# Electric field due to infinite sheet of charge

the electric field intensity due to a charged body decreases as the distance from it increases. why is then the electric field intensity due to an infinite sheet of charge independent of the distance from it? if it is beacause the distance of the test charge from d sheet is negligibly small as compared to its size, is the same not applicable for infinite line of charge? but intensity due to an infinite line of charge is inversely proportional to distance from it?

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A simple explanation: The electric field of a point charge is inversely proportional to r^2. For a line charge, it is infinite in one of the directions so you will never be far enough away from it to observe the 1/r^2, but you can observe the 1/r. You are basically replacing one of the r's with L, and then you have Q/L = $\lambda$ (charge per unit length).

For a 2d sheet, it is infinite in 2 directions. So that completely cancels out the 1/r^2 and you are left with a constant. Or replacing both r's in r^2 with A = area to get $Q/A = \sigma$ (charger per unit area).

This is a very basic explanation.

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has this got something to do with the divergence theorem? pls can someone simplfy dat? i have no idea abt dat theorem

Take the infinite line charge in the z-direction as an example. If you cut out an infinitesmal plane in the x-y direction, then you can solve the electric field in that plane alone with charge, $\lambda dx$. And you can say it is periodic above and below this plane to get back the full solution.

So you have a 2D poisson equation for a point charge. This ends up giving you a 1/r term for the electric field. So the infinite line charge acts like a point charge in 2 dimensions.

You can do the same for an infinite sheet charge in the x-y plane, by taking a 1D space in the z-direction and saying it is periodic in the x and y-directions. Solve for the E-field in 1D space, you get a constant. So an infinite sheet charge acts like a point charge in 1 dimension.