# Electric Flux and Gauss's law

• psysicsfeet
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.

## 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.

∮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.

psysicsfeet said:
thank you SammyS. can you give me some hints for second question?
What is the electric field within the conducting material itself?

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?

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.