Electric fields of a spherical shell

In summary: A volume at the surface does have a contribution, but it is essentially zero.In summary, the conversation discusses Gauss's Law and the magnitudes of electric fields at different distances from a spherical shell with a uniformly distributed charge. There is also a question about denoting electric field as ε and where to find electric field formulas and proofs. The main points are clarified, including the fact that the charge on a surface is effectively zero due to the charge being uniformly distributed over a volume. Different distances can be solved for using the same steps, and a search engine can be used to find electric field formulas and proofs.
  • #1
MathewsMD
433
7
My class hasn't delved into Gauss's Law much besides describing conductors at electrostatic equilibrium to have no net electric field or force within itself.

For the picture, the question is:

What are the magnitudes of the electric fields at:

1) r = a
2) r = 3/2 a
3) r = b

Firstly, if r = 0, E = 0, correct?

At r = a, is it correct to assume that the charge on the smaller spherical shell is also 0 since it has no electric fields from it? Is this the correct explanation?

For r = 3/2a, I was having a bit of trouble and was wondering if we are to find

##Q = ρ\frac {4}{3}π [(1.5a)^3 - a^3] ##

Then:

##E = kQ/(1.5a)^2## ?

Would the same steps for r = 3/2a be followed for r = b? Any help regarding how to answer this question, with an explanation (please!), and possibly links to links that you find very informative with tho topic would be greatly appreciated!

Also, does anyone know where I could find a list of electric field formulas for various objects (infinitely thin and long sheets, spheres, quadrupoles, etc.) and accompanying proofs?

And finally, is it incorrect to denote electric field as ε? I've seen both around but E more commonly and was uncertain if it's common convention or not.
 

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  • #2
MathewsMD said:
Firstly, if r = 0, E = 0, correct?
Yes
At r = a, is it correct to assume that the charge on the smaller spherical shell is also 0 since it has no electric fields from it?
What "smaller spherical shell"? There is only one shell. Do you mean the inner surface of the shell?
The charge is uniformly distributed through a volume. That means the charge on a surface is effectively zero.
For r = 3/2a, I was having a bit of trouble and was wondering if we are to find

##Q = ρ\frac {4}{3}π [(1.5a)^3 - a^3] ##

Then:

##E = kQ/(1.5a)^2## ?
Yes
Would the same steps for r = 3/2a be followed for r = b?
Yes. In fact, you could save yourself a bit of work by solving for a radius which is an arbitrary multiple of a, r = ca, then plug in c = 1.5, c = 2.
Also, does anyone know where I could find a list of electric field formulas for various objects (infinitely thin and long sheets, spheres, quadrupoles, etc.) and accompanying proofs?
Only by doing what you can do - use a search engine.
And finally, is it incorrect to denote electric field as ε? I've seen both around but E more commonly and was uncertain if it's common convention or not.
Pass.
 
  • #3
haruspex said:
That means the charge on a surface is effectively zero.

Do you mind expanding on this point? If charge is uniformly distributed, won't it be the same everywhere, including on the surface of the inner sphere?
 
  • #4
MathewsMD said:
Do you mind expanding on this point? If charge is uniformly distributed, won't it be the same everywhere, including on the surface of the inner sphere?
Since it is a finite charge uniformly distributed over a volume, the charge in a given portion is proportional to the volume of that portion. An area has no volume, so contains no charge.
 
  • #5


I appreciate your curiosity and willingness to learn about electric fields. To answer your questions, let's first review the concept of electric fields and Gauss's Law.

Electric fields are created by charged particles and are present in the space surrounding these particles. The strength of an electric field is measured by the force it would exert on a test charge placed at a specific point in space. The direction of the electric field is always in the direction of the force it would exert on a positive test charge.

Gauss's Law states that the net electric flux (or flow) through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space. In simpler terms, this means that the electric field at a point is proportional to the amount of charge enclosed by a surface surrounding that point.

Now, let's apply these concepts to the scenario you described. In the case of a spherical shell, we can assume that the charge is uniformly distributed on the surface of the shell. Therefore, at a distance r = a from the center of the shell, the electric field would be zero. This is because the electric field from one side of the shell would be cancelled out by the electric field from the other side, resulting in a net electric field of zero.

At r = 3/2a, we can use Gauss's Law to calculate the electric field. As you correctly stated, we can find the charge enclosed by the surface at r = 3/2a by using the volume charge density and the volume of the spherical shell. Then, we can use the formula E = kQ/r^2 to calculate the electric field at that point.

For r = b, we can follow the same steps as above to calculate the electric field. However, the charge enclosed by the surface would be different as the radius of the shell is now b. Therefore, the electric field at this point would also be different.

Regarding your question about denoting electric field as ε, it is not incorrect to use this symbol. In fact, ε is often used to represent the permittivity of a material, which is a constant that relates the electric field to the charge present in that material.

To find a list of electric field formulas for various objects, I would recommend looking at a textbook or online resources specifically focused on electromagnetism. There are also many videos and tutorials available online that can help you understand and visualize the concept of electric fields better.

I
 

1. What is an electric field of a spherical shell?

The electric field of a spherical shell is a type of electric field that is created by a spherical shell of charge. It is a vector field that describes the direction and strength of the electric force that would be experienced by a charged particle placed in the vicinity of the spherical shell.

2. How is the electric field of a spherical shell calculated?

The electric field of a spherical shell can be calculated using the formula E = kQ/r^2, where E is the electric field, k is the Coulomb's constant, Q is the total charge of the spherical shell, and r is the distance from the center of the shell. This formula applies to points both inside and outside of the spherical shell.

3. Does the electric field of a spherical shell depend on the distance from the center?

Yes, the electric field of a spherical shell does depend on the distance from the center. As the distance increases, the electric field strength decreases. This is because the electric field is spread out over a larger area as the distance increases, resulting in a weaker electric field.

4. How does the electric field of a spherical shell change with the amount of charge on the shell?

The electric field of a spherical shell is directly proportional to the amount of charge on the shell. This means that as the charge on the shell increases, the electric field strength also increases. However, the electric field is inversely proportional to the square of the distance from the center, so as the charge increases, the electric field strength decreases with distance.

5. Can the electric field of a spherical shell be zero?

Yes, the electric field of a spherical shell can be zero at certain points. This occurs when the distance from the center of the shell is equal to the radius of the shell. At this point, the electric field strength is cancelled out by the contributions from opposite sides of the spherical shell, resulting in a net electric field of zero.

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