Finding the electric field magnitude in each region of a sphere

[Joa Boaz]

Homework Statement

A charged insulating spherical shell has winner radius of a/3 an an outer radius of a. The cross section is as shown on the picture. The outer shell has a non-constant volume charge density of ρ = 6*α*r^3. Find the electric field magnitude in each region (outside the object, in the shell itself, and in the cavity contained in the shell). Find the electric potential in each region, and sketch a graph of the electric potential as a function of radius for r = 0 to r = 2a and label the potential at r = a/3, a, and 2a. The reference point for the potential is that V = 0 to r = ∞

Homework Equations

Not sure but,

E = (1/4∏*ε) * Q/r^2

E = (1/4∏*ε) * (Q/r^3) * r

The Attempt at a Solution

q(inside) = -Q(sphere) = - 4/3 * ∏ * (a^3) ρ = (-4/3)*∏*((a^6)/729)*6α = (-8/729)*(a^6)*∏*α
q(outise) = - q(inside) = (8/729)*∏*α

σ(outer) = q(outside)/4∏*a^2 = (2/729)*(a^4)*∏*α

Not sure if the above is correct, but I am not sure what I have to do after this. Please can anyone help, thank you.

 

[vanhees71]

I'm not so sure about your notation. In such cases it's much easier to integrate the electrostatic Poisson equation,
$$\Delta \Phi=-\frac{1}{\epsilon_0} \rho.$$
In this case you have spherical symmetry and thus $\Phi=\Phi(r)$, where $r=|\vec{r}|$ is the distance from the origin (the center of your spheres). Just look up the Laplace operator in spherical coordinates and solve the differential equation in the different regions. The constants are determined by the appropriate boundary conditions.

 

[Joa Boaz]

Thank you for your help. But I am not sure if I am allowed to use the Laplace operator since the professor has not gone over the Laplace transform. Is there any other way to tackle the above question without using Laplace transform?

Thank you.

 

[vanhees71]

It's not about the Laplace transform but the Laplace operator. In Cartesian coordinates it reads
$$\Delta \Phi=\partial_x^2 \Phi + \partial_y^2 \Phi + \partial_z^2 \Phi.$$

 

[vela]

Homework Statement

A charged insulating spherical shell has winner radius of a/3 an an outer radius of a. The cross section is as shown on the picture. The outer shell has a non-constant volume charge density of ρ = 6*α*r^3. Find the electric field magnitude in each region (outside the object, in the shell itself, and in the cavity contained in the shell). Find the electric potential in each region, and sketch a graph of the electric potential as a function of radius for r = 0 to r = 2a and label the potential at r = a/3, a, and 2a. The reference point for the potential is that V = 0 to r = ∞

Homework Equations

Not sure but,

E = (1/4∏*ε) * Q/r^2

E = (1/4∏*ε) * (Q/r^3) * r

The Attempt at a Solution

q(inside) = -Q(sphere) = - 4/3 * ∏ * (a^3) ρ = (-4/3)*∏*((a^6)/729)*6α = (-8/729)*(a^6)*∏*α
q(outise) = - q(inside) = (8/729)*∏*α

σ(outer) = q(outside)/4∏*a^2 = (2/729)*(a^4)*∏*α

Not sure if the above is correct, but I am not sure what I have to do after this. Please can anyone help, thank you.

It's not very clear at all what you're doing. As far as I can tell, you found a similar example in the book or your notes and just plugged in the values from this problem into the equations from the example. That approach doesn't require any real understanding, and it won't help you in the long run. In this case, you seemed to have picked an example that doesn't apply to the described situation.

Try again, this time justifying why you're taking each step. To find the electric field, use Gauss's law. You have spherical symmetry here, so use it. Because the charge density is non-uniform, you'll need to integrate to find the enclosed charge.

 

[Joa Boaz]

It's not about the Laplace transform but the Laplace operator. In Cartesian coordinates it reads
$$\Delta \Phi=\partial_x^2 \Phi + \partial_y^2 \Phi + \partial_z^2 \Phi.$$

I do really appreciate your help. It seems that my level of physics is very low because I just don't see how I can use partial derivatives on this problem. But I really do appreciate your help.

 

[Dr Transport]

This problem is solvable by Gauss's Law since it is spherically symmetric.