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If we take two thin sheets of opposite areal charge density with seperation $l$ then everywhere above and below both the sheets the electric field is zero as it cancels out So it makes sense to conclude that the potential everywhere outside is zero too as work done in moving the charge is zero

but...

Now we take a limiting scenario. We take two discs instead with same seperation and opposite areal charge density $\sigma$ and $-\sigma$, and we calculate the electric field on the axis at distance x $$ E = \frac{\sigma}{2\epsilon_0}\left(\frac{x+l}{\sqrt{(x+l)^2+r^2}}-\frac{x}{\sqrt{(x)^2+r^2}}\right) \, . $$ We then integrate the electric field from $x$ to infinity to get the potential difference in between. The indefinite integral is $$ V = \frac{\sigma}{2\epsilon_0}\left(\sqrt{(x+l)^2+r^2}-\sqrt{(x)^2+r^2}\right) \, . $$ Now we evaluate it from infinity to $x$, then we get $$ V = \frac{\sigma}{2\epsilon_0}(l-(\sqrt{(x+l)^2+r^2}-\sqrt{(x)^2+r^2}) \, , $$ and as now $r$ approaches infinity like a sheet, the potential becomes $$V = \frac{\sigma l}{2\epsilon_0} \, .$$ The potential is not zero and is independent of the distance inbetween.

Also I checked in a book and the answer is $V = \frac{\sigma l}{2\epsilon_0}$. Please explain that why is the result different

1: See, in the second case we took the potential difference between a point at infinity and a point at $x$ distance from the centre of axis of the set of disc whose radius approaches $\infty$ to make it as a set of planes in the limiting scenario and we got a finite potential difference $\sigma / (2\epsilon_0)$, but if we take two infinitely extended set of such charged planes, then we get the electric field to be zero, which results in the potential difference between a point at infinity to $x$ to be zero. There is an apparent contradiction. And not only that the potential difference between infinity to $x$ is finite but in the second case the potential difference between any two points is finite which is contradicted in first scenario where we get zero potential difference everywhere outside the set of planes.

2: Also there should be a potential difference between a points on different sides of the sheets as we move the charge through the space inbetween where the electric field is non zero, but whith te second method we end up at same potential outside, but this is not so.

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    $\begingroup$ Please add at what position is the electric field in the case of disc is calculated $\endgroup$ Commented Jun 24 at 16:55
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    $\begingroup$ I remember stumbuling upon a similar question in IE IRODOV, electrostats question 44 and 45, although this is quiet different. $\endgroup$ Commented Jun 25 at 17:47

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Original answer

So it makes sense to conclude that the potential everywhere outside is zero too as work done in moving the charge is zero

This is not correct. It's ok to conclude the potential is constant outside the plates based on your argument, but not zero. That's because the work $W$ needed to move a particle of charge $q$ from $a$ to $b$ is $W = q\int_a^b \vec{E} \cdot d \vec{\ell} = - q \int_a^b \vec{\nabla} V \cdot d \vec{\ell} = -q\left(V(b) - V(a)\right)$ (where $\vec{E}$ is the electric field, $V$ is the electric potential, and $d\vec{\ell}$ is an infinestimal line element along the path from $a$ to $b$). As you can see from the last equation, the work needed to move a particle from point $a$ to point $b$ is given by the difference of the potentials at $a$ and $b$ (multiplied by the charge of the particle). $W=0$ implies $V(a)=V(b)$, not $V(a)=0$.

The potential is a constant $V_0$ on one side of the plates, and a different constant $V_0 + \frac{\sigma \ell}{2 \epsilon_0}$ on the other side of the plates. (Often we just take $V_0=0$ for convenience). Between the plates, the potential linearly extrapolates between these two values, since the field (negative gradient of potential) is constant between the plates.

Additional answer

Restating problem

Based on some discussion in comments, I realized the above answer did not address the OP's question. Here is my way of phrasing their question as I understand it and how I'd approach it.

