# Electric field of a line of charge

1. Feb 7, 2014

### andylie

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

The figure below shows a finite line charge with linear charge density of λ and total length L. The point P shown is a distance s away from its end.

Please calculate a formula for the electric field at point P, in terms of λ, L and s.

Then use the following values to find it numerically.
λ = +3 μC/m, L = 5 m, s = 4 m

2. Relevant equations

Electric field to point P in the x direction is going to be dE*cosΘ y direction is going to be -dE*sinΘ

the equation for is E=∫(k*λdy/r^2) , where λ=Q/L, r=(s^2+L^2)^(1/2)
limit of integral is from 0 to 5

3. The attempt at a solution

i did get the answer for +x direction, but i couldnt get the answer for -y direction

for the x-dir, my eq start from ∫(Kλdy/r^2)cosθ. where cos=s/r

so it become ∫(kλdy/r^2)(s/r) since, k,λ,s, is a constant, the integration become

kλs∫dy/(s^2+y^2)^(3/2) using table of integration for ∫dy/(s^2+L^2)^(3/2), then
kλs(1/s^2)(y/(y^2+s^2)^(1/2)) s on the numerator and denominator cancels each other
(kλ/s)(5/(5^2+4^2)^(1/2))
((9*E9)(3*E-6)(5)/4)*(5/(5^2+4^2)^(1/2))=5270.8644

for the y-dir, its negative.
limit of integration is 0 to 5 of L value
my eq start from ∫(Kλdy/r^2)sinθ. where sin=L/r

so it become -∫(kλdy/r^2)(L/r) since, k,λ,L, is a constant, the integration become
-kλL∫dy/(s^2+y^2)^(3/2) using table of integration for ∫dy/(s^2+L^2)^(3/2), then
-kλL(1/s^2)(y/(y^2+s^2)^(1/2)
-(kλL)(5/(5^2+4^2)^(1/2))
-((9*e9)(3*E-6)(5)/(4^2))*(5/(5^2+4^2)^(1/2))= -6588.58058

but for the y-dir value is wrong. can someone help me?

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Last edited: Feb 7, 2014
2. Feb 7, 2014

### BvU

Is it ok if I can't make out head or tails from your story ? And the sheet from your notebook also is somewhat difficult to follow. So I propose we pick up what is understandable and take it from there.

I agree with $E_x=\int_0^L \frac{ k \lambda}{r^2} \cos\theta\ dy$ but not with $r^2=s^2+L^2$. Also not with considering $\cos\theta = s/r$ as a constant.
For the distance you have $r^2=y^2+s^2$ and $\cos\theta = s/r(y)$.

• the $1/r^2$ is missing in the first line
• the integral sign is missing in the second line
• the third line and onwards look ok provided one reads y instead of L in the proper places

You should not be needing a table of integration to do the integral, but ok. I get 5270 too. Can't decipher the units. Should be N/C.

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On with the y component. I vaguely distinguish a minus sign. This time you need a sine and you grab L/r. Nice try, but that's only at the end when y = L .
again, $r^2=y^2+s^2$ and now $\sin\theta = y/r(y)$.

This enough to get you going again ?

3. Feb 8, 2014

### andylie

I already turned it my assignment, i didnt get that one correct, but thank you though.
:D