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# Using Gauss' Law To Find Capacitance

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Steps to calculating capacitance

1. Imagine the capacitors are charged to +Q and -Q

2. Use Gauss' Law to find the electric field between them

3. $V = \int E\cdot dl$

4. $C =\frac{Q}{V}$

So we know from the derrivation in the previous post that the electric field between two parallel plates with +Q and -Q (or $+ \sigma$ and $- \sigma$ ) is:

$E=\frac{\sigma }{\epsilon _{0}}$

Because we did the derrivation for the this in the previous post we can skip steps 1 and part of step 2.

Step 2.5:

$E=\frac{\sigma }{\epsilon _{0}}$

$\sigma =\frac{Q}{A}$

$E=\frac{Q }{\epsilon _{0}A}$

$Q=E\epsilon _{0}A$

Step 3:

$V=\int E\cdot dl$

$V=El=Ed$

(since l is just the distance d between the plates)

Step 4:

$C=\frac{Q}{V}$

$Q=E\epsilon _{0}A$ (from step 2.5)

$V=Ed$ (from step 3)

$C=\frac{\epsilon _{0}A}{d}$

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