Gauss Law problem

  • Thread starter mborn
  • Start date
  • #1
30
0
Hi, I have this problem in my Physics Book:

The electric field everywhere on the surface of a hollow sphere of radius 0.75m is found to be 890 N/C and points radially toward the center of the sphere,
(a) What is the net charge within the sphere's surface?
(b) What can you tell about the nature of the charge and its distribution inside the sphere?

I have answered the first part, it is 5.56x10^-8 C
but for the second part, I can't follow the answer (I have it). It says it is negative and uniformly distributed. But I know that:

1- Gauss' law says nothing about charge distribution on a surface. My be due to symmetry, but again, Gauss' law says nothing about charge distribuion.
2- There can be no charges on the inner surface, this will lead to an electric field at the center. The electric field lines will be directed to the center.

What is wrong in my understanding?
 

Answers and Replies

  • #2
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
7,083
18
Where does it say in the question that you must only use Gauss' Law to make a statement about the charge ditribution ?

Anyway, saying that the charge distribution is uniform would be incorrect. The right answer should be that :
[tex]\rho (r, \theta , \phi) = \rho _0 f(r) [/itex]
 
  • #3
30
0
But, it said in Halliday that uniform distribution occurs -only- for the spherical case.
so it is solved if this is to be taken as a rule, Yet, how negative?
 
  • #4
Doc Al
Mentor
45,140
1,439
mborn said:
Yet, how negative?
You are given that the field points inward.
 
  • #5
30
0
yes, it is negative on the upper surface, but I say there should be no charge on the inner surface
 
  • #6
cepheid
Staff Emeritus
Science Advisor
Gold Member
5,192
38
I think you're making this problem more difficult than it is. There is no charge on the surface of the sphere...we don't even know whether it's a conductor or a dielectric, and such considerations are irrelevant, because it need not even be a physical surface! It's just a Gaussian surface, one at which we are measuring the properties of the local electric field in order to determine some information about the distribution of charge enclosed by said surface. Gauss's law can also used to solve the opposite problem (given some info about the charge enclosed, what can you deduce about the flux of the electric field, and therefore the electric field itself?)

In this case, given the properties of the local electric field (more specifically, the field at all points on the surface of the sphere), what information does Gauss's Law (below) tell us about the charge enclosed?

[tex] \oint{\vec{E} \cdot d\vec{a}} = \frac{Q_{enc}}{\epsilon_0} [/tex]

It gives us two pieces of information:

1. The field is radially inward...an inward field suggests that Qenc is some cluster of negative charge. More rigorously, since [itex] \vec{E} \ \ \text{and} \ \ d\vec{a} [/itex] point in opposite directions, (radially inward and outward, respectively) it follows that [itex] \vec{E} \cdot d\vec{a} [/itex] is negative: The net flux through the surface is negative, therefore so is Qenc.

2. The electric field is radial and its strength is constant at all points on the sphere. Since all of these points are equidistant from the centre, it seems that the magnitude of the electric field, depends only on radius i.e. it is a function of r. What would produce such a nice, radially inward E(r)? A point charge at the centre of the Gaussian sphere certainly would, and IIRC, so would any uniform volume charge distribution centred in the Gaussian sphere.

There are your two answers...that the charge is negative, and that it is distributed uniformly.
 
  • #7
30
0
Thanks all, may be he is talking about an INCLOSED charge, and in this case it is what you all said. But, if this charge is ON the spherical shell, then I have a right to confuse a little! :smile:
 

Related Threads on Gauss Law problem

  • Last Post
Replies
1
Views
8K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
841
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
10
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
0
Views
3K
  • Last Post
Replies
2
Views
2K
Top