# Homework Help: Sphericall shell having hole

1. Jan 8, 2014

### Tanya Sharma

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

An insulating hollow spherical shell of radius R has a small hole on it .The shell has a charge Q uniformly distributed over it. What is the electric field strength at the center of this hole?

2. Relevant equations

3. The attempt at a solution

If the hole weren't present ,the electric field due to the shell just above the surface would have been kQ/R2 .

Now I am not sure how does the hole influences the electric field .

Another doubt I am having is shouldn't the question ask about the electric field just above the hole instead of at the center of the hole ?

Could someone help me with the problem ?

2. Jan 8, 2014

### rude man

With problems like this, always look for some superposition argumernt.

In this case, what is the field if there were no hole? (To answer this you may want to assume a finite thickness for the shell and compute the field in the middle of the shell, radially speaking).

Then, take an uncharged shell of same size but place negative charge of the same volumetric charge density where the hole was before.

Then the net field is the superposition of the two fields.

3. Jan 8, 2014

### Tanya Sharma

Do I have to consider the hole as a cylinder ?

4. Jan 8, 2014

5. Jan 8, 2014

### haruspex

No, it's a thin shell, so to plug the hole all you need is a disc.

6. Jan 8, 2014

### Tanya Sharma

Do I need to use the electric field strength of the circular disk = σ/2ε ? But that is the result we have for an infinite sheet of charge .

7. Jan 8, 2014

### ehild

Nothing is said about the size of the hole. It is just a problem, so you do not need to do complicated things. Assume that the hole is small enough that the part of the surface removed can be considered a disk, and big enough that the field very close to it above the centre is the same as for an infinite disk.

It would be interesting to solve the problem directly, by integration.

ehild

8. Jan 8, 2014

### Tanya Sharma

Hi ehild...

Thanks for chipping in :)

So just above the center the electric field strength would be simply σ/ε - σ/2ε .The net result being σ/2ε . Is it correct ?

But does that mean the electric field just below the hole(within the shell is also σ/2ε radially outward ) ?

9. Jan 8, 2014

### ehild

I think so. Only give it in terms of Q and R.

The electric field lines are continuous and can end only on charges. There is no charge on the hole, so the field lines can not change direction...
I drew some of the charges with some of the field lines emerging from them. So what do you think? Which field lines reach the hole and go through it?

ehild

#### Attached Files:

• ###### spherehole.jpg
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10. Jan 8, 2014

### Tanya Sharma

Q/(8πεR2)

Every field line emerging within the shell has to pass through the hole . Isn't it ?

11. Jan 8, 2014

### ehild

Not every. Those only inside the cone. (I forgot to show the hole)
ehild

#### Attached Files:

• ###### spherehole2.JPG
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12. Jan 8, 2014

### rude man

If you think of the piece 'cut out' from the shell, which is a short, thin right circular cylinder with charge density equal and opposite to that of the shell, what is the field in the middle of that?

13. Jan 8, 2014

### Saitama

Hi ehild! :)

Please look at the attachment, I think of solving the problem by considering small rings on the surface of sphere.

Let the radius of sphere be R, then the radius of ring is $R\sin\alpha$. The distance of hole from the centre of this ring is $R(1-\cos\alpha)$. The charge contained on this ring is $dq=\sigma dA=\sigma 2\pi R^2\sin\alpha \,\, d\alpha$ where $\sigma$ is the surface charge density of sphere.

Calculating $\sigma$:

Assume that the hole subtends an angle of $2\theta$ (negligibly small) at the centre of sphere. Then the area of remaining sphere is $4\pi R^2-2\pi R^2(1-\cos\theta)=\pi R^2(4-\theta^2)$. Hence,
$$\sigma=\frac{Q}{\pi R^2(4-\theta^2)}$$

Lets find the electric field at the hole. I use the result of electric field on the axis of ring at a distance z from the centre. The result is
$$E=\frac{kqz}{(z^2+r^2)^{3/2}}$$
In the case of selected ring, $z=R(1-\cos\alpha)$ and $r=R\sin\alpha$. Substituting and simplifying:
$$dE=\frac{k\sigma \pi}{\sqrt{2}}\frac{\sin\alpha}{\sqrt{1-\cos\alpha}}d\alpha$$
Integrating the above with limits for $\alpha$ from $\theta$ to $\pi$, we get:
$$E=2k\sigma \pi(1-\sin(\theta/2))=\frac{2kQ}{R^2}\frac{1-\sin(\theta/2)}{4-\theta^2}$$

