Finding E, D and P in a Dielectric Slab

[zhillyz]

Homework Statement

For a dielectric slab of relative permeability er fills the space between z = ±a in the xy-plane, and contains uniform density of free charge ρf per unit volume. Find E, D and P as functions of z.

Homework Equations

P = εoXeE
D = εrεoE
∇×P=εoXe(∇×E) = 0
∇×D=εo(∇×E)+∇×P=0

The Attempt at a Solution

Well I have been looking online and through lecture notes and have found dielectric equations for E, D and P but I am not sure what to do with them. I have found examples to do with dielectric slabs between capacitors with thickness 2d and I assume this is what my question must be based from(z = ±a, z(thickness) = 2a). I am sure that I have all the relevant equations but with no values to enter into them. If P, D and E are all vectors in the x,y and z axis is it maybe a case of calculating the curls and looking at the z component?


Not sure what to do from here, please help.

 

[TSny]

Hi zhillyz. I think it would be easiest to start with finding D as a function of z. What equation could you use to do this? (It's not in your list of relevant equations.)

 

[zhillyz]

Emmm ∇.D = ρ-ρb = ρf = d^3q/dxdydz? There are other electric displacement formulae but none that seem to have relevance to the axis z like this one.

The question does mention free charge density which would make this formula look promising.

So if D = εoE + P

then ∇.D = εo∇.E+∇.P = d^3/dxdydz

am I going along the correct lines?

 

[TSny]

See if you can use ∇$\cdot$D = ρ_f or it's integral counterpart (Gauss' Law): $\oint$D$\cdot$dA = Q_f,enclosed. Symmetry will tell you the direction of D and also where D = 0.

 

[zhillyz]

∮D⋅dA = Qf,enc

So using a gaussian pillbox with the centre of the dielectric slab being the centre of the box and the top of the box being inside +a and bottom inside -a of the z-axis. No flux through bottom or sides of box so, ∮dA = A(area of top of box)*z.

D*A = Qfenc*A*z
D = Qfenc*z

Then E = D/(K*εo)..

Am I getting it now?

 

[TSny]

∮D⋅dA = Qf,enc

So using a gaussian pillbox with the centre of the dielectric slab being the centre of the box and the top of the box being inside +a and bottom inside -a of the z-axis. No flux through bottom or sides of box so, ∮dA = A(area of top of box)*z.

D*A = Qfenc*A*z
D = Qfenc*z

Then E = D/(K*εo)..

Am I getting it now?

That's the right idea. But, if you choose the center of your box at z = 0, then there will be flux through the bottom as well as the top of the pillbox. You might want to consider putting the bottom of the box at z = 0 since you should be able to argue that D will be zero at z = 0.

If you're letting the symbol Q_f,enc stand for free charge enclosed, then how would you express that in terms of the free charge density ρ_f?

Can you also get an expression for D for values of z outside the slab?

You are on the right track for getting E from D. Look's like you switched notation from ε_r to K.

 

[zhillyz]

Thank you I think I understand now, yes I did switch notations by accident as in my research of the question some people use the symbol K.

Basically the you only get flux through boundaries, so the slab is infinitely wide along the plane of the xy-axis and will not have boundaries on the side, the pillbox is placed conveniently with its base in the centre of the slab and top surface just over the top of the slab leaving only the flux through the top to be worked out.

Surface integral will only be area A of the top of pill box,

∮D⋅dA = Qf,enc
D⋅A=A⋅z⋅ρf
D=ρf⋅z
E=D/εrεo=ρf⋅z/εrεo [-a < z < +a]

P = εoXeE = ρf⋅a⋅Xe/εr = [ρf⋅a⋅εr - 1]/εr

But you won't get an accurate electric field over the boundary because of the surface charge so need to measure outside which would leave there no need for εr. Soo..

E=ρf⋅a/εo

 

[TSny]

Looks good to me, except for a typo in the placement of a bracket in the expression for P.

 

[zhillyz]

Thank you very kindly for your help :).