilon_0}=\frac{\sigma L^2}{\epsilon_0}\] \[E=\frac{\sigma}{\epsilon_0}\]
since we have two plates do we use \(2\sigma\)?
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yeah yeah ... that;s the way ... let me add few things ... |dw:1362266439292:dw|
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