What Happens to the Electric Field Equation at Different Limits?

[vorcil]

Homework Statement

Hi, I need to figure out what happens to this equation in the limits

E = \frac{1}{4\pi\epsilon_0} \frac{2\lambda L}{z \sqrt{z^2+L^2}}

in the two different cases

that z>>L

and when L -> infinity

(note this equation was derived from finding the electric field da distance z, above the midpoint of a straight line segment of length 2L, which carries a uniform line charge of \lambda

The Attempt at a Solution

for the case when, z>>L I can see how the L term becomes insignificant in the square root on the bottom,
and so the equation would just become
\frac{1}{4\pi\epsilon_0} \frac{2\lambda L}{z^2}

but for the case when L approaches infinity, what do I do?
the squareroot of a L^2 +z^2 == L?
does that mean the L can just be canceled out?
\frac{1}{4\pi\epsilon_0} \frac{2\lambda L}{z *L}
and the equation becomes?
\frac{1}{4\pi\epsilon_0} \frac{2\lambda}{z}

i'm not sure if I'm allowed to since the Z was the distance from the midpoint of the line,

the first one makes sense since it just becomes a point charge of 2lambda L
but the second case, I'm not too sure what it becomes

 

[lanedance]

consider the form of the field produced by an infinite conductor, which should agree with your 2nd result

 

[lanedance]

it helps to re-write the expression in the following form so it is clear when you take the limit

first as follows
E <br /> = \frac{1}{4\pi\epsilon_0} \frac{2\lambda L}{z \sqrt{z^2+L^2}} <br /> = \frac{\lambda}{ 2 \pi\epsilon_0} \frac{ L}{z^2} \frac{1}{\sqrt{1+\frac{L^2}{z^2}}} <br /> <br /> \approx \frac{\lambda}{ 2 \pi\epsilon_0} \frac{ L}{z^2} (1-\left(\frac{L}{z}\right)^2 + O\left(\frac{L}{z}\right)^4)<br /> <br />