The potential due to a charged disk of radius $r$ and charge density $\sigma$ at position $z$ along the axis of symmetry (where the disk is at $z=0$) is (http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potlin.html) $$ V_{\rm disk}(z; 0, \sigma, r) = \frac{\sigma}{2\epsilon_0}\left(\sqrt{z^2+r^2} - |z|\right) $$ I think the OP may have missed this absolute value (which is needed by the reflection symmetry across the plane of the disk), which led to incorrect conclusions.

Then the potential for two parallel disks, one with density $\sigma$ at $z=0$ and one at $-\sigma$ at $z=\ell$ is \begin{eqnarray} V_{\rm parallel\ disks}(z; \sigma, \ell, r) &=& V_{\rm disk}(z; 0, \sigma, r) + V_{\rm disk}(z; \ell, -\sigma, r) \\ &=& \frac{\sigma}{2\epsilon_0} \left(\sqrt{z^2 + r^2} - \sqrt{(z-\ell)^2 + r^2} - |z| + |z - \ell| \right) \end{eqnarray} We also know the potential for two parallel plates with densities $\pm \sigma$ of infinite size (at $z=0$ and $z=\ell$) is $$ V_{\rm infinite\ plates}(z; \sigma, \ell) = \begin{cases} 0, z< 0; \\ -\frac{\sigma z}{\epsilon_0}, 0 < z < \ell \\ -\frac{\sigma \ell}{\epsilon_0}, z > \ell \end{cases} $$ Now, we expect that the limit of infinitely large parallel disks should match the case of infinite parallel plates, up to an overall offset $V_0$. But what precisely does this mean?

Taking the limit, wrong way

One thought is that we want to understand the potential far away from the plates. So lets take $z \rightarrow \infty$ first. Then $$ \lim_{z\rightarrow \infty} V_{\rm parallel\ disks}(z; \sigma, \ell, r) = \lim_{z\rightarrow \infty} \frac{\sigma}{2\epsilon_0} \left(\sqrt{z^2 + r^2} - \sqrt{(z-\ell)^2 + r^2} - |z| + |z - \ell| \right) = 0 $$ By the same logic, we also have $$ \lim_{z\rightarrow -\infty} V_{\rm parallel\ disks}(z; \sigma, \ell, r) =0 $$ However, this doesn't match our expectation of $V_{\rm infinite\ plates}(z; \sigma, \ell)$, which should have two different values as $z\rightarrow +\infty$ and $z\rightarrow-\infty$. What is going on?

We want to look at both $z$ and $r$ becoming large. Here, we took the limit $z\rightarrow \infty$ for fixed $r$, and found $V=0$ everywhere, since the potential is zero in every direction as you move away from the plates at finite $r$. But that isn't really the situation we are interested in, since we cannot get infinitely far away from infinite parallel plates in directions parallel to the plates.

In other words, the limits $z\rightarrow \infty$ and $r\rightarrow\infty$ do not commute. The reason there is a subtlety in this case is that the limit $r\rightarrow\infty$ changes the asymptotic boundary conditions; for finite $r$ you can get infinitely far away from the plates in any direction, but for infinite $r$ you cannot.

Instead, we should first take the limit $r\rightarrow\infty$ for fixed $z$, in order to guarantee we stay at a fixed distance from the plates as they become larger. Then, we can take the limit $z\rightarrow\infty$ safely.

Taking the limit, correctly

What we should really do is take the limit $r\rightarrow\infty$ for fixed $z$. Physically this corresponds to focusing in "nearby" the plates, then making the plates bigger and bigger, so that the edges of the plates become less and less important.

