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Old May 13th 2014, 01:50 AM   #1
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Behavior at conductor surface?

I'm feeling a bit confused about how electric fields actually behave at the surface of a conductor. I mean, I understand the mathematical derivation that says that the tangential component is zero and so on, but take this as a very specific example:

Say we overlay a room with a typical Cartesian coordinate system and set up an electric field going exactly straight from -x to +x for y > 0. Now imagine taking a conducting square box, with one side parallel to the x-axis, and moving it from some -y to some +y. Will the electric field actually bend at the top of the box and twist around 90 degrees? Is this measurable and, just roughly, how far do these distortions spread? That is, are they normal to the conductor just precisely at the surface and then a thousandth of a millimeter out they're back to pointing in their original direction or how does that work?

Any help in sorting this out would be greatly appreciated!
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Old May 13th 2014, 05:53 AM   #2
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Yes, your concept of the bending electric field is correct, assuming that the metal box is in equilibrium. If the electric field had a component parallel to the surface then there would be a net force acting on charges in the metal box, which would cause them to move and hence the box is not in equilibrium. The actual shape of the field lines - how sharply they turn from horizontal to vertical- is dependent on the geometry, charge density on the box and strength of the external field, and can be calculated from Gauss's Law with appropriate boundary conditions:


Last edited by ChipB; May 13th 2014 at 07:10 AM.
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Old May 13th 2014, 06:29 AM   #3
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Ah yes of course, you actually get the shape of the field from that. Duh. Making the distinction that a non-normal component is associated with a net force cleared it up completely; my text was a bit unclear on that point and I simply didn't think far enough ahead. Thanks a bunch!
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