How Does Dielectric Susceptibility Affect Charge Force Calculation?

[teastu]

Can anyone help with this-a charge 10 microcoul. is at 10 cm from a large dielectric block of susceptibility 10, find force on charge.

 

[Andrew Mason]

Can anyone help with this-a charge 10 microcoul. is at 10 cm from a large dielectric block of susceptibility 10, find force on charge.

I think this is how to approach it:

1. Use Gauss' law to find the field of q at 10 cm without the dielectric present (this is the applied field E_a that induces an opposing field in the dielectric by polarizing the polar molecules in the dielectric).

2. Determine the magnitude of the induced field of the dielectric from the susceptibility \chi_e using:

P = \epsilon_0\chi_eE_a = \frac{E_i - E_a}{4\pi}

3. Apply Gauss' law to the dielectric surface to find the induced charge from the polarized dielectric.

4. Determine the force is using Coulomb's law.

AM

 

[teastu]

Thanks, but I can't still do it. 1, Won't the field due to the charge be different at different points of the large dielectric?
2, Can you give the equation quoted in SI units?
3, Please tell me how to apply Gauss's law here.
Can this be done with the method of images?

 

[Andrew Mason]

Thanks, but I can't still do it. 1, Won't the field due to the charge be different at different points of the large dielectric?
2, Can you give the equation quoted in SI units?
3, Please tell me how to apply Gauss's law here.
Can this be done with the method of images?

1. Gauss' law:

\oint E\cdot dA = 4\pi r^2E = q/\epsilon_0

so that gives you E from the charge q.

2. In MKS the induced field is:

E_i = \epsilon_0E(1 + \chi_0) where E is from 1

so you can work out the induced field from that.

3. Apply Gauss' law to the dielectric surface to find the induced charge per unit area:

\oint E_i\cdot dA = E_i A = q_i/\epsilon_0 = \sigma_iA/\epsilon_0

So from 1 2, and 3.: E_i = \sigma_i/\epsilon_0 = \frac{q}{4\pi r^2}(1 + \chi_0)

So that gives you the induced charge density in the dielectric.

4. Determine the force (per unit area of dielectric) using Coulomb's law.

AM

 

[Nakis]

Hi,

I am working on a similar problem, and there is something I don't understand in the approach above: the induced field (which is actually the displacement, right?) does not have the same units as the applied field - actually Ei has units C/m2! so that you can't get a charge density...

Can anyone help?

-Nakis

 

[Andrew Mason]

Hi,

I am working on a similar problem, and there is something I don't understand in the approach above: the induced field (which is actually the displacement, right?) does not have the same units as the applied field - actually Ei has units C/m2! so that you can't get a charge density...

Can anyone help?

-Nakis

I see that there is an error in my earlier post. It should be:

2. In MKS the induced field is:

E_i = E(1 + \chi_0) where E is from 1

[note: this can also be viewed in terms of the field from a displacement charge density, D:

\vec D = \epsilon_0\vec{E}(1 + \chi_0) = \epsilon_0\vec{E}_i]


3. Apply Gauss' law to the dielectric surface to find the induced charge per unit area:

\oint E_i\cdot dA = E_i A = q_i/\epsilon_0 = \sigma_iA/\epsilon_0

So from 1, 2 and 3.: E_i = \sigma_i/\epsilon_0 = \frac{q}{\epsilon_04\pi r^2}(1 + \chi_0)

See: http://en.wikipedia.org/wiki/Electric_displacement_field

AM