Dielectric sphere - Find the electric field

[Qyzren]

Consider a dielectric sphere of radius R and dielectric constant e_r. The sphere contains
free charges that have been uniformly distributed with density rho.

(a) Show that electric field inside the sphere is given by
E = [rho*r /(3*e_0*e_r)]* r hat

where r hat is the unit vector pointing in the radial direction.

(b) Show that the electric field outside the sphere is
E = [rho*R³]/[3*e_0*r²] * r hat

(c) Calculate the potential V at the centre of the sphere compared to that at infinity.

Any tips/hints that can help start me off will be appreciated.

 

[Claude Bile]

Start with Gauss' law. Try to calculate how E varies with the radius of your spherical Gaussian surface.

Claude.

 

[Qyzren]

well gauss' law is just integral of E.dA = Q/e_0 since we have a sphere the SA is just 4*pi*r^2 so it gives E = Q/[4*pi*e_0*R^2].

How does the density rho and dielectric constant e_r affect/influence this?

Thanks everyone for helping.

 

[Reshma]

well gauss' law is just integral of E.dA = Q/e_0 since we have a sphere the SA is just 4*pi*r^2 so it gives E = Q/[4*pi*e_0*R^2].

How does the density rho and dielectric constant e_r affect/influence this?

Thanks everyone for helping.

Use the Guass's law in presence of dielectrics.
Within the dielectric the total charge density can be written as:
$$\rho = \rho_b + \rho_f$$
where [itex]\rho_b[/tex] is the bound charge and $\rho_f$ is the free charge density resp.
The Gauss's law will be modified accordingly. Read up on the electric displacement vector and you can easily find the solution.

 

[olgranpappy]

well gauss' law is just integral of E.dA = Q/e_0 since we have a sphere the SA is just 4*pi*r^2 so it gives E = Q/[4*pi*e_0*R^2].

How does the density rho and dielectric constant e_r affect/influence this?

It influences it because you don't know what Q is. As you have written it Q is the source of the electric field E, i.e. the true and total Q (both bound and otherwise). But you don't know Q because you only know the *free* charge density.

Luckily, gauss' law will work for the electric displacement too with
$$\int \vec D \cdot \vec dA = Q_{\textrm{free}}^{\textrm{enclosed}}$$

 

[olgranpappy]

...and you also have the definition (for a linear dielectric) that
$$\epsilon_r\epsilon_0 \vec E = \vec D$$

So, use what's in my first post to solve for D and then just divide to get E

...I might not have all the factors right since I am used to using Guassian units... but I think I got the placements of the epsilons correct...