Potential difference in a 2 disk system (Capacitor)

[Zack K]

Homework Statement

A capacitor consists of two large metal disks placed a distance $s$ apart. The radius of each disk is R $(R \gg s)$ and the thickness of each disk is $t$. The disk on the left has a net charge of $+Q$ and the disk on the right has a net charge of $-Q$. Calculate the potential difference $V_2-V_1$. Where location 1 is inside the left disk and at its center and location 2 is in the center of the air gap between the disks. Explain briefly.(I've uploaded a diagram)

Homework Equations

$E_{capacitor}=\frac {Q}{2A\epsilon_o}[1-\frac {z}{\sqrt{R^2+z^2}}]$ and to a first approximation, since R is much greater than z, $E_{capacitor}=\frac {Q}{2A\epsilon_o}[1-\frac {z}{R}]$. Where z is the distance of reference from the disk system and A is the area of the disk.
$\Delta V=-\int_1^2 Edz$

The Attempt at a Solution

To me, the only distance that matters is the distance from between the capacitors to the edge of the left disk since the electric field inside a metal is 0(in equilibrium), so the potential difference would be 0. In that case we can use $$\Delta V=-\int_s^\frac{s}{2} E*dz$$That step is definitely where I messed up. But anyways, using $$E_{capacitor}=\frac {Q}{2A\epsilon_o}[1-\frac {z}{R}]$$Then $$\Delta V=-\int_s^\frac{s}{2}\frac {Q}{2A\epsilon_o}[1-\frac {z}{R}]dz=-\frac {Q}{2A\epsilon_o}\int_s^\frac{s}{2}[1-\frac {z}{R}]$$After integrating and grouping terms I get $$\Delta V=\frac {Q}{2A\epsilon_o}[\frac {5s^2}{8R}-\frac {s}{2}]$$This feels horribly wrong and I'm not confident with the answer

 

[Delta2]

You can further simplify the expression you found, given that $s \ll R$ then $\frac{s^2}{R}\to 0$ so all you left with is $\Delta V=-\frac{Q}{2A\epsilon_0}\frac{s}{2}$ which I believe is much more appealing and it is fully compatible with the typical simplifying assumption that is done in cases like this and is to consider the field homogeneous in between the capacitor plates.

 

[Delta2]

Actually I need to give a bit better explanation on why we can drop the term $s^2/R$. It is $A=\pi R^2$ therefore it will be

$$\Delta V=\frac{Q}{2\pi\epsilon_0R}(\frac{5s^2}{8R^2}-\frac{s}{2R})$$
and from those two terms $s^2/R^2$ goes faster to zero than $s/R$.

 

[haruspex]

Are you sure about that expression for the field in a capacitor? Is there a 2 in the denominator?
I don't see it at http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html.

It is much easier if you use R>>s from the start, allowing you to use the infinite plate expression for the fields.
The inner surfaces of each plate (where all the charge sits) generate fields each side of magnitude $\frac{Q}{2A\epsilon_0}$. In going from position 1 to position 2, the potential from the left plate rises for distance t/2 then falls for distance s/2. The potential from the right plate falls for distance (s+t)/2. Net drop is $\frac{Qs}{2A\epsilon_0}$.

 

[Zack K]

Are you sure about that expression for the field in a capacitor? Is there a 2 in the denominator?
I don't see it at http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html.

It is much easier if you use R>>s from the start, allowing you to use the infinite plate expression for the fields.
The inner surfaces of each plate (where all the charge sits) generate fields each side of magnitude $\frac{Q}{2A\epsilon_0}$. In going from position 1 to position 2, the potential from the left plate rises for distance t/2 then falls for distance s/2. The potential from the right plate falls for distance (s+t)/2. Net drop is $\frac{Qs}{2A\epsilon_0}$.

Sorry there shouldn't be a 2 you're right.