Quick electric field of a line segment integral question

In summary, the conversation discusses finding the electric field above a midpoint of a wire carrying uniform charge, and the solution involves an integral with a substitution. The question focuses on why one method of substitution does not work and the solution is to express x in terms of u before proceeding with the integration.
  • #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
 
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  • #2
StephenD420 said:
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
You're not quite done yet, if u substitution is to work. You still have an x in your expression, and you'll have to express it in terms of u, if you wish to proceed.
=
-2/2x*u^(-1/2)
That's not quite right there. You've treated x as a constant number. But it's not a constant number. It's a function of u in this case. So you'll need to express that x in terms of u, before the integration.

And if that doesn't make the integral easier to evaluate than the original integral, you'll have to find some other way to proceed (perhaps such as your x = ztanθ idea).
 
  • #3
ahhh of course!

thanks.
 

1. What is the equation for calculating the electric field of a line segment?

The equation for calculating the electric field of a line segment is E = (kλ)/r, where k is the Coulomb's constant, λ is the linear charge density, and r is the distance from the line segment.

2. How do you determine the direction of the electric field for a line segment?

The direction of the electric field for a line segment can be determined using the right-hand rule. Point your thumb in the direction of the charge flow and your fingers will curl in the direction of the electric field.

3. What is the significance of the linear charge density in the electric field equation?

The linear charge density represents the amount of charge per unit length along the line segment. It determines the strength of the electric field, with a higher linear charge density resulting in a stronger electric field.

4. Can the electric field of a line segment be negative?

Yes, the electric field of a line segment can be negative. This indicates that the direction of the electric field is opposite to the direction of the charge flow.

5. How does the distance from the line segment affect the electric field?

The electric field is inversely proportional to the distance from the line segment. This means that as the distance increases, the electric field decreases. This relationship follows the inverse-square law.

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