We expect that in this limit, up to an overall constant offset in potential $V_0$ (which does not affect any physical results), that the parallel disks should give us the same potential as the infinite parallel plates.

$$ \lim_{r\rightarrow \infty} \left(V_{\rm parallel\ disks}(z; \sigma, \ell, r) + V_0\right) = V_{\rm infinite\ plates}(z; \sigma, \ell) $$

Let's check this expectation. We start by evaluating the limit $r\rightarrow\infty$ \begin{eqnarray} \lim_{r\rightarrow \infty} V_{\rm parallel\ disks}(z; \sigma, \ell, r) + V_0 &=& \lim_{r\rightarrow\infty} \frac{\sigma}{2\epsilon_0} \left((r - r) + \frac{z^2 - (z-\ell)^2}{2r} - |z| + |z-\ell|\right) + V_0 \\ &=& -\frac{\sigma}{2\epsilon_0} \left(|z| - |z-\ell|\right) + V_0 \end{eqnarray} where we used the expansion $\sqrt{z^2+r^2}=r+\frac{z^2}{2r} + \cdots$ valid for large $r$. Here, we see those crucial absolute value signs appear, which will be very relevant for matching with $V_{\rm infinte\ plates}.$

We can handle the absolute values in cases.

Case 1: $z > \ell$

If $z > \ell$, then both $z$ and $z-\ell$ are positive, so $|z|-|z-\ell|=\ell$. Then $$ \lim_{r\rightarrow \infty} V_{\rm parallel\ disks}(z; \sigma, \ell, r) + V_0 = -\frac{\sigma\ell}{2\epsilon_0} + V_0 \ \ (z>\ell) $$ To match with the parallel plate case in this range, we should take $$ V_0=-\frac{\sigma \ell}{2\epsilon_0} $$ That gives the expected result \begin{eqnarray} \lim_{r\rightarrow \infty} V_{\rm parallel\ disks}(z; \sigma, \ell, r) + V_0 &=& -\frac{\sigma\ell}{\epsilon_0} \\ &=& V_{\rm infinite\ plates}(z; \sigma, \ell), \ \ (z > \ell) \end{eqnarray}

Case 2: $0 < z < \ell$

If $0 < z < \ell$, then $z>0$ but $z-\ell<0$ so $|z|-|z-\ell|=z+z-\ell=2z-\ell$. Then

\begin{eqnarray} \lim_{r\rightarrow \infty} V_{\rm parallel\ disks}(z; \sigma, \ell, r) + V_0 &=& -\frac{\sigma}{2\epsilon_0} \left(2z - \ell\right) + V_0\\ &=& -\frac{\sigma z}{\epsilon_0} - \frac{\sigma\ell}{2\epsilon_0} + V_0 \\ &=& -\frac{\sigma z}{\epsilon_0} \end{eqnarray}

where the last two terms cancel because of our choice of $V_0$.

So we've shown \begin{eqnarray} \lim_{r\rightarrow \infty} V_{\rm parallel\ disks}(z; \sigma, \ell, r) + V_0 &=& -\frac{\sigma z}{\epsilon_0} \\ &=& V_{\rm infinite\ plates}(z; \sigma, \ell), \ \ (0 < z < \ell) \end{eqnarray}

Case 3: $z < 0$

This case is similar to case 1, but now both $z$ and $z-\ell$ are negative, so $|z|-|z-\ell|=-\ell$.

Then $$ \lim_{r\rightarrow \infty} V_{\rm parallel\ disks}(z; \sigma, \ell, r) + V_0= \frac{\sigma\ell}{2\epsilon_0} + V_0 = 0 $$

since we've taken $V_0=-\frac{\sigma \ell}{2\epsilon_0}$.

That means \begin{eqnarray} \lim_{r\rightarrow \infty} V_{\rm parallel\ disks}(z; \sigma, \ell, r) + V_0 &=& 0 \\ &=& V_{\rm infinite\ plates}(z; \sigma, \ell), \ \ (z < 0) \end{eqnarray}

Summary

Therefore, we've explicitly verified our expectation the the potential on the symmetry axis of two parallel circular disks of radius $r$, in the limit $r\rightarrow\infty$, matches the potential of two infinite parallel plates.

We also saw we had to be careful to take the limit $r\rightarrow\infty$ first, and then the limit $z\rightarrow \pm \infty$, not the other way around. Mathematically, this subtlety arises because the boundary conditions for infinite plates are different than for finite plates. Physically, what we are saying is that the limit we are interested in is to zoom in very close to the center of the disks, and then make their radius much bigger than their separation so we can ignore effects of the edges of the disk. Moving very far away from the disks will not help us understand that situation.

