# Limits and infinity problem

1. Aug 21, 2010

### vorcil

1. The problem statement, all variables and given/known data

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 $$\lambda$$

3. 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

2. Aug 21, 2010

### lanedance

Re: Limits

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

3. Aug 21, 2010

### lanedance

Re: Limits

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

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