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1. Homework Statement
Two semi-infinite grounded conductive planes meet at right angles. In the region b/w the conductors, there is the plane with angle 45° having surface charge density σ. Using the method of images, find the field distribution in this region.
(There is a picture included, essentially showing the first quadrant, bounded by the two grounded planes, with the non-grounded plane at 45° above the x-axis).
2. Homework Equations
V=kq/r
E*dA=σ*dA/ε
3. The Attempt at a Solution
First of all, I am making the assumption that determining the field means electric field and not potential. Most prior examples I have seen with method of images (MoI) usually works with getting potential, but maybe this is a different case?
For MoI, there is a plane with +σ going up through 45°, so, mirroring how I would do this if this were a point charge, I put planes with -σ at 135° and -45° and then one with +σ at -135°. So basically two perpendicular, oppositely charged, infinite planes.
I looked at this problem a very long time trying to calculate potential, but then I thought, can you just use Gauss's Law for each and confine your solution to the first quadrant? It seemed incredibly simple after that, but sometimes I make an incorrect assumption that oversimplifies the situation-- is this okay to assume?
E=σ/2ε from the positive plane, and the direction is normal to the plane (-i/√2 +j/√2) and the direction from the negative plane is -i/√2 -j/√2 so that the after summing these components the field would be pointing in the -i direction. Is that the basic idea or have I made one too many assumptions?
Thanks!
Two semi-infinite grounded conductive planes meet at right angles. In the region b/w the conductors, there is the plane with angle 45° having surface charge density σ. Using the method of images, find the field distribution in this region.
(There is a picture included, essentially showing the first quadrant, bounded by the two grounded planes, with the non-grounded plane at 45° above the x-axis).
2. Homework Equations
V=kq/r
E*dA=σ*dA/ε
3. The Attempt at a Solution
First of all, I am making the assumption that determining the field means electric field and not potential. Most prior examples I have seen with method of images (MoI) usually works with getting potential, but maybe this is a different case?
For MoI, there is a plane with +σ going up through 45°, so, mirroring how I would do this if this were a point charge, I put planes with -σ at 135° and -45° and then one with +σ at -135°. So basically two perpendicular, oppositely charged, infinite planes.
I looked at this problem a very long time trying to calculate potential, but then I thought, can you just use Gauss's Law for each and confine your solution to the first quadrant? It seemed incredibly simple after that, but sometimes I make an incorrect assumption that oversimplifies the situation-- is this okay to assume?
E=σ/2ε from the positive plane, and the direction is normal to the plane (-i/√2 +j/√2) and the direction from the negative plane is -i/√2 -j/√2 so that the after summing these components the field would be pointing in the -i direction. Is that the basic idea or have I made one too many assumptions?
Thanks!