Electric Flux and Gauss's law

In summary: So, within the inner shell, what is the electric field everywhere? and within the outer shell, what is the electric field everywhere?In summary, the first question involves finding the net charge inside a cube immersed in an electric field, using the equation ∮E*dA = Q/E0. The second question involves finding the charge on the outer surface of a conducting, concentric sphere system, taking into consideration the electric field within the conducting material itself.
  • #1
psysicsfeet
5
0

Homework Statement


1)
A large cube has its bottom face on the x-z plane and its back face on the x-y plane. The corners on the x-axis are at (3.39 m,0,0) and (12.3 m,0,0). The cube is immersed in an electric field pointing in the positive x-direction, and given by:

E = (91.2x^2 - 2.9)i, x is the distance along the x-axis in m, and E is in N/C.

Find the net charge Q inside the cube, in μC.

2)
There are two hollow, conducting, concentric spheres, with air between the spheres:

-Sphere 1 is the inner sphere; it has inner radius a = 2.35 mm, outer radius b = 8.53 mm, and carries charge Q1 = -8.41 pC.

- Sphere 2 is the outer sphere; it has inner radius c = 3.58 cm, outer radius d = 7.92 cm, and carries charge Q2 = 5.84 pC.

- At the common center is point charge Q = -7.21 pC.

Find the charge on the outer surface of sphere 2, in pC.

Homework Equations


∮E*dA = Q/E0

The Attempt at a Solution


Q 1) I try to use equation ∮E*dA = Q/E0 to solve first question, but what I am confusing is the variable E is not a constant, should I plug in the value of x to solve it? and for the surface area since it is a cube, I should count as 6r^2? r is the slice length of the cube.
Q 2) I guess it is a simple question but I just can not figure out the relationship between two spheres.

thank you for help.
 
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  • #3
  • #4
psysicsfeet said:
thank you SammyS. can you give me some hints for second question?
What is the electric field within the conducting material itself?
 
  • #5
SammyS said:
What is the electric field within the conducting material itself?
within the conducting material itself. do you mean inside the inner radius of inner cycle? is it kQ/a^2?
 
  • #6
psysicsfeet said:
within the conducting material itself. do you mean inside the inner radius of inner cycle? is it kQ/a^2?
I mean: within the conducting material itself, for either of the shells in this problem.
 

1. What is electric flux?

Electric flux is a measure of the amount of electric field passing through a given area. It is defined as the dot product of the electric field and the area vector.

2. How is electric flux calculated?

Electric flux is calculated by taking the dot product of the electric field and the area vector. This can be represented by the equation Φ = E⃗ ⋅ A⃗, where Φ is the electric flux, E⃗ is the electric field, and A⃗ is the area vector.

3. What is Gauss's law?

Gauss's law states that the electric flux through a closed surface is equal to the enclosed electric charge divided by the permittivity of free space. Mathematically, it can be represented by the equation Φ = Qenc/ε0, where Φ is the electric flux, Qenc is the enclosed electric charge, and ε0 is the permittivity of free space.

4. How is Gauss's law used in electrostatics?

Gauss's law is used in electrostatics to calculate the electric field of a charge distribution. By using Gauss's law, we can find the electric field at a point by knowing the charge enclosed by a Gaussian surface surrounding that point.

5. What is a Gaussian surface?

A Gaussian surface is a hypothetical surface used in Gauss's law to enclose a charge distribution. It is chosen to simplify the calculation of electric flux and to make use of the symmetry of the charge distribution. A Gaussian surface can be any closed surface, but it is often chosen to be a sphere, cylinder, or cube for easier calculations.

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