- #1
issacnewton
- 1,026
- 36
here is what I am doing. I am trying to argue that the electric field is zero at the
center of the sheet using only symmetry arguments.
consider an insulator in the shape of infinite sheet of thickness 2a and with uniform charge density [itex]\rho[/itex]. Now let me specify the coordinate system. The sheet is from
z=-a to z=a. Now we need to give convincing arguments for the statement that
the electric field at z=0 is zero. Before I answer that, let's notice the
symmtries present in the charge distribution. There is reflection symmetry
at z=0. There is rotational symmetry about the z axis. And there is translational
symmetry if we move z axis parallel it itself.
So let's assume that [itex]\mathbf{E}\neq 0[/itex] at z=0. So it will , in general case, have
x,y,z components. But if it has z component then it violates relfection symmetry
about z=0. So let's remove that component. So we are left with x,y components. The net
[itex]\mathbf{E}[/itex] will have certain direction in the x-y plane. But since charge has
rotational symmetry around z axis, even this particular direction of [itex] \mathbf{E}[/itex]
is not possible. There is one last possiblity left. We can let [itex] \mathbf{E}[/itex] emerge
from (0,0,0) and then it will radially go outwards in the x-y plane. In this way, the
[itex] \mathbf{E}[/itex] is symmetric about z axis. But since the charge is symmetric about
any axis parallel to the z axis, this choice of [itex] \mathbf{E}[/itex] is not possible.
So the only logical option left is the case where [itex] \mathbf{E}=0[/itex] at z=0.
I hope my argument is right. I have not solved any differential equation so far
to arrive at this conclusion. I have assumed that the electric field obeys the
same symmetry as the charge distribution. If we look at the Gauss's law in
differntial form,
[tex]\vec{\nabla}\cdot \mathbf{E} =\frac{\rho}{\epsilon_o}[/tex]
I think its possible to see that the symmetry in charge distribution should result
in symmetry in electric field. But I don't know how to prove this mathematically.
I think we might be able to prove this using group theory, I am not sure.
What do you think about this ?
It should be possible to apply this line of reasoning to other Maxwell equations
as well.
I have not seen the above reasoning in introductory physics books. Please comment and
let me know any flaws.
center of the sheet using only symmetry arguments.
consider an insulator in the shape of infinite sheet of thickness 2a and with uniform charge density [itex]\rho[/itex]. Now let me specify the coordinate system. The sheet is from
z=-a to z=a. Now we need to give convincing arguments for the statement that
the electric field at z=0 is zero. Before I answer that, let's notice the
symmtries present in the charge distribution. There is reflection symmetry
at z=0. There is rotational symmetry about the z axis. And there is translational
symmetry if we move z axis parallel it itself.
So let's assume that [itex]\mathbf{E}\neq 0[/itex] at z=0. So it will , in general case, have
x,y,z components. But if it has z component then it violates relfection symmetry
about z=0. So let's remove that component. So we are left with x,y components. The net
[itex]\mathbf{E}[/itex] will have certain direction in the x-y plane. But since charge has
rotational symmetry around z axis, even this particular direction of [itex] \mathbf{E}[/itex]
is not possible. There is one last possiblity left. We can let [itex] \mathbf{E}[/itex] emerge
from (0,0,0) and then it will radially go outwards in the x-y plane. In this way, the
[itex] \mathbf{E}[/itex] is symmetric about z axis. But since the charge is symmetric about
any axis parallel to the z axis, this choice of [itex] \mathbf{E}[/itex] is not possible.
So the only logical option left is the case where [itex] \mathbf{E}=0[/itex] at z=0.
I hope my argument is right. I have not solved any differential equation so far
to arrive at this conclusion. I have assumed that the electric field obeys the
same symmetry as the charge distribution. If we look at the Gauss's law in
differntial form,
[tex]\vec{\nabla}\cdot \mathbf{E} =\frac{\rho}{\epsilon_o}[/tex]
I think its possible to see that the symmetry in charge distribution should result
in symmetry in electric field. But I don't know how to prove this mathematically.
I think we might be able to prove this using group theory, I am not sure.
What do you think about this ?
It should be possible to apply this line of reasoning to other Maxwell equations
as well.
I have not seen the above reasoning in introductory physics books. Please comment and
let me know any flaws.