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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. png.latex? V = \int E\cdot dl

4. png.latex? 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 png.latex? + \sigma and png.latex?- \sigma ) is:

png.latex? 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:

png.latex? E=\frac{\sigma }{\epsilon _{0

png.latex? \sigma =\frac{Q}{A}

png.latex? E=\frac{Q }{\epsilon _{0}A}

png.latex? Q=E\epsilon _{0}A

Step 3:

png.latex? V=\int E\cdot dl

png.latex? V=El=Ed

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

Step 4:

png.latex? C=\frac{Q}{V}

png.latex? Q=E\epsilon _{0}A (from step 2.5)

png.latex? V=Ed (from step 3)

png.latex? C=\frac{\epsilon _{0}A}{d}

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