# Electric field at centre of hole

1. Nov 24, 2013

### Pranav-Arora

1. The problem statement, all variables and given/known data
A hollow insulating sphere of radius R is charged uniformly to a charge of Q. There is a small hole on this sphere. What is the electric field strength at the centre of this hole?

2. Relevant equations

3. The attempt at a solution
I think I have to use Gauss's law here but what should be the Gaussian surface here? I am thinking of a small cylinder around the hole. Correct?

Please see the attachment.

$E_1$ is the electric field just above the hole (outside) and $E_2$ is the electric field inside.

From Gauss's Law, $E_1S-E_2S=0 \Rightarrow E_1=E_2$.

Am I going in the right direction? What should be the next step?

#### Attached Files:

• ###### E-field.png
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2. Nov 24, 2013

### Simon Bridge

It is the electric field for the entire shell, but subtract the field due to the bit where the hole was.

3. Nov 25, 2013

### Pranav-Arora

Umm....I don't get this. Which electric filed are you talking about? $E_1$ or $E_2$? How do I calculate the field to be subtracted?

Thanks!

4. Nov 25, 2013

### Simon Bridge

There is only one electric field.
You want the electric field at a position in the middle of a hole in a shell of charges?

What would the the electric field at that position be if there was no hole?
How big is the hole?

5. Nov 26, 2013

### Pranav-Arora

Yes.
It would be $\displaystyle \frac{kQ}{R^2}$ where Q is the charge on sphere.

Very small compared to the dimensions of the sphere.

I am still lost, I can't really think of any equations to begin with.

6. Nov 26, 2013

### Simon Bridge

So the hole is small enough that the bit of the shell that is removed to make the hole can be approximated by a disk?

You know Coulomb's Law right?
If you have a disk of charge, radius a and thickness t, what is the electric field in the center of the disk?

(Note - electric field is a vector ... you left off the direction.)

Last edited: Nov 26, 2013
7. Nov 26, 2013

### Pranav-Arora

I think I have to assume the hole as a disk of negative charge density $-\sigma$ where $\sigma$ is the charge density of sphere. If I place a small disk of negative charge density on a surface of positive charge density, it acts as a hole.

Electric field due to disk will be in direction opposite to that of the sphere. Hence, net electric near the hole is:
$$\frac{\sigma}{\epsilon_0}-\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{2\epsilon_0}=\frac{Q}{8\pi \epsilon_0 R^2}$$
Though the above answer is correct, I would like to know if I solved the problem correctly?

8. Nov 26, 2013

### Simon Bridge

You just used the superposition principle - which is a reasonable way to go about this.
Notice that adding (the field due to) a negative charge density is the same as subtracting a positive charge density? Thinking of the hole in the shell as an absence rather than a presence usually looks better on paper.

Can you express the result in words?
What is the effect of putting a small hole in the shell about a position on the shell, on the strength of the electric field at that position? i.e. does it make the field bigger or smaller? how much bigger or smaller? Is this surprising?

9. Nov 27, 2013

### Pranav-Arora

The strength of electric field decreases and becomes half. Surprising? I am not sure, I am not good at interpreting the final results, can you show me why this is interesting? :)

Thanks a lot Simon Bridge.

10. Nov 27, 2013

### haruspex

I don't think of it that way. Well away from the hole, if you plot the field against the radius it jumps from 0 to its max value as you cross the shell. The same plot through the hole shows the spike smoothed out - it becomes nonzero before reaching the radius of the sphere, reaches half max at or near the middle of the hole, continues to increase for a while beyond the hole (but never reaching the max it reaches elsewhere), before falling away again. That's all as one would expect.

11. Nov 28, 2013

### Simon Bridge

What your result is telling you is that half the electric field close to the surface is due to the charge in the very small region close by.
You don't think that's interesting?

Admittedly, all that charge was very close to the point of interest.

At the risk of confusing you: there are several ways of thinking about this problem.

If you treat any point at r=R to be infinitesimally outside the sphere, for example, then Gauss' law will have you treat all the charge as at the centroid.
For a small hole, the centroid will be close to the center of the equivalent complete sphere. What does that do to your calculation?

If you treat the hole diameter d as removing a disk from the surface - d<<R means that the removed section is approximately flat.
The field at the center of the hole would be the field, there, due to the entire sphere, minus the field at the center of a flat disk.
The field in the center of a flat disk is....

You could say that the center of the hole is at distance $C=R\cos\alpha = \sqrt{R^2-d^24}$ $\alpha$ being half the apex angle of a cone with the hole as it's base, point at the origin. In which case, C<R so there is no charge inside, so the field is zero.

All these ways appear to contradict the "correct" answer. But they each make some kind of assumption...

The exercises you are being given are not arbitrary - they are supposed to teach you stuff beyond how to do the calculation.
But the prof is probably happy that you could do the calculation, but, especially in exams, it can be helpful (i.e. to the final grade) to write a short bit in more regular language about what the answer is telling you.