# Homework Help: Electric field in an off-center hole of an uniformly charged sphere

1. Mar 29, 2008

### yiuyan

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

Hi, imagine an insulating uniformly charged sphere. Its electric field points from the center outwards radially. Now cut out another smaller sphere from inside, such that their centers do not coincide. What is the electric field inside the hollow like?

2. Relevant equations
Principle of superposition

3. The attempt at a solution
I figured that by principle of superposition I could subtract the vectors of the old E-field by the E-field due to the removed portion. Is this correct? And it doesn't help in visualizing the pattern too.

2. Mar 29, 2008

### merryjman

This is a pretty classic Gauss's Law problem; I've seen it in many textbooks and a friend of mine even saw it on his graduate level EM exam. You are right on the money about the superposition, but doing it your way would be a real headache in terms of finding the field due to this asymmetric weird shape. Instead, superimpose a second sphere of opposite charge density, so that the charge inside is still zero. This will allow you to use Gauss's Law for each sphere.

You should find that one component of the E-field cancels anyway, due to symmetry, and the other component will be some comibation of the charge density $$\rho$$, the radius of the small sphere, some integers, and the vacuum permittivity$$\epsilon_{0}$$

3. Mar 29, 2008

### yiuyan

Thank you for the prompt reply, but I don't really understand it.

How does superimposing a sphere of opposite charge density changes things? Gauss Law is concerned with the charge a surface encloses; if the charges cancel out eventually, there shouldn't be a difference right. Also, if Gauss Law is to be used, where should the surface be drawn at? I didn't use Gauss Law much on this one due to its asymmetry.

4. Mar 30, 2008

### merryjman

I assume you know how to use Gauss's Law to find the field within a dielectric (nonconducting) sphere. It is a somewhat difficult calculation, but most textbooks show you how to do it. So do this separately for each sphere, the real positive one and the superimposed negative one. Make your surface a spherical shell concentric with the real sphere. Unless you have some information on the relative sizes of the spheres it will be somewhat messier than I hinted above; the version I have seen tells you the radii of the two spheres.

5. Mar 31, 2008

### yiuyan

Sorry I can't really see how Gauss' Law would come in handy in this case. Correct me if I'm wrong, but what Gauss' Law relates is the net flux that comes out of the Gaussian surface is proportional to the magnitude of charge enclosed. It doesn't say anything about the direction or strength of an E-field at any particular point on the surface. Of course unless it is a highly symmetrical one, then we can make necessary assumptions. But in this case, since the smaller sphere is off center, Gauss' Law wouldn't tell us much about the E-field inside the hollow right?

6. Mar 31, 2008

### merryjman

Gauss's Law absolutely DOES say something about the strength of the E-field, since electric flux is directly proportional to the E-field. That's the reason you use Gauss's Law in the first place - to express the electric field in terms of enclosed charge. For example, you've surely by now used Gauss's Law to find the electric field outside a long charged rod, or an infinite charged sheet, or many other cases...in this situation, you will be finding the electric field within the sphere using GL. As for the direction, usually the field lines are radially inward or outward, like for a point charge.

It doesn't matter that the other sphere is off-center. Use Gauss's Law separately for each sphere and then use the superposition principle for a region inside the smaller one. If you give more specific information; i.e. radii of the spheres and their arrangement, or a picture, we might be able to help more.

7. Apr 1, 2008

### yiuyan

Here is one I've drawn. Sphere with radius R has a smaller hollow sphere inside with radius r (r<R). The distance between their centers is L, and the bigger sphere has uniform charge density rho. How is the E-field inside the hollow like?

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8. Apr 1, 2008

### Staff: Mentor

The field everywhere is just the superposition of two field contributions: (1) the field from the solid sphere of charge of radius R; (2) the field from the sphere of opposite charge of radius r.

Use Gauss's law to find those field contributions. (You can't use Gauss's law directly on the original charge distribution since it lacks symmetry--but you can certainly use it for the spherical distributions.)

9. Apr 1, 2008

### merryjman

First, use Gauss's Law for the large sphere, and ignore the hole. Derive an expression for the electric field anywhere inside the sphere. Your result should be linear so that as r increases the field increases. Next, use Gauss's Law again for the small negative sphere (BTW I would pick a different letter for its radius, like a or something, so that you don't mix up r from Gauss's Law) and derive an expression for the field inside the negative sphere only. Then simply add your two results together, and evaluate for some spot inside the hole. The only tricky thing will be relating a, R, and L to the r for the spot you've chosen.

10. Apr 1, 2008

### yiuyan

Sorry if I mis phrased my question. I had the idea of getting the E-field in the hole too, what I'm having trouble over is how it looks like. I tried drawing the lines of the big sphere, and the lines of the smaller one and adding them up vectorially, but it was tedious and beyond me. It is possible to do it mathematically, but the result did not help in visualizing the pattern much, only the radial line that joins the two centres. Is there another way?

11. Apr 1, 2008

### merryjman

I tried to make a VERY rough sketch of what the field lines might look like. Since the field inside get stronger as you approach the surface, I tried to increase the density of field lines near the surface. Inside the hollow, the field is in a uniform direction since the horizontal component of the field cancels. Finally, if you were very far from this thing it should resemble a point charge and all field lines are radially outward and never cross. I hope all these things were conveyed in the picture.

#### Attached Files:

• ###### hollowed out dielectric sphere.bmp
File size:
214.8 KB
Views:
700
12. Apr 4, 2008

### yiuyan

Thanks for your answers merryjman, and for taking the time for the sketch. I agree with you that from a very far distance, the E-field would be as if they originated from a point charge. But I can't see how the E-field inside the hollow can be horizontal.
I've made another sketch, a rather similar one. The bigger dark gray sphere has uniform positive charge density rho, and so its E-field is nice and points radially outward. Now the smaller sphere has uniform charge density rho too, but negative now. Such that by putting it in, we effectively create a hollow sphere with no enclosed charge. Now if we do the same addition for the E-fields, by superposition; what I see is that there is only one radial component of the E-field, and that is the one which passes through both centers. The remaining E-field have to be added, and that is the part that causes confusion. Do correct me where I went off track. Thanks.

#### Attached Files:

• ###### untitled.JPG
File size:
17.7 KB
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13. Feb 17, 2011

### rusevm

http://webapps3.ucalgary.ca/~dppvan/phys259/examples/Sphere%20with%20a%20hole%20in%20it.pdf [Broken]

It illustrates everything in a very accessible way. Credits to University of Calgary for uploading that file.

Last edited by a moderator: May 5, 2017
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