Calculating Electric Field in Hollow Metal Sphere

[dtesselstrom]

Homework Statement

A hollow metal sphere has inner radius a and outer radius b. The hollow sphere has charge +2Q. A point charge +Q sits at the center of the hollow sphere.
Determine the magnitude of the electric field in the region r < a.
Determine the magnitude of the electric field in the region a < r < b.
Determine the magnitude of the electric field in the region r >b.
Solve in Q/(pi*Eo*r^2) units

Homework Equations

Net Flux=Q/Eo
Net Flux= integral(E*dA)

The Attempt at a Solution

Not really sure how to approach. I figured it was fairly straight forward since the units you solve for pretty much get rid of the calculation. I thought it would be 1 when r<a 2 for when a<r<b and 0 for r>b but none of them work and I've used numbers around that. So I guess I need some explaining on this because I can't figure out any of the problems similar to this when it comes to net flux. So if anyone can also provide some explanation to what I am doing here that would be appreciated.

 

[anathema]

Hello,

First, let's review the concept of electric field and flux. Electric field is a vector quantity that describes the force per unit charge at a given point in space. It is given by the equation E = kQ/r^2, where k is the Coulomb's constant and r is the distance from the point charge. Flux, on the other hand, is a measure of the amount of electric field passing through a given surface. It is given by the equation Flux = E*A, where E is the electric field and A is the area of the surface.

Now, let's apply these concepts to the problem at hand. We have a hollow metal sphere with charge +2Q, and a point charge +Q at the center. We are asked to determine the magnitude of the electric field in three different regions: r<a, a<r<b, and r>b. Here's how we can approach each region:

1. For r<a: Since the point charge is at the center, the electric field at any point within the hollow sphere will be directed towards the center. This means that the electric field will be constant and have the same magnitude at all points within the sphere. Using the equation E = kQ/r^2, we can calculate the electric field at any point within the sphere. However, we need to keep in mind that the electric field due to the point charge and the charged hollow sphere will add up. So, the electric field at any point within the sphere will be given by the vector sum of the electric fields due to the point charge and the hollow sphere. In this case, since the point charge is at the center, the electric field due to the point charge will be zero (since r=0). Therefore, the electric field at any point within the sphere will be given by the electric field due to the hollow sphere, which is E = k(2Q)/r^2 = 2kQ/r^2. In terms of Q/(pi*Eo*r^2) units, this will be 2/pi.

2. For a<r<b: In this region, we are still within the hollow sphere, but outside of the point charge. Therefore, the electric field at any point within this region will be given by the electric field due to the hollow sphere only. Using the same equation as above, we can calculate the electric field at any point within this region. However, since we are now at a distance r

 

[m2003]

I would approach this problem by first looking at the basic principles of electric fields and flux. The electric field is a vector quantity that describes the force per unit charge experienced by a test charge in the presence of other charges. It is given by the equation E = kQ/r^2, where k is the Coulomb's constant, Q is the charge of the source, and r is the distance from the source.

In this problem, we have a hollow metal sphere with a charge of +2Q distributed on its surface. This charge will create an electric field both inside and outside the sphere. The point charge +Q at the center of the sphere will also contribute to the electric field.

To determine the electric field in the different regions, we can use Gauss's law. This law states that the net electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space (εo). Mathematically, this can be written as Φ = Q/εo.

In the region r<a, the electric field will only be due to the point charge +Q at the center of the sphere. Using the equation E = kQ/r^2, we can calculate the electric field at any distance r from the center. However, since we are solving in Q/(πεo*r^2) units, we can simply write E = Q/(πεo*r^2) as the magnitude of the electric field.

In the region a<r<b, the electric field will be due to both the point charge +Q and the charge +2Q on the surface of the sphere. Using Gauss's law, we can write Φ = Q/εo = E*A, where A is the area of the surface of the sphere. Solving for E, we get E = Q/(εo*A). Since A = 4πr^2, we can rewrite this as E = Q/(4πεo*r^2).

In the region r>b, the electric field will only be due to the charge +2Q on the surface of the sphere. Using Gauss's law, we can write Φ = Q/εo = E*A, where A is the area of the surface of the sphere. Solving for E, we get E = Q/(εo*A). However, since the charge +2Q is distributed on the surface of the sphere, the