# Electric field in a parallel plate capacitor [duplicate]

I am trying to calculate electric field between two parallel plates () using Gauss's law and using field due to sheet of charge. The fields using both the methods is not matching. I am not able to point out the mistake. I am attaching the derivations

• More on capacitors and factors of 2: physics.stackexchange.com/q/110480/2451 and links therein. – Qmechanic Mar 25 '17 at 13:09
• I have posted the actual derivation, but it will be helpful if you showed how you calculated the answer by your method. – Mitchell Mar 25 '17 at 15:24
• @Bhavya Sharma I added the derivation – user31058 Apr 4 '17 at 23:42

As you know that the electric field due to an infinite plane is given by $E=\frac{\sigma}{2\epsilon_{\circ}}$.

Between the two plates, there are two different fields. One due the positively charged plate and another due the negatively charged plate. So using the superposition principle, the electric field between the plates will be given by the addition of the fields due to both the plates, thus,

$E=\frac{\sigma}{2\epsilon_{\circ}}+\frac{\sigma}{2\epsilon_{\circ}}$

$E=\frac{\sigma}{\epsilon_{\circ}}$

This electric field will be directed from the positive plate to the negative plate. For an infinitely large plate the electric field is independent​ of the distance of the point where electric​ field is to be calculated.

In the region outside the plate, electric field will be $0$.

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

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

$C=\frac{Q}{\frac{\sigma}{\epsilon_{\circ}}d}$

But, $\sigma=\frac{Q}{A}$ , where A is the area of the plates.

Therefore,

$C=\frac{A\epsilon_{\circ}}{d}$

To be precise,

$C=\frac{A\epsilon}{d}$,

Where, $\epsilon=\epsilon_{r}\epsilon_{\circ}$.