- #1
kmm
- 188
- 15
I understand that taking the boundary condition that the electric field changes [itex] \frac{\sigma}{\epsilon_{0}} [/itex] when you cross a charged surface and the fact that the electric field is zero inside a conductor that the electric field outside the conductor should be [itex] \mathbf{E} = \frac{\sigma}{\epsilon_{0}} \hat{\mathbf{n}} [/itex].
What's troubling me is considering electric fields other than the field due to some patch. In the case of a conductor, the electric field [itex] \mathbf{E} = \frac{\sigma}{\epsilon_{0}} \hat{\mathbf{n}} [/itex] is due to the patch and other areas, even external ones. This was because of the boundary condition which was derived for a single patch of a charged surface. The field must come from other areas, not just the patch.
Although, when deriving the boundary condition, the Gaussian pillbox used over the surface was made small enough as not to include areas with different values of [itex] \sigma [/itex] and not to include any curving of the surface. (Wouldn't that mean the boundary condition would be different if we did include these other regions?) Yet, by invoking the boundary condition for a conductor, we now have to include these other fields.
Assuming the "other" fields are the same on top and bottom and using a positive charge distribution, I took [itex] \mathbf{E}_{above} = \mathbf{E}_{other} + \mathbf{E}_{patch} [/itex] and [itex] \mathbf{E}_{below} = \mathbf{E}_{other} + \mathbf{E}_{patch} [/itex]. Above, [itex] \mathbf{E}_{patch} = \frac{\sigma}{2 \epsilon_{0}} \hat{\mathbf{n}} [/itex] and below [itex] \mathbf{E}_{patch} = - \frac{\sigma}{2 \epsilon_{0}} \hat{\mathbf{n}} [/itex] so when we take [itex] E_{above} - E_{below} [/itex] the "other" fields cancel and we get [itex] \frac{\sigma}{\epsilon_{0}} [/itex], as expected. It's not surprising to me that the external fields canceled though because [itex] E_{above} - E_{below} [/itex] was obtained from gauss's law and any external fields through a Gaussian surface won't contribute to the flux. This makes me wonder if the other fields could change the boundary condition. It also makes me question applying the boundary condition to a conducting surface. I don't think I'm right, I'm just not sure what I'm thinking wrong.
What's troubling me is considering electric fields other than the field due to some patch. In the case of a conductor, the electric field [itex] \mathbf{E} = \frac{\sigma}{\epsilon_{0}} \hat{\mathbf{n}} [/itex] is due to the patch and other areas, even external ones. This was because of the boundary condition which was derived for a single patch of a charged surface. The field must come from other areas, not just the patch.
Although, when deriving the boundary condition, the Gaussian pillbox used over the surface was made small enough as not to include areas with different values of [itex] \sigma [/itex] and not to include any curving of the surface. (Wouldn't that mean the boundary condition would be different if we did include these other regions?) Yet, by invoking the boundary condition for a conductor, we now have to include these other fields.
Assuming the "other" fields are the same on top and bottom and using a positive charge distribution, I took [itex] \mathbf{E}_{above} = \mathbf{E}_{other} + \mathbf{E}_{patch} [/itex] and [itex] \mathbf{E}_{below} = \mathbf{E}_{other} + \mathbf{E}_{patch} [/itex]. Above, [itex] \mathbf{E}_{patch} = \frac{\sigma}{2 \epsilon_{0}} \hat{\mathbf{n}} [/itex] and below [itex] \mathbf{E}_{patch} = - \frac{\sigma}{2 \epsilon_{0}} \hat{\mathbf{n}} [/itex] so when we take [itex] E_{above} - E_{below} [/itex] the "other" fields cancel and we get [itex] \frac{\sigma}{\epsilon_{0}} [/itex], as expected. It's not surprising to me that the external fields canceled though because [itex] E_{above} - E_{below} [/itex] was obtained from gauss's law and any external fields through a Gaussian surface won't contribute to the flux. This makes me wonder if the other fields could change the boundary condition. It also makes me question applying the boundary condition to a conducting surface. I don't think I'm right, I'm just not sure what I'm thinking wrong.
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