I evaluated the series expansion at $\theta=0$ using wolfram alpha. http://www.wolframalpha.com/input/?i=series+expansion+(1-sin(x/2))/(4-x^2)+at+x=0

Neglecting the terms containing $\theta$, we are simply left with $1/4$ i.e

$$E=\frac{2kQ}{R^2}\frac{1}{4}=\frac{Q}{8\pi \epsilon_{\circ} R^2}$$

Does this look good? :)

#### Attached Files:

• ###### E-field at hole.png
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Last edited: Jan 8, 2014
14. Jan 8, 2014

### ehild

Hi Pranav,

You are a HERO! I haven't check the derivation in detail yet, but it looks good ...

ehild

15. Jan 8, 2014

### rude man

I got the same result by superposition:

Take the arbitrarily thin (thickness = 2Δr) unblemished shell, slice it into two essentially equal volumes radially (one shell has radius = r - Δr, the other r + Δr). By Gauss the field at radius r is E = Q/8πε0R2.

Then, recognize that a cylindrical section of the shell equal to the hole diameter and thickness, and stuffed with volumetric charge density equal and opposite to that of the unblemished shell, has zero E field at its middle (that is true for any uniform charge density BTW).

So, bottom line, by superposition the E field in the hole center is that of the unblemished shell at its radial center.

P.S. 'unblemished' means without a hole in it.

Last edited: Jan 9, 2014
16. Jan 8, 2014

### haruspex

It's also interesting to take the opposite approach - avoid even needing to know the solution for an infinite disc.
Let the field just outside the (complete) sphere be S and the field at distance d from the centre of the disc radius r be f(r, d) away from the disc. So the field just outside the hole will be S - f(r, d). Since there's no field inside a complete sphere, the field just inside will be -(-f(r,d)) = f(r, d). Since there's no charge in the vicinity of the centre of the hole, the field must vary continuously through the hole. Taking the limit as d tends to 0 we have f(r, 0) = S/2.

17. Jan 8, 2014

### Tanya Sharma

What about the field lines emerging from charges marked in green ? Do they get cancelled by their diametrically opposite charges (marked in purple) ?

Another doubt I am having is that -

The field lines are imaginary lines which

1) terminate on charges
2) terminate at infinity
3) can get cancelled , just as in this case ,except the field lines from the diametrically opposite charges to the hole get cancelled

Is my understanding correct ?

#### Attached Files:

• ###### spherehole2.JPG
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9.3 KB
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18. Jan 8, 2014

### ehild

Sorry, I intended to made them red. No idea how they became green .... They were still red in the first picture.
There are the same number of field lines emerging from all dots, both inward and outward. But the net inward lines are cancelled by the lines from the opposite charges, and the outward field lines doubled. Because of the missing charges at the hole, the field lines from the opposite charges go through the sphere surface.

ehild

19. Jan 9, 2014

### Tanya Sharma

I made them green , in order to be specific .

1) Does that mean there are regions in the sphere where electric field is zero and another region where it is σ/2ε ?
2) Does that mean electric field just above the hole is σ/2ε whereas just above the rest of the shell's surface is σ/ε ?

Last edited: Jan 9, 2014
20. Jan 9, 2014

### ehild

Ah it was you!! I was lazy to draw all the field lines...

It is σ/2ε just on the area of the hole and on the opposite side.
It is (approximately) valid, in case of a very small hole.
See Pranav's derivation.

ehild

21. Jan 9, 2014

### Saitama

Thank you ehild.

Btw, there was no need to use wolfram alpha, simple binomial approximation would also do. I had

$$\frac{1-\sin(\theta/2)}{4-\theta^2}\approx \frac{1}{2}\frac{2-\theta}{(2-\theta)(2+\theta)} = \frac{1}{4}\left(1+\frac{\theta}{2}\right)^{-1}$$

22. Jan 9, 2014

### haruspex

No. Again, treat the holed sphere is a complete sphere plus a negative disc. For points within the sphere, we know the whole sphere generates no field, so we can throw that away. Thus the field will be just that provided by the small disc.
Sort of, but the σ/2ε only applies directly above the centre of the hole, and the σ/ε is only exact for a whole sphere. If you start just above the centre of the hole and move around the sphere keeping a constant distance from the sphere's centre, the field will increase continuously from σ/2ε to something close to σ/ε. Near the edge of the hole the gradient of the field gets arbitrarily steep.