# Electric field through an aperture

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1. May 3, 2017

### Buffu

1. The problem statement, all variables and given/known data

Solve by integration method :-
From a spherical shell of radius $a$, a circular disk of radius $b$ has been removed. The shell has a surface charge density $\sigma$. Find the Electric field through that aperture.

2. Relevant equations

3. The attempt at a solution

We put the midpoint of the aperture at the origin.

Then I got,

$\displaystyle E = \int_\text{Region} {\sigma dA \over (R(A))^2}$.

Where is $R$ is function that depends on area to output the distance between the small charge and the origin.

The problem is that I am unable to express $R$ in terms of $A$.

After thinking for a lot of time I have not done much,

Any hints on how to move from here ?

2. May 3, 2017

One question I have: Is the radius "b" small compared to the radius "a"? If that is the case, you can assume "b" is a plane-shaped circular disc. The next part in proceeding would be to simply overlay a $- \sigma$ surface charge on this small area to negate the $+ \sigma$. Superimposing the two solutions (of the whole sphere and that of the disc) very close to the surface should give you $E$ in the aperture. (You can assume close to the surface that the disc with $-\sigma$ looks like an infinite sheet of charge. At the microscopic level, that's what i would look like if you are right next to the surface.) Notice that $E$ is different on both sides of the shell and also on both sides of the small disc. $\\$ I don't think they are looking for the general solution to this problem (for arbitrary "b") which, unless I'm missing something, could be a very difficult calculation. You might ask your instructor for further clarification. $\\$ And a follow-on: I worked the problem in this manner and did get an answer for the electric field $E$ in the aperture which is continuous across the plane at $r=a$.

Last edited: May 3, 2017
3. May 3, 2017

### Buffu

I actually would like to find for arbitary $b$ but I think if that is too difficult I am fine for the case $a >> b$.

By "overlaying" did you mean that I imagine that the aperture is filled with a disk of $-\sigma$ surface charge density ?

I did not understand what does it mean to superimpose the two solutions. Which solution are you referring ?

4. May 3, 2017

The original spherical shell before any drilling has $+ \sigma$ on its surface. The solution $\vec{E}1=\vec{E}(\vec{r})$ is well known everywhere. That's the first solution. $\\$ Instead of drilling, just apply a surface charge density $- \sigma$ to a small area of radius "b". For this small disc, you can compute $E$ everywhere, but you only need it in close proximity on either side of the disc. The electric field for this second charge distribution is the second solution, which will get added/superimposed to the first solution. $\\$ For just very near the surface this second solution is simpler: You need the solution for $E$ on either side of an infinite plane of surface charge $-\sigma$. Computing this $E2$ on both sides of the sheet of charge ( $E2_{in}$ and $E2_{out}$ ) will give you what you need to get the answer. That along with the first solution ($E1_{out}=\frac{+Q}{4 \pi \epsilon_o a^2}$ and $E1_{in}=0$) will give you the solution (to the $E$ field) in the aperture. $\\$ $Et_{out}=E1_{out}+E2_{out}$ and $\\$ $Et_{in}=E1_{in}+E2_{in}$. $\\$ It turns out $Et_{out}=Et_{in}$. The $Et$ across the empty (uncharged) space in the aperture is continuous. $\\$ The subscripts "in" and "out" refer to just "inside" and just "outside" the surface charge. And $"Et"$ represents the "total" electric field from both solutions.

Last edited: May 3, 2017
5. May 4, 2017

### Buffu

$E_{1, in} = 0$

Since this is very close to the surface $E_1 out = 4\pi\sigma$

$E_{2,in} = -2\pi \sigma$ -ve sign to denote direction.

$E_{2,out} = 2\pi\sigma$.

$E_{in-t} = -2\pi \sigma$

$E_{out-t} = 6\pi\sigma$

$E_{t} = 4\pi \sigma$ going out of the sphere.

Last edited: May 4, 2017
6. May 4, 2017

You have the "sign" on the $-\sigma$ for the disc of radius "b", but also, are you using c.g.s. units? The part I wrote for you is in SI / MKS. If you want to work it in c.g.s., that is ok too, but please tell what units you are using.$\\$ Editing: Looks like you did your first post in c.g.s., so I should have used that. $\\$ For a surface charge density of $- \sigma$, $E2_{in}=+2 \pi \sigma$ and $E2_{out}=-2 \pi \sigma$. $\\$ As you correctly said $E1_{in}=0$ and $E1_{out}=+ 4 \pi \sigma$. It's a simple matter to compute $Et$. $\\$ Perhaps it would be better written as $\sigma=+\sigma_o$ on the shell with radius "a", and $\sigma=-\sigma_o$ on the disc with radius "b".

Last edited: May 4, 2017
7. May 4, 2017

### Buffu

Sorry I would be clear from next time.

It is nice that I got the right answer but I did not understand how.

One thing I did not understand is why did replacing hole with a infinite sheet of charge of equal and opposite charge density is same ?

