# Electric Field due to Continuous Line Charge

1. Dec 9, 2009

### darkpsi

1. The problem statement, all variables and given/known data
Find the electric field a distance z above the midpoint of a straight line segment of length
2L, which carries a uniform line charge λ.

2. Relevant equations
1) my textbook says :
E(r) = 1/4πεoV λ(r')/r2 r dl'

2) and this also works? :
E(r) = 1/4πεoV λ(r')/r3 r dl'

both where r is the unit vector from the charge to the point we are caluclating the field at

3. The attempt at a solution
I can integrate once I get the setup down just fine. I just wanted to know when and why this two are the same. For example, in my book it says for the setup for this problem :
E(r) = 2 * 1/4πεoV(λdx/r2)cosθz

For one thing, how did that cos appear and since it equals z/r couldn't 2) have just been used from the start? And if 2) can be used could I apply it to surface and volume integrals?

2. Dec 10, 2009

### ApexOfDE

General formula for E is:

dE = 1/4πεo (dq/r^2)$$\hat{r}$$

However, r = r.$$\hat{r}$$ (r_hat is unit vector)
So you can write"

dE = 1/4πεo (dq/r^3).r

--
Electric field of a point above distance Z from a line is a cumulative Electric field caused by every charge in straight line.

Lets say an angle formed by E caused by one charge and a vertical axis is theta. From there, you can get:
E_x = E * sin(theta)
E_y = E * cos(theta)

And because E in x-axis will automatically blow up, you just need to find E in y-axis.

Hope this helps

3. Dec 10, 2009

### darkpsi

Oh now I see. I wasn't even thinking that because r_hat wasn't pointing vertically that the z component wasn't all of the electric field. I was thinking the x-fields cancelled so the field in the z direction was exactly equal to the electric field, but its actually only cos(theta) of it. Thanks a bunch