Electric field inside/outside (uniformly charged sphere)

In summary: If you insulate the sphere with a material that doesn't allow charges to move, like metal for example, then the electric field inside will be zero.
  • #1
bolzano95
89
7
Thread moved from the technical forums, so no Homework Template is shown
A sphere of radius a carries a total charge q which is uniformly distributed over the volume of the sphere.

I'm trying to find the electric field distribution both inside and outside the sphere using Gauss Law.

We know that on the closed gaussian surface with spherically symmetric charge distribution Gauss Law states:
[tex]\frac{q}{ε_0}= \oint \vec{E} \cdot d\vec{A}[/tex]
1. Outside of sphere:
Logically, the charge outside of a sphere will be always on the Gaussian surface and it doesn't change, therefore the electric field outside of a sphere:
[tex]E=\frac{q}{4πε_0r^{2}}[/tex]
2. Inside of sphere:
Because the charge is symmetrically distributed on the surface and if I image a little sphere with radius r<a inside the sphere with radius r, the little sphere will have less charge on its surface.
[tex]E=\frac{q \ r}{4πε_0a^{3}}[/tex]
Is this explanation sufficient?
The problem I'm having is that in textbook is written and drawn that the electrical field inside a charged spherical shell = 0.
Isn't sphere = shell (just taking smaller shell for point 2)?
What am I missing?
 
Last edited:
Physics news on Phys.org
  • #2
Is the sphere conducting? What effect does the answer have on the charge distribution?
 
  • #3
Logically I would think that if I have a conducting sphere the charge is located also inside of a sphere (for example the sphere is made of copper and inside there are also charged particles, but for the insulating one (Iike a thin shell made of metal, but inside is filled with insulator) I suppose the charge will be distributed only on the outer surface, therefore the electric field inside will be 0.

But here I have a problem:
For sphere written up, it is assumed that we have an insulating sphere of radius a.
(http://farside.ph.utexas.edu/teaching/302l/lectures/node30.html)
But by my thinking the electric field inside will be 0, but if I look in the previous post, I see that it is not.

What am I missing?
 
  • #4
If the charges can move, and they are like charges, where will they go to? It may be easiest to imagine just two free excess charges to start with then add more.

If the charges cannot move, they obviously won't go anywhere.

Which of these is the insulator and which the conductor?
 
  • #5
charges can move = conductor
charges cannot move = insulator
 
  • #6
Right. So the question must be about an insulator because it says uniform charge throughout the volume. If you go back and look at the references giving zero field inside you'll see they're talking about conductors.
 

1. What is an electric field?

An electric field is a physical phenomenon that describes the influence of electrically charged objects on other objects within their proximity. It is a vector quantity, meaning it has both magnitude and direction.

2. What is a uniformly charged sphere?

A uniformly charged sphere is a spherical object that has a constant charge distribution, meaning the charge is evenly distributed throughout the surface of the sphere.

3. What is the electric field inside a uniformly charged sphere?

The electric field inside a uniformly charged sphere is zero. This is because the electric field lines within a uniformly charged sphere cancel each other out due to the symmetry of the charge distribution.

4. What is the electric field outside a uniformly charged sphere?

The electric field outside a uniformly charged sphere can be calculated using Coulomb's law. It is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance from the center of the sphere.

5. How does the electric field change as you move away from a uniformly charged sphere?

The electric field outside a uniformly charged sphere decreases as the distance from the center of the sphere increases. This is because the electric field lines spread out and become weaker as they move away from the source of the electric field.

Similar threads

  • Introductory Physics Homework Help
Replies
17
Views
384
  • Introductory Physics Homework Help
Replies
2
Views
877
Replies
1
Views
122
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
907
  • Introductory Physics Homework Help
Replies
14
Views
624
  • Introductory Physics Homework Help
Replies
9
Views
246
  • Introductory Physics Homework Help
Replies
26
Views
553
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
1K
Back
Top