Also I would like to know that is there a way to do this with integration ? I can try summing the field by indiviual particles of the sphere at midpoint of aperture ? I am unable to set up the integral.

8. May 4, 2017

Let's do it for a disc of radius "b" at the midpoint of the disc and a distance $z$ above the surface. I will do it for disc of surface charge density $\sigma$. The radial components will cancel and we will get the z-component with a factor of $cos(\theta)=z/R$ where $R=\sqrt{r^2+z^2}$. Using inverse square law: $E=\int\limits_{0}^{2 \pi} \int\limits_{0}^{b} (\frac{ \sigma }{R^2})(\frac{z}{R}) r \, dr \, d \phi$ . The result I get upon integrating is $E=2 \pi \sigma (1-(\frac{z}{\sqrt{b^2+z^2}}))$. Notice as $z$ approaches zero, $E=2 \pi \sigma$ (approximately)=the infinite sheet of charge result, even if $b$ is quite small.

Last edited: May 4, 2017
9. May 4, 2017

### Buffu

Is you disk placed something like the picture in this thread ?

10. May 4, 2017

11. May 4, 2017

### Buffu

That then answers why I can replace the disk with a infinite surface with charge but one thing remains is why we can replace a hole with a infinite surface of -ve charge ?

In other words why is making a hole in sphere is equivalent to placing a infinite surface of -ve charge in close vicinity of sphere?

12. May 4, 2017

We only replaced a small area of radius $b$ by the surface charge of $- \sigma$, and since we only needed to compute the electric field at z=0, we simulated it (the disc) by an infinite sheet. Certainly at any large finite distance, these two give very different results. It was only because we were only interested in very small z that the disc could be replaced by the infinite sheet of charge.

13. May 4, 2017

### Buffu

But my question was why evaluating electric field at aperture by summing all the individual electric fields by each particle of sphere is same as placing a disk of $-\sigma$ at that place and calculating on E either sides ?

14. May 4, 2017

The solution of the charged sphere, $E=4 \pi \sigma$ at the surface of the sphere is added to the solution of the disc of surface charge $- \sigma$. The disc of surface charge $- \sigma$ will negate the charge of the sphere over the radius $b$ and thereby be the same electrically as an aperture there.

15. May 4, 2017

### Buffu

Will this still work if our removed disk had some curve to it ? like a small hemisphere ?

16. May 4, 2017

The more general problem, as far as I can tell, with curvature and much larger radius $b$ where you solve for the electric field at an arbitrary location is extremely difficult. If they want you to compute the electric field at the center of the aperture when the surface of the aperture is somewhat large and thereby has a curve to it, I think the integrals are probably workable, but only at the center of the aperture. In that case, you would simply use $+ \sigma$ over the surface of radius $a$, and determine in polar coordinates the limit on the $\theta$ integral for an aperture of radius $b$. I think there are probably much more useful E&M exercises than this one, but I think the integral might be workable for this one point. $\\$ Editing... Additional comment: The center of the aperture is still a somewhat difficult point to write out the integral for $E$. A preferred point to evaluate the integral for the electric field, in the case of a larger aperture, would be at the center of the sphere. $\\$ Additional note: The general case could be readily solved with numerical computations to evaluate the integrals, but I think the numerical method using the computer is usually done in courses other than an E&M course.

Last edited: May 4, 2017
17. May 4, 2017

### Buffu

I think I have evaluated the integral you were talking in your edit, that the total field at the centre of the sphere by the remaining charges,

I got,

$$\vec E_s = -2\pi\sigma \sqrt{1 - {b^2 \over r^2}}$$,

I know $E_s + E_d = 0$ for the centre of sphere.

Where $E_d$ is electric field by the disk that is removed at the centre.

How can I find the E field at the centre of aperture with this ?

18. May 4, 2017

Just for the fun of it, (I just solved it), I'll show you the solution: At the center of the sphere: $E_z =\int\limits_{0}^{2 \pi} \int\limits_{0}^{\theta_o} (\frac{\sigma}{a^2}) a^2 cos(\theta)sin(\theta) d \theta \, d \phi$ where $\theta_o=cos^{-1}(-\frac{\sqrt{a^2-b^2}}{a^2})$. ( $sin(\theta_o)=b/a$.)(Evaluating the integral gives you a term $(1/2)sin^2(\theta)$.) The result is $E_z=\pi \sigma \frac{b^2}{a^2}$ and the result is symmetric. The part of the charged sphere that remains can be a small piece on the left of radius "b", or a large piece with a small aperture on the right of radius "b". Both geometries yield the same answer. Notice $\pi b^2$ is the area of the circle for a small flat disc.

Last edited: May 4, 2017
19. May 4, 2017

### Buffu

Oh I did not multiply by $\cos \theta$.

The $E_z$ you found is at the centre of the sphere and not at the centre of the aperture. Are the two same ?

20. May 4, 2017

For a small aperture, we previously found that $E_z =2 \pi \sigma$ in the aperture, independent of the size $b$, provided $b << a$, and that result was across the whole aperture. This result at the center of the sphere is considerably different from that result.