Electric Field Inside a Charged Insulating Sphere

In summary, the question asks for the magnitude of the electric field at a distance r from the center of a charged insulating shell with inner radius a and outer radius b. Using Gauss' law, it can be concluded that the electric field inside the cavity is 0, as there is no charge enclosed within a Gaussian surface placed inside the cavity. This is due to the symmetry of the sphere, where the electric field should cancel out in all directions. Therefore, the magnitude of the electric field at a distance r from the center of the shell where r < a is 0.
  • #1
yango_17
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1

Homework Statement


A charged spherical insulating shell has inner radius a and outer radius b. The charge density on the shell is ρ.
insulator.gif

What is the magnitude of the E-field at a distance r away from the center of the shell where r < a?

Homework Equations


Gauss' Law

The Attempt at a Solution


I read through the Gauss' law chapter in my textbook, and it stated that the internal electric field of such a charged insulator is zero. Therefore, I know the answer is 0 but I do not know why. Any explanation as to why this answer is correct would be appreciated. Thanks.
 
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  • #2
Well, Gauss's law states that: [tex]\oint \mathbf{E} \cdot \ d \mathbf{A} = \frac{Q_{enc}}{\epsilon_0}[/tex] What is the enclosed charge if you use a Gaussian surface inside the cavity?
 
  • #3
0. Is it because there is no charge in the cavity?
 
  • #4
Correct. If you imagine your Gaussian surface to be a sphere inside the cavity centered at the center of the larger sphere, no charge is inside (so Qenc = 0). All the charge is in the insulated sphere in this problem.

Now, the integral being equal to 0 is not a necessary condition to declare that the electric field is 0. For instance, [itex]\int _{-1} ^1 x \ dx = 0[/itex], so this is not enough to declare that there is no electric field inside.
Edit: changing the argument because I saw a flaw in it. Simply put: by symmetry, the electric field should cancel out everywhere inside the sphere.

If it didn't cancel out, which direction would get precedence? Why should any direction of a perfectly symmetrical sphere get the privilege of having the electric field not cancel out?
 

What is a charged spherical shell?

A charged spherical shell is a hollow, spherical object with a net electric charge. This charge is distributed evenly across the surface of the shell.

What causes a charged spherical shell to have a net charge?

The net charge on a spherical shell is caused by an excess or deficit of electrons on its surface. This can occur through the process of induction, where an external electric field causes a redistribution of charges on the surface.

How does the electric field inside and outside a charged spherical shell differ?

Inside a charged spherical shell, the electric field is zero. This is because the charges on the surface cancel each other out, resulting in no net electric field. Outside the shell, the electric field follows the same pattern as that of a point charge, with the same magnitude and direction.

What happens to the electric field inside a charged spherical shell if a point charge is placed at its center?

If a point charge is placed at the center of a charged spherical shell, the electric field inside the shell will still be zero. This is because the charges on the surface will still cancel out the electric field of the point charge.

What is the relationship between the charge on a spherical shell and the electric potential at its surface?

The electric potential at the surface of a charged spherical shell is directly proportional to the net charge on the shell. This means that as the charge increases, so does the electric potential. This relationship is described by the equation V = kQ/r, where V is the electric potential, k is the Coulomb constant, Q is the net charge on the shell, and r is the radius of the shell.

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