- #1
Roodles01
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A problem involving E, D & P. I have the answer but not sure why for the P bit.A bit of clarification if someone could, please.
A thick sheet of polystyrene, of relative permittivity ε = 2.5, is placed
normal to a uniform electric field of magnitude 500Vm−1 in air.
Find the magnitudes of the electric field E, the electric displacement D and the polarization P inside the material.
OK div D = ρf
and D = εε0E
Boundary conditions Doutside = Dinside → εoutEout = εε0Ein
so
Ein = Eout / ε
Ein = 500/2.5
Ein = 200 Vm-1
Dout = Din = ε0 Eout
Dout = 8.85x10-12 x 200
Dout = 1.77x10-9
Fine so far . . .
When I get down to calculating P I have to use
Din = ε0Ein + Pin
why, when everything else has the suffix "in" do I use ε0? Shouldn't it be just plain ε?
Thanks in advance.
A thick sheet of polystyrene, of relative permittivity ε = 2.5, is placed
normal to a uniform electric field of magnitude 500Vm−1 in air.
Find the magnitudes of the electric field E, the electric displacement D and the polarization P inside the material.
OK div D = ρf
and D = εε0E
Boundary conditions Doutside = Dinside → εoutEout = εε0Ein
so
Ein = Eout / ε
Ein = 500/2.5
Ein = 200 Vm-1
Dout = Din = ε0 Eout
Dout = 8.85x10-12 x 200
Dout = 1.77x10-9
Fine so far . . .
When I get down to calculating P I have to use
Din = ε0Ein + Pin
why, when everything else has the suffix "in" do I use ε0? Shouldn't it be just plain ε?
Thanks in advance.