An infinitely long sheet of charge of width L lies in the xy -plane between x=L/2 and x=L/2 . The surface charge density is . Derive an expression for the electric field E at height z above the centerline of the sheet. (Assume that z0 .) Express your answer in terms of the variables , L , z , unit vector k^ , and appropriate constants. Solution The sheet is kept on the xy plane with a width of L along the x axis. We would consider a section of the sheet at a distance of x from the origin of thickness dx. Since it extends infinitely along the y axis. We have an infinite rod with at distance of x from origin with linear charge density of dx Now, the electic field at z due to this infinitely long strip would be dx / 2o * sqrt(z^2 + x^2) However, by symmetry we can say that the net electric field will point along the z axis as the components in the xy plane would eventually cancel out. Hence dE = z dx / 2o * (z^2 + x^2) We integrate the above to obtain: E = ( / 2o)[atan(x/z)](from -L/2 to + L/2) or, E = ( / 2o)[atan(L/2z) - atan(-L/2z)] = ( /o)[atan(L/2z)] NOTE: There are two ways of approaching the problem, either you can assume a small section of thickness dy at distance y and then proceed, or you can assume an infinitely long strip at some distance x and then go ahead. However, it needs to be mentioned that the first way gives a little tricky integral whereas the second one, as above, is fairly straightforward. The integral above is of the form dx / (a^2 + x^2) = (1/a) atan (x/a) Further, while subtracting the max and min value for the integral, the negative. you would need the relation tanx = tan-x Hence the difference of atan of negative x and postiive x ends up getting added..

1. An infinitely long sheet of charge of width L lies in the xy -plane between x=L/2 and x=L/2 .
The surface charge density is . Derive an expression for the electric field E at height z above the
centerline of the sheet. (Assume that z0 .)
Express your answer in terms of the variables , L , z , unit vector k^ , and appropriate constants.
Solution
The sheet is kept on the xy plane with a width of L along the x axis.
We would consider a section of the sheet at a distance of x from the origin of thickness dx. Since
it extends infinitely along the y axis. We have an infinite rod with at distance of x from origin
with linear charge density of dx
Now, the electic field at z due to this infinitely long strip would be dx / 2o * sqrt(z^2 + x^2)
However, by symmetry we can say that the net electric field will point along the z axis as the
components in the xy plane would eventually cancel out.
Hence dE = z dx / 2o * (z^2 + x^2)
We integrate the above to obtain:
E = ( / 2o)[atan(x/z)](from -L/2 to + L/2)
or, E = ( / 2o)[atan(L/2z) - atan(-L/2z)] = ( /o)[atan(L/2z)]
NOTE: There are two ways of approaching the problem, either you can assume a small section
of thickness dy at distance y and then proceed, or you can assume an infinitely long strip at some
distance x and then go ahead.
However, it needs to be mentioned that the first way gives a little tricky integral whereas the
second one, as above, is fairly straightforward. The integral above is of the form dx / (a^2 + x^2)
= (1/a) atan (x/a)
Further, while subtracting the max and min value for the integral, the negative. you would need
the relation tanx = tan-x
Hence the difference of atan of negative x and postiive x ends up getting added.