Dielectric sphere - Find the electric field

In summary, the electric field inside the sphere is given by:E = [rho*r /(3*e_0*e_r)]* r hatwhere r hat is the unit vector pointing in the radial direction.The electric field outside the sphere is given by:E = [rho*R³]/[3*e_0*r²] * r hatwhere r hat is the unit vector pointing in the radial direction.The potential V at the centre of the sphere compared to that at infinity is:V = E*r²where r is the radius of the sphere.
  • #1
Qyzren
44
0
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.
 
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  • #2
Start with Gauss' law. Try to calculate how E varies with the radius of your spherical Gaussian surface.

Claude.
 
  • #3
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.
 
  • #4
Qyzren said:
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:
[tex]\rho = \rho_b + \rho_f[/tex]
where [itex]\rho_b[/tex] is the bound charge and [itex]\rho_f[/itex] 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.
 
  • #5
Qyzren said:
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
[tex]
\int \vec D \cdot \vec dA = Q_{\textrm{free}}^{\textrm{enclosed}}
[/tex]
 
Last edited:
  • #6
...and you also have the definition (for a linear dielectric) that
[tex]
\epsilon_r\epsilon_0 \vec E = \vec D
[/tex]

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...
 

1. How is the electric field calculated for a dielectric sphere?

The electric field for a dielectric sphere can be calculated using the formula E = kQ/r^2, where k is the Coulomb constant, Q is the charge of the sphere, and r is the distance from the center of the sphere.

2. What is the difference between the electric field inside and outside of a dielectric sphere?

The electric field inside a dielectric sphere is weaker than the field outside due to the presence of the dielectric material, which reduces the overall electric field strength. This is known as polarization.

3. How does the electric field change as the dielectric constant of the sphere changes?

The electric field inside the dielectric sphere increases as the dielectric constant increases. This is because the dielectric material can store more charge and therefore have a greater impact on the overall electric field.

4. Does the size of the dielectric sphere affect the electric field?

Yes, the electric field inside a dielectric sphere increases as the size of the sphere increases. This is because the charge is spread out over a larger area, resulting in a weaker electric field.

5. Can the electric field inside a dielectric sphere ever be zero?

Yes, if the dielectric constant of the sphere is equal to the surrounding medium, there will be no change in the electric field and it will be zero inside the sphere.

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