Charge on disk with non uniform surface charge density

[songoku]

Homework Statement

A charged disk has radius R with its center at the origin. The disk has a non-uniform surface charge density $\sigma=\frac{a}{x}$, where $a$ is positive constant and $x$ is the distance from the origin. Calculate the total charge on the disk.

Relevant Equations

$Q=\int \sigma dA$

$$Q=\int_{0}^{R} \left(\frac{a}{x} \times 2\pi x \right)dx$$
$$=2\pi aR$$

But the answer key is $4\pi aR$. Where is my mistake?

Thanks

 

[BvU]

I get that too. Perhaps the answer key also has the same charge on the other side ?

$\$

 

[songoku]

I get that too. Perhaps the answer key also has the same charge on the other side ?

$\$

You mean like the front part of the disk and also the back part of it?

But should we consider both sides? Wouldn't we get the total charge on all parts of the disk using integration?

 

[BvU]

$\sigma$ is on a surface. There are two surfacces...

$\$

 

[kuruman]

$\sigma$ is on a surface. There are two surfacces...

$\$

If there are two surfaces, they must be separated by some distance $h$ in which case you have a cylinder and not a disk even if $h<<R$. I think it's fair to say that disks are two-dimensional structures while cylinders are three-dimensional. Otherwise, we would have to worry about when a disk stops being a disk and becomes a cylinder. It seems to me that the answer key is wrong.

To @songoku: Is there justification for the answer $4\pi a R$ in the answer key?

 

[haruspex]

If there are two surfaces, they must be separated by some distance $h$ in which case you have a cylinder and not a disk even if $h<<R$. I think it's fair to say that disks are two-dimensional structures while cylinders are three-dimensional. Otherwise, we would have to worry about when a disk stops being a disk and becomes a cylinder. It seems to me that the answer key is wrong.

To @songoku: Is there justification for the answer $4\pi a R$ in the answer key?

An alternative view is that the context is the 3D world we experience, in which case a 2D disc is an idealisation and just means an arbitrarily short cylinder. As you may be tired of reading, I treat an idealisation in physics as being the limit of realistic models; in this case, solve for a cylinder then take the limit as its length tends to zero.
The result here is the extra factor of 2.

 

[Gordianus]

Usually, a disk has zero thickness. I think the answer key is wrong.

 

[kuruman]

An alternative view is that the context is the 3D world we experience, in which case a 2D disc is an idealisation and just means an arbitrarily short cylinder. As you may be tired of reading, I treat an idealisation in physics as being the limit of realistic models; in this case, solve for a cylinder then take the limit as its length tends to zero.
The result here is the extra factor of 2.

OK, let's proceed according to your idealization. Let's assume that the surface charge density is a monolayer of atoms. Also, let's assume uniform charge density $\sigma$ so as not to worry about integrals. A cylinder of radius $R$ and length $L$ will have total charge
$Q=2\times \sigma(\pi R^2)+\sigma 2\pi R L.$
In the limit $L\rightarrow 0$, this becomes $Q=2\times \sigma(\pi R^2)~$ which suggests that both sides contribute equal amounts of charge. That is OK.

However, I think that what we have is a ridiculously thin cylinder but not a disk. A disk of charge would consist of a single layer of charged atoms. If there are $N$ atoms each of charge $q$, forming a disk of radius $R$, then $\sigma=\dfrac{Nq}{\pi R^2}.$ If I put a factor of $2$ in the denominator, I would be counting each atom twice. I think that the granularity introduced by the quantization of charge is more realistic than modeling charge as a continuous fluid.

 

[haruspex]

However, I think that what we have is a ridiculously thin cylinder but not a disk. A disk of charge would consist of a single layer of charged atoms. If there are $N$ atoms each of charge $q$, forming a disk of radius $R$, then $\sigma=\dfrac{Nq}{\pi R^2}.$ If I put a factor of $2$ in the denominator, I would be counting each atom twice. I think that the granularity introduced by the quantization of charge is more realistic than modeling charge as a continuous fluid.

Where are the charges in relation to the atoms?
I would agree that, at best, the question is ambiguous.

 

[kuruman]

Where are the charges in relation to the atoms?

I mentioned atoms to keep the picture realistic. One can imagine discrete point charges spread over a 2D surface.

I would agree that, at best, the question is ambiguous.

Ditto.

 

[Gavran]

A disk is the region in a plane bounded by a circle. See https://en.wikipedia.org/wiki/Disk_(mathematics)#.
So, the correct answer is $2\pi aR$.

 

[haruspex]

A disk is the region in a plane bounded by a circle. See https://en.wikipedia.org/wiki/Disk_(mathematics)#.
So, the correct answer is $2\pi aR$.

That is fine in a 2D space, but physics, at this level, is in a 3D space.
E.g., in a 2D space, the field from a point charge would follow an inverse law, not an inverse square law.

 

[songoku]

To @songoku: Is there justification for the answer $4\pi a R$ in the answer key?

I am sorry for late reply. There is no justification. The answer key just writes $4\pi a R$, no working and no explanation.

Thank you very much for all the help and explanation BvU, kuruman, Gordianus, haruspex, Gavran