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  • $\begingroup$ See, in the second case we took the potential difference between a point at infinity and a point at x distance from the centre of axis of the set of disc whose radius approaches ♾️ to make it as a set of planes in the limiting scenario and we got a finite potential difference $\frac{\sigma}{2\epsilon_0}$, BUT if we take two infinitely extended set of such charges planes We get the electric field to be zero, which results in the potential difference between a point at infinity to x to be zero. There is an apparent contradiction $\endgroup$ Commented Jun 25 at 16:53
  • $\begingroup$ @MorphingScience (1) "BUT if we take two infinitely extended set of such charges planes We get the electric field to be zero" --> Just to clarify, you mean you have two parallel plates, and the electric field is zero outside the plates, and non-zero between the plates, correct? (2) "which results in the potential difference between a point at infinity to x to be zero." --> OK but the point at infinity need not have zero potential. For the parallel plates, there is a potential difference between points at $x\rightarrow -\infty$ and $x\rightarrow +\infty$; they can't both be zero. $\endgroup$ Commented Jun 25 at 17:12
  • $\begingroup$ I am talking about the potential $difference$ between a point at infinity to a point at x on the $same side$,in this scenario we get different solution for the two ways, I am seeking clarification on this apparent contradiction $\endgroup$ Commented Jun 25 at 17:39
  • $\begingroup$ @MorphingScience OK, I think I see what you are saying now. What you are basically saying is that you have two disks of radius $r$ and separation $\ell$, and you can compute the potential on the line connecting them. Then if you take the limit $r\rightarrow\infty$ of that answer, you don't get the correct answer for two infinite plates. This actually is a little subtle to do correctly. I'll try to write up an answer in a little bit when I have some time but if I understood correctly this is a good question that has an answer but it's not easy. $\endgroup$ Commented Jun 25 at 17:43
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    $\begingroup$ @cameronpoe I think I might have overcomplicated the problem in this case in my comments. I had a chance to work out the algebra for what I believe the OP is asking and there's no need to worry about boundary conditions. Although that always lurks in the background for infinite charge distributions. Your answer is also very nice, btw. $\endgroup$ Commented Jun 25 at 20:42
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The electric field is zero everywhere outside the plates (i.e. in the region that is not between the plates), but the potential is not zero everywhere outside the plates.

Let's say our plates are oriented so that their faces are perpendicular to the $z$-axis, and the negatively charged plate is at $z=0$ and the positively charged plate is at $z=l$ (so they are separated by a distance $l$).

One can show using Gauss' law that for two infinite plates with surface charge density $\sigma$, the electric field in the region $0<z<l$ is given by: $$ \vec{E} = -\frac{\sigma}{\epsilon_0} \hat{z} $$ and outside this region $\vec{E} = 0$.

We can choose any point to be our reference point, and then set that reference point to have $V=0$. Let's take $z=-\infty$ to have $V=0$. Next, the equation for finding the potential difference between any two points is then: $$ V(B) - V(A) = - \int_{A}^{B} \vec{E} \cdot d\vec{l} $$

Starting from $z=-\infty$, any path that stays in the region $z<0$ will cause $V(B) - V(z=-\infty) = 0$ since the electric field is 0 for $z<0$, so potential is zero in the region $z<0$.

Any path in the region $z>l$ will have $V(B) - V(A) = 0$ since the electric field is zero in this region. The potential in this region is therefore some constant, but we do not yet know what that constant is.

If we now try integrating from $z=0$ to $z=l$, we have: $$ V(l) - V(0) = - \int_0^l - \frac{\sigma}{\epsilon_0} dz = \frac{\sigma l}{\epsilon_0} $$ Therefore, the potential below the plates is $0$, the potential above the plates is $\frac{\sigma l}{\epsilon_0}$, and the potential between the plates linearly increases with distance (one can show this too).

