Timeline for Why is the field inside a conducting shell zero when only external charges are present?
Current License: CC BY-SA 4.0
9 events
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| Aug 11, 2020 at 8:09 | comment | added | Void | @JohnRennie What I mean is that there are no charges in the cavity surrounded by the conductor (-> edit). | |
| Aug 11, 2020 at 8:05 | history | edited | Void | CC BY-SA 4.0 |
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| Aug 8, 2020 at 8:53 | comment | added | John Rennie | How can you assume that there are no charges in the closed surface of the conductor? there are certainly induced charges on the outer surface and it isn't obvious why there would not also be induced charges on the inner surface? | |
| Aug 7, 2020 at 20:46 | comment | added | najkim | @Void I think you dropped a negative in your Poisson Equation. See here. | |
| Aug 7, 2020 at 17:18 | comment | added | A. Jahin | Yes, sorry always forget what is the triangle. Thanks! | |
| Aug 7, 2020 at 17:11 | comment | added | user258881 | @A.Jahin $\Delta$ is the Laplacian. Thus $\nabla^2 \phi=\rho/\varepsilon_0$, which is completely valid. | |
| Aug 7, 2020 at 16:54 | comment | added | A. Jahin | how do you write $ \vec{\nabla} \phi = \rho / \epsilon_0$ LHS is a vector and the RHS is a scalar, or do I miss understand something? | |
| Aug 7, 2020 at 16:03 | history | edited | Void | CC BY-SA 4.0 |
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| Aug 7, 2020 at 15:56 | history | answered | Void | CC BY-SA 4.0 |