- #1

WrongMan

- 149

- 15

## Homework Statement

there is a square on the XoY plane, centered at the origin (just outlines of the square) it has a charge Q (Q>0) and side 2L, i must evaluate the electric field along the z axis. see attached image

## Homework Equations

E=k*q/r^2

## The Attempt at a Solution

So first i divided the square into 4 lines, and noticed that due to the simmetry there is only a field in the z direction, and all the lines of the square contribute the same to that field.

So:

dE= k(Q/8L)

_{1}* dx/(x

^{2}+y

^{2}+z

^{2})

_{2}*(z/(x

^{2}+y

^{2}+z

^{2}))

^{½})

_{3}

where subscrip1 is Q/8L because the full Q is for the whole square, so we divide it by 4 to get the charge on the line and divide again by 2L to get the charge density.

subscript 2 is the distance of the charge to the line.

subscript 3 is cosine of E with Ez so we get only the z component

Now the integral, i get an not-so-straightforward integral, so if there is a simplification i could do here please advise.

so i get

E= k(Q/8L)z*∫dx/((x

^{2}+y

^{2}+z

^{2})

^{3/2}))

and i integrate either over -L to L or from 0 to L and multiply it by 2.

and i get

E=kQz*1/((y

^{2}+z

^{2})*(L

^{2}+y

^{2}+z

^{2})

^{½})

This result makes sense to me physically, as field is 0 when z is 0 (makes sense since field would no longer have a z component) and also goes to 0 when z goes to infinity.

So the total field is 4*E. Am i correct? any simplification suggestions for the integral

EDIT: oh and y should be L since its fixed.

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