- #1
AcidRainLiTE
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Suppose we apply a uniform field to an infinite conducting slab (i.e. like an infinite parallel plate capacitor, but the interior is included as part of the conductor). What is the resulting field?
The simple answer is that a surface charge develops on the boundary planes of the conductor so as to cancel the field inside. Outside, the field is unaffected by the surface charges. Since the field inside the conductor is zero, the charge configuration is static.
But, it seems that the field on the boundary of the conductor is not zero (and hence, a small charge [itex]d\sigma[/itex] on the boundary would move). To see this, let a small charge [itex]d\sigma[/itex] be located at a point p on the left plate. Then the field acting on [itex]d\sigma[/itex] due to the rest of the left plate is zero, so the field acting on it is due only to the external field and the right plate. But the right plate alone is insufficient to cancel the external field. Hence, there is a non-zero field acting on [itex]d\sigma[/itex].
I am not looking for a physical resolution of this problem. I do not care what actually physically happens (i.e. that there are no true surface charges, that the surface charge dies off exponentially, etc.). I am, rather, interested in whether there is a consistent solution within the idealized model. So, I can restate my question more precisely as follows:
Suppose we have a conducting slab, C, where the slab fills the closed region [itex]R_C = \{(x,y,z)|-1 \le x \le 1\}[/itex]. My question is whether there exists a static charge distribution (along with whatever boundary conditions you need) such that:
(1) E(x,y,z) is constant at all points outside [itex]R_C[/itex] and is perpendicular to the conductor surface
(2) E(x,y,z) is well-defined at every point in space
(3) If [itex](x,y,z) \in R_C[/itex], then E(x,y,z) = 0.
The simple answer is that a surface charge develops on the boundary planes of the conductor so as to cancel the field inside. Outside, the field is unaffected by the surface charges. Since the field inside the conductor is zero, the charge configuration is static.
But, it seems that the field on the boundary of the conductor is not zero (and hence, a small charge [itex]d\sigma[/itex] on the boundary would move). To see this, let a small charge [itex]d\sigma[/itex] be located at a point p on the left plate. Then the field acting on [itex]d\sigma[/itex] due to the rest of the left plate is zero, so the field acting on it is due only to the external field and the right plate. But the right plate alone is insufficient to cancel the external field. Hence, there is a non-zero field acting on [itex]d\sigma[/itex].
I am not looking for a physical resolution of this problem. I do not care what actually physically happens (i.e. that there are no true surface charges, that the surface charge dies off exponentially, etc.). I am, rather, interested in whether there is a consistent solution within the idealized model. So, I can restate my question more precisely as follows:
Suppose we have a conducting slab, C, where the slab fills the closed region [itex]R_C = \{(x,y,z)|-1 \le x \le 1\}[/itex]. My question is whether there exists a static charge distribution (along with whatever boundary conditions you need) such that:
(1) E(x,y,z) is constant at all points outside [itex]R_C[/itex] and is perpendicular to the conductor surface
(2) E(x,y,z) is well-defined at every point in space
(3) If [itex](x,y,z) \in R_C[/itex], then E(x,y,z) = 0.
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