- #1
StephenD420
- 100
- 0
Question: Find the electric field a distance 'z' above the midpoint of a straight line segment of length 2L, which carries uniform charge λ
Solution: Let me work out the steps and show you where my problem is(I have avoided showing some of the steps).
The coordinate axis has been set up taking the midpoint of the wire as the origin. Magnitude of the electric field is given by:
E=1/4piϵ0∫2λz/(z^2+x^2)^3/2 dx
so the integral that I have a question on is
∫1/(z^2+x^2)^3/2 dx
Now I know you can use x=ztan(theta)
but why can I not do
u=z^2+x^2
du=2xdx
dx=du/2x
so
∫u^(-3/2) du/2x
=
-2/2x*u^(-1/2)
so the integral would be equal to
-1/(x*(z^2+x^2)^(1/2))
which, of course is not right, but why??
Thanks.
Stephen
Solution: Let me work out the steps and show you where my problem is(I have avoided showing some of the steps).
The coordinate axis has been set up taking the midpoint of the wire as the origin. Magnitude of the electric field is given by:
E=1/4piϵ0∫2λz/(z^2+x^2)^3/2 dx
so the integral that I have a question on is
∫1/(z^2+x^2)^3/2 dx
Now I know you can use x=ztan(theta)
but why can I not do
u=z^2+x^2
du=2xdx
dx=du/2x
so
∫u^(-3/2) du/2x
=
-2/2x*u^(-1/2)
so the integral would be equal to
-1/(x*(z^2+x^2)^(1/2))
which, of course is not right, but why??
Thanks.
Stephen