I'm not sure where your textbook is getting the factor of $\frac{1}{2}$.

Addendum 1: Taking the limit $R\rightarrow \infty$ of the field

Thinking about the disks, let's first see that the electric field between two disks in the limit $R\rightarrow \infty$ gives the same field as the infinite parallel plate capacitor.

Let's write the field between the disks in a more symmetric way. The field a distance $z$ above (and coaxial with) a disk of radius $R$ is: $$ \vec{E} = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{z}{\sqrt{z^2 + R^2}}\right)\hat{z} $$

Therefore, if the disks are separated by a distance $l$, and we set the positive charged disk at $z=l/2$ and the negatively charged disk at a distance $z=-l/2$, the field is $$ \vec{E} = -\frac{\sigma}{\epsilon_0}\hat{z} + \frac{\sigma}{2\epsilon_0}\left(\frac{l/2 - z}{\sqrt{(l/2 -z)^2 + R^2}} + \frac{l/2 + z}{\sqrt{(l/2+z)^2 + R^2} }\right)\hat{z} $$

We next want to compute the limit $$ \lim_{R\rightarrow \infty} \frac{l/2 \pm z}{\sqrt{(l/2 \pm z)^2 + R^2}} $$

This clearly evaluates to 0. Therefore, the field between the disks in this limit is just $-\frac{\sigma}{\epsilon_0}\hat{z}$, the same as a parallel plate capacitor. Therefore, the potential difference between the plates will be the same as computed above for the parallel plate capacitor separated by a distance $l$.

We also want to calculate that the electric field goes to 0 above and below the plates in the limit. It turns out that the electric field both above and below the disk for finite radius is given by the same expression when we write these symmetrically: $$ \vec{E} = \frac{\sigma}{2\epsilon_0}\left(\frac{l/2 - z}{\sqrt{(l/2 -z)^2 + R^2}} + \frac{l/2 + z}{\sqrt{(l/2 + z)^2 + R^2}}\right)\hat{z} $$ which in the limit $R\rightarrow \infty$ gives $\vec{E}=0$.

Addendum 2: Finding the potential then taking the limit

I think I understand the confusion, so let's focus just on the potential below the disks. For finite disk radius, it is not zero. Integrating the electric field given in the last equation above yields $$ V(z) = \frac{\sigma}{4\epsilon_0}\left(\sqrt{4R^2 + (l+2z)^2} - \sqrt{4R^2 + (l-2z)^2}\right) + C $$ where I found this using Mathematica (I'm sure you can do it by hand). The factor of $C$ is there because I have no yet evaluated the bounds $V(z=-l/2) - V(z=-\infty)$. When you take the limit of this expression for $R\rightarrow \infty$, it is of the form $\infty - \infty$, and plugging this limit into Mathematica yields 0 since the $R$s have the same coefficients. Therefore the potential difference in this region is 0.

However, what if you evaluate the bounds first? Then $$ V(z\rightarrow -\infty) = -\frac{l\sigma}{2\epsilon_0} + C $$ and $$ V(z=-l/2) = C $$ If we define redefine the potential to be 0 at $z=-\infty$, then the potential difference from $-\infty$ to $z=-l/2$ is $$ \Delta V = \frac{\sigma l}{2 \epsilon_0} $$ This holds regardless of $R$, so as we take $R\rightarrow \infty$, we get the same expression. This matches what is in OP's question.

What went wrong though? The issue is the order in which we take the limits. The issue with the parallel plate capacitor is even at $z=-\infty$, we are claiming the electric fields are of equal magnitude. However, this is only due the infinite size of the plates that we are able to say this. For finite disks, the disk at $z=l/2$ will always have a weaker field than the disk at $z=-l/2$ when we evaluate the field at $z=-\infty$.

My understanding is that in order to get the limits to match, we must take the radii of the disks to go to infinity at different rates such that the electric field at $z=-\infty$ is the same for each. If not, we are always supposing that the upper disk has a weaker field, even in the infinite case, which is not true!

