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gnegnegne
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Hi everyone!
I have to solve a problem using Poisson's equation.
There are two parallel infinite conductor planes in vacuum. The distance between them is [itex]d[/itex] and they are both kept at a potential [itex]V=0[/itex]. Between them there is a uniform volume density charge [itex]\rho_0>0[/itex] infinite along the directions parallel to the planes and its width is [itex]d/2[/itex] (it's a parallelepiped that starts at x=0 and ends at d/2). I have to find the the potential and the electric field in the region between the planes.
For [itex]0<x<d/2[/itex], solving the Poisson's equation (I integrated [itex]\frac{\partial^2 V}{\partial x^2}=-\frac{\rho_0}{\epsilon_0}[/itex]):
[tex]V_1(x)=-\frac{\rho_0 x^2}{2 \epsilon_0}+Ax+B[/tex]
For [itex]d/2<x<d[/itex], integrating [itex]\frac{\partial^2 V}{\partial x^2}=0[/itex]:
[tex]V_2(x)=Cx+D[/tex]
Setting the two boundary conditions at x=0 and at x=d I obtain [itex]D=-Cd[/itex] and [itex]B=0[/itex], therefore:
[tex]V_1(x)=-\frac{\rho_0 x^2}{2 \epsilon_0}+Ax[/tex]
[tex]V_2(x)=C(x-d)[/tex]
I can get ride of one of the two unknowns using the condition that V(x) has to be continuous at x=d/2, however I'm missing another condition. What other condition should be considered? Intuitively I would impose the condition that the electric field has to be continuous too, but I'm not 100% sure.
I have to solve a problem using Poisson's equation.
There are two parallel infinite conductor planes in vacuum. The distance between them is [itex]d[/itex] and they are both kept at a potential [itex]V=0[/itex]. Between them there is a uniform volume density charge [itex]\rho_0>0[/itex] infinite along the directions parallel to the planes and its width is [itex]d/2[/itex] (it's a parallelepiped that starts at x=0 and ends at d/2). I have to find the the potential and the electric field in the region between the planes.
For [itex]0<x<d/2[/itex], solving the Poisson's equation (I integrated [itex]\frac{\partial^2 V}{\partial x^2}=-\frac{\rho_0}{\epsilon_0}[/itex]):
[tex]V_1(x)=-\frac{\rho_0 x^2}{2 \epsilon_0}+Ax+B[/tex]
For [itex]d/2<x<d[/itex], integrating [itex]\frac{\partial^2 V}{\partial x^2}=0[/itex]:
[tex]V_2(x)=Cx+D[/tex]
Setting the two boundary conditions at x=0 and at x=d I obtain [itex]D=-Cd[/itex] and [itex]B=0[/itex], therefore:
[tex]V_1(x)=-\frac{\rho_0 x^2}{2 \epsilon_0}+Ax[/tex]
[tex]V_2(x)=C(x-d)[/tex]
I can get ride of one of the two unknowns using the condition that V(x) has to be continuous at x=d/2, however I'm missing another condition. What other condition should be considered? Intuitively I would impose the condition that the electric field has to be continuous too, but I'm not 100% sure.
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