# Electric Field due to a line of charge

1. ### stunner5000pt

Have a look at the diagram

Find the elctric field a distance z above one end of a astraight line segment of lenght L which carries uniform line charge lambda. Check that your formula is consistent with what you would expect for the case z >> L

SOlution:
$$\lambda= q/L$$
$$dq = \lambda dx$$

For the electric Field in the horizontal (points to the left and is negative)

$$dE_{x} = \frac{1}{4 \pi \epsilon_{0}} \frac{dq}{(z^2 + x^2)} \sin \theta$$

Subsituting what we know about sin theta and dq

$$dE_{x} = \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda dx}{(z^2 + x^2)} \frac{x}{\sqrt{z^2 + x^2}}$$

integrating x = 0 to x = L

$$E_{x} = \int dE_{x} = \frac{\lambda}{4 \pi \epsilon_{0}} \int_{x=0}^{x=L} \frac{xdx}{(z^2+x^2)^{\frac{3}{2}}} = \frac{\lambda}{4 \pi \epsilon_{0}} \left[ \frac{-1}{\sqrt{z^2 + x^2}} \right]_{x=0}^{x=L} = \frac{\lambda}{4 \pi \epsilon_{0}} \left[ \frac{-1}{\sqrt{z^2 + L^2}} + \frac{1}{z} \right]$$

$$E_{x} = \frac{\lambda}{4 \pi \epsilon_{0}} \left[ \frac{1}{z} -\frac{1}{\sqrt{z^2 + L^2}} \right]$$

ok so suppose z >> L then the electric field is zero?? Shouldnt it reduce to that of a point charge and not zero?? this should be regardless of whether i solve for x or z right??

for the Z direction i got
$$E_{z} = \frac{\lambda}{4 \pi \epsilon_{0}z} \frac{L}{\sqrt{L^2 + z^2}}$$
unlike the last one this one does reduce to the equatiopn for a point charge but is off my a factor of 1/z ...

combining the two yields

$$\vec{E} = \frac{\lambda}{4 \pi \epsilon_{0}} \left[ \frac{1}{z} -\frac{1}{\sqrt{z^2 + L^2}} \right] \hat{x} + \frac{\lambda}{4 \pi \epsilon_{0}z} \frac{L}{\sqrt{L^2 + z^2}} \hat{z}$$

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Last edited: Sep 21, 2006

### Staff: Mentor

If you get far enough away, the field from anything (finite) goes to zero! But you need to understand how it approaches zero. In this case, think about it: You are very far from a point charge--What's the direction of the field?

It's not off. You just need to examine its behavior as z goes to infinity. Use a binomial expansion:
$$(L^2 + z^2)^{-1/2} \approx (1/z)(1 - \frac{L^2}{2z^2})$$

3. ### Peppe

6

The electric potential:

$$V = \frac{ \lambda }{4\epsilon_0 \pi} ln\left(\frac{L+\sqrt{z^2+L^2}}{z}\right)$$

Calculating the electric field from the V's gradient, it has x component always at zero.
Where the problem?

I'm sorry for my poor english, good evening mate!

4. ### lepton5

26
Yeah, Griffith's book is good.

From your calculation for $$\vec{E}$$ (yes i think it's correct for both direction), for z >> L, the electric field of component - x will dissapear (since $$\frac{1}{z} - \frac{1}{z} = 0)$$ and leaves for us the z - component as below:

after simplifying, $$\vec{E} = \frac{\lambda L}{4 \pi \epsilon_{0}z^2}\hat{z}$$ for z>>L.

the case is the same as above example from griffith's book (that is for E at a distance z above midpoint of line L). I quoted it : From far away the line

"looks" like a point charge $$q = 2\lambda L$$.

I think x and y component will be gone since you do grad $$\nabla$$-operation to potential (the potential only depend on z). so,

$$\vec{E} = -\nabla V = -\frac{\lambda}{4\pi \epsilon_{0}}\frac{\partial}{\partial z}\ln (\frac{L + \sqrt{z^2 + L^2}}{z})$$.

suppose $$k = \sqrt{z^2 + L^2}$$,

it will give $$\vec{E} = \frac{\lambda}{4 \pi \epsilon_{0}}(\frac{z}{k(L + k)} - \frac{1}{z})\hat{z}$$

btw, i cant find this problem in griffith's book....