# Question on capacitance

I am recently trying to learn electrics on my own but am a bit confused with regard to capacitance. By Gauss law, I understand that the electric field from a single parallel plate results in:

$$E = \frac{Q}{2A\epsilon}$$

This is because we must take into account the flux of both sides of a parallel plate.

However, in a capacitor, with one plate +Q and another -Q we only take into account one side, hence:

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

I understand that the above equation is due to a closed Gaussian surface on the positive plate, however, I do not understand why we don't take into account the charge carried by the negative plate. If electric flux is about quantifying the number of electric field lines through an area, wouldn't the negative plate add to the number of field lines? Hence instead of:

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

Why is this wrong?:

$$C = \frac{2Q}{|\Delta V|}$$

adding on the charge of the negative plate.

• Hi and welcome to the Physics SE! The equations become much easier to read, search and edit when mathjax is used. This one has already been edited for you, but it'd be great if you could use it yourself in your next posts. Dec 25 '18 at 16:00
• Hi ok! is html not preferred? Dec 28 '18 at 0:21
• Not really, mathjax is more powerful and has a better output. Dec 28 '18 at 0:39
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