# Derivation of Displacement current term in Ampere's Law

I have a quick question: In deriving the displacement current term for Ampere's Law, my book has the line:

$$\Phi_E= \int_S \mathbb{E} \cdot \hat{n} da= \int_S \frac{\sigma}{\epsilon_0} da = \frac{Q}{A \epsilon_0} \int_S da= \frac{Q}{\epsilon_0}$$

My question is: Here the electric field is the electric field between two conducting plates (capacitor) neglecting edge effects. $\sigma$ is the charge density of a plate given by: $\frac{Q}{A}$ where A is the capacitor plate area. My book substitues $\sigma= \frac{Q}{A}$ and arrives at: $$\frac{Q}{A \epsilon_0} \int_S da = \frac{Q}{A \epsilon_0} A = \frac{Q}{\epsilon_0}.$$ Why do the "A"'s cancel? One $A$ is the area of the capacitor plate and the other area is the area of the gaussian surface. In general, these area's will not be equal. The image to go along with the derivation is similar to the one shown below. The flux is only has a nonzero part where there is a nonzero electric field. Next if we assume the $\sigma$ is the charge density on the plate, then when you replace the $\bf E$ integral with a $\sigma$ integral then the area there is just over the area of the capacitor $C$ because that is where the electric field is nonzero, and equals $\sigma/\epsilon_0.$

$$\Phi_E= \int_S \mathbf E \cdot \hat{\mathbf n}\,\mathrm da= \int_C \frac{\sigma}{\epsilon_0} \mathrm da$$

I have never understood why this sort of derivation in textbooks is not made conceptually easier by choosing an appropriate bulging amperian surface as the one shown below. Because the surface is flat and has a plane perpendicular to the electric field in the region of the electric field the flux is easy to calculate.

$\Phi_E= \int_S \mathbf E \cdot \hat{\mathbf n}\,\mathrm da= E \; A = \dfrac{\sigma}{\epsilon_0} A = \dfrac {Q}{ \epsilon_0}$

• I would be interested to know why my comment has attracted a negative grade. When one first met this topic it is difficult enough without having to cope with a bulging surface of arbitrary shape. I think that one of the most important things about the amperian surface is that one can choose any shape and therefore make it the most convenient. Is that not what is done for Gauss's law in a lot of simple examples? I agree that @Timaeus' answer is very good but isn't the last step in finding the area C just equivalent to changing the shape of the amperian surface? – Farcher Mar 10 '16 at 8:03
• I don't know why your answer got a downvote, you've got some weirdness in your boldfaces, even a bold face Q for a scalar electric flux. A bulging surface on the right isn't a problem because if you put a flat surface next to the plate on the right then then flux through that flat surface equals the flux through the right bulging surface because the two surfaces together enclose a region to the right of the plate with zero charge. And such techniques should be covered before you get to Ampère-Maxwell. – Timaeus Mar 10 '16 at 16:40
• @Timaeus The boldface is because I just copied your formulae. I have now changed them. I understand about the bulging surface but what interests me is why when Gauss' law is discussed with regard to a capacitor textbooks invariable have cylinders drawn and yet when they discuss Ampere they generalise the bulging surface. – Farcher Mar 10 '16 at 16:50
• The boldface in my answer is becasue someone else edited it (I used vector arrows). I totally agree that your surface is better that the book surface, but you don't have to pull anything out of the integral, just use Gauss' Law. I think they want to connect to an arbitrary loop around the current, but then they should address loops narrower than the capacitor as well. – Timaeus Mar 10 '16 at 17:13
• @Timaeus Thank you for this and all you very informative and clear comments. – Farcher Mar 10 '16 at 18:16