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  • $\begingroup$ I do not think you can have a potential difference without an electric field since "electric potential is the amount of work needed to move a test charge from a reference point to a specific point in a static electric field"; a potential differential is the work the static field would do on a charge moved between the points. $\endgroup$ Commented Jun 25 at 11:30
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    $\begingroup$ @Peter-ReinstateMonica What part of this answer is your comment about? I think the answer makes it clear that there cannot be a potential difference between two points if the electric field is zero in a path between them. $\endgroup$ Commented Jun 25 at 15:41
  • $\begingroup$ You start the answer with "The electric field is zero everywhere outside the plates" (emphasis mine). I'm not sure what "outside the plates" means, and I think an image would be good (I assume it's like classic school capacitor plates); but to my understanding you have to be "outside the plates" to get from one to the other (and you can also do that without being between the plates), and because work is done to a charge on the way the field there cannot be zero. That "zero" statement can only be meant as an approximation that gets better with distance, not as an absolute statement. $\endgroup$ Commented Jun 25 at 15:56
  • $\begingroup$ @Peter-ReinstateMonica Suppose $x$ is a coordinate that extends from $-\infty$ to $\infty$ and is perpendicular to the plates (which we assume are infinite in extent in the two directions perpendicular to the $x$ axis), so the left edge of the left plate is at $x=0$ and the right edge of the right plate is at $x=L$. Then I think "outside the plates" refers the regions $x<0$ and $x>L$. It does not refer to space that might be inside the plates (if they have a finite width), nor to space between the plates. $\endgroup$ Commented Jun 25 at 16:38
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    $\begingroup$ @cameronpoe Ooooh -- two plates of infinite size may be a classic way to introduce the capacitor -- so classic in fact that neither the question nor your original answer mentioned infinity ;-). Yes, now "outside" makes sense. $\endgroup$ Commented Jun 25 at 21:06
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If we take two thin sheets of opposite areal charge density with seperation l then everywhere above and below both the sheets the electric field is zero as it cancels out So it makes sense to conclude that the potential everywhere outside is zero too as work done in moving the charge is zero

You have to be careful here. It is only true that "everywhere above and below both the sheets the electric field is zero" if the thin sheets are infinitely large. If they are finite, then the field is non-zero everywhere. In fact, if they are finite then the field at large distances is approximately a dipole field.

If they are finite and the field is non-zero everywhere then the work done in moving the charge is non-zero regardless of whether you take a path through the plates or around the plates.

If they are infinite then the field is 0 everywhere outside the plates, but there is no path around the plates. Every path from one side to the other necessarily goes through the plates. So, although the field is zero outside the plates the work is non-zero along any path because a portion of the path necessarily goes through the plates where the field is non-zero.

Fixing this error resolves the contradiction.

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(a) I've taken the liberty of correcting a typo in your expression for the indefinite integral you've called V.

(b) Your evaluation of the integral at $x=\infty$ assumes that $x\gg r$. But you then use your definite integral for the case when "$r$ approaches infinity like a sheet". This is equivalent to assuming that $r\gg x$, contradicting the earlier assumption. This contradiction is, I believe, responsible for the discrepancy to which you draw attention.

Suppose instead that we first apply the condition $r\gg x$ to your indefinite integral, because, however far we are from the discs we shall always require that they appear to us as infinitely wide. We then have, using the truncated binomial expansion, $$V=\frac{\sigma r}{2\epsilon_0}\left[\left(1+\tfrac 12{\frac{(x+l)}{r^2}}^2\right)-\left(1+\tfrac 12 \frac {x^2}{r^2}\right)\right]$$ which, if $l\ll x$, boils down to $$V=\frac {\sigma l x}{2\epsilon_0 r}$$ But however large the value of $x$, we must always have $x\ll r$ for the reason given above. The indefinite integral is therefore zero for all values of $x$. The potential at any point on the axis not between the plates is therefore zero.

This agrees with the simple result that because the resultant electric field is zero at all points on the axis not between the plates, no work is done by the field on a test charge going to infinity from any such point.

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