Flux Through Gaussian Surface

In summary, the conversation discusses the use of Gauss's Law to find the charge enclosed within a spherical Gaussian surface located between two metal spheres and outside a larger metal sphere. The flux through these surfaces is given, and by using Gauss's Law, the total charge and surface charge densities of each sphere can be determined.
  • #1
tomrja
18
0

Homework Statement


A metal sphere of radius a is surrounded by a metal shell of inner radius b and outer radius R, as shown in the diagram below. The flux through a spherical Gaussian surface located between a and b is 1.20Q/εo and the flux through a spherical Gaussian surface just outside R is 0.80Q/εo.

a) What is the total charge on the inner sphere? (Express your answer as a multiple of Q. For example, if the total charge is 0.2Q, then input 0.2).

b) What is the surface charge density of the inner sphere? (Express your answer as a multiple of Q/a2.)

c) What is the total charge on the inner surface of the outer sphere? (Express your answer as a multiple of Q.)

d) What is the surface charge density of the outer surface of the outer sphere? (Express your answer as a multiple of Q/R2.)


Homework Equations



[tex]\phi[/tex]= integral(E dot dA)= Qenclosed/epsilon not

The Attempt at a Solution



I tried to find the charge enclosed but no charges were given. I don't know how the flux between a and b and the flux through a spherical Gaussian surface come into play. Thanks for your help!
 

Attachments

  • prob19a.gif
    prob19a.gif
    5.1 KB · Views: 1,090
Physics news on Phys.org
  • #2
What does Gauss's Law tell you?
 
  • #3
Flux is equal to the charge enclosed within the surface divided by a constant. This is Gauss' Law.
So for the surface between a and b, say the radius of the surface is r,

[tex] \phi_E = 1.20Q/ \epsilon_0 = q_{enc}^r / \epsilon_0 [/tex]

where [tex] q_{enc}^r [/tex] is the charge enclosed within the radius r

Then you also have for the gaussian surface outside R:

[tex] \phi_E = 0.80Q/ \epsilon_0 = q_{enc}^R / \epsilon_0 [/tex]

These relations should help
 

1. What is a Gaussian surface?

A Gaussian surface is an imaginary surface that is used in Gauss's law to calculate the electric flux through a closed surface. It is typically chosen to be a symmetrical shape, such as a sphere or cylinder, to simplify the calculations.

2. How is electric flux through a Gaussian surface calculated?

The electric flux through a Gaussian surface is calculated using Gauss's law, which states that the flux is equal to the charge enclosed by the surface divided by the permittivity of free space. This can be expressed mathematically as Φ = Q/ε0, where Φ is the flux, Q is the enclosed charge, and ε0 is the permittivity of free space.

3. What is the significance of a closed Gaussian surface?

A closed Gaussian surface is necessary for the application of Gauss's law because it allows for the calculation of the total electric flux passing through a given area. The closed surface ensures that the electric field vectors are perpendicular to the surface, simplifying the calculation of the flux.

4. Can a Gaussian surface be any shape?

No, a Gaussian surface must be a symmetrical shape in order to apply Gauss's law. This ensures that the electric field is constant over the entire surface, making the calculation of the flux more accurate.

5. How is the direction of the electric flux through a Gaussian surface determined?

The direction of the electric flux through a Gaussian surface is determined by the direction of the electric field and the orientation of the surface. The flux is considered positive if the electric field and surface are in the same direction, and negative if they are in opposite directions.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
3K
  • Introductory Physics Homework Help
Replies
11
Views
395
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
503
  • Introductory Physics Homework Help
Replies
17
Views
404
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
21
Views
667
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top