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SadStudent
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Homework Statement
We have some uniformly polarized cylinder with some polarization [itex] P\hat{z} [/itex].
The The cylinder's center is on [itex]\hat{z}[/itex] and it's on the [itex]xy[/itex] surface.
The cylinder has radius [itex]R [/itex] and height [itex] d [/itex]where [itex]R \gg d[/itex] .
If we were to ground the [itex]xy[/itex] plane what would the electric field be inside the cylinder and outside (disregarding edge effects)?
Homework Equations
Well if the there was no grounded [itex]xy [/itex] surface the electric field would be [itex]-\frac{\overrightarrow{P}}{\varepsilon_{0}}[/itex] inside and outside it would look like a dipole, but now there's is, so I'm not quite sure.
The Attempt at a Solution
I've tried thinking about it in two ways but I couldn't find proof to either of them
- Charges from [itex]\infty [/itex] arrive to "cancel out" the negative surface charge that's on the bottom of the cylinder so it's like a single surface but with field lines only half the magnitude, therefore it's [itex]-\frac{\overrightarrow{P}}{2\varepsilon_{0}}[/itex]
- Or we can try and find an "image charge" or an "image object" for example if we put an identical cylinder under this cylinder, the upper surface charge on the "image" cylinder will cancel the bottom surface charges of the real one but then we have [itex]-\frac{2\overrightarrow{P}}{\varepsilon_{0}}[/itex] because the [itex]\overrightarrow{P} [/itex] is proportional to the distance.
But this is all hand-wavy and there might be a simple way to show which one preserves zero potential on the [itex]xy [/itex] surface but I don't know how to show it mathematically.
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