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.

  • $\begingroup$ 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. $\endgroup$
    – stafusa
    Dec 25 '18 at 16:00
  • $\begingroup$ Hi ok! is html not preferred? $\endgroup$
    – D. Soul
    Dec 28 '18 at 0:21
  • $\begingroup$ Not really, mathjax is more powerful and has a better output. $\endgroup$
    – stafusa
    Dec 28 '18 at 0:39

Electric field lines start on the positive charges and finish on the negative charges so the electric field intensity between the charged plates is due to both sets of charges.
When you do your derivation for the field due to a positively charged infinite plate what you perhaps ignore/neglect is the fact that those electric field lines which originate on the positive charge have to terminate on negative charges (at infinity).
Bringing the negative charges closer to the infinite positive plate does not have an effect on the electric field strength as the electric field is still uniform.

With the parallel plate capacitor derivation you choose to have a finite dimension for the plates and make their separation much smaller than this the ensure a uniform electric field between the plates.


Visualise two parallel plates at distance d, carrying a surface charge. The electric field is the sum of that of both plates. In some places the fields cancel and at others they add up. In this way you will find the solution.


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