How Does Charge Density Affect Electric Field Strength Between Two Cables?

[Electra]

1. As a member of a team of storm physicists, you are attempting to replicate lightning by chargeing two long cables stretched over a canyon. one cable will attain a highly positive and uniform density of -$\lambda$ and the other will attain the same amount of charge density, but opposite in sign(i.e.-$\lambda$. Since the appareance of lightning directly depends on the electric field strength created by charge separation, it is important to derive an expression for electric field strength at all points between the two cables (albeit near the midpoint of the wires).

Given Data:
$\lambda$= +/- 40 $\mu$C
D=30.0m
r=15.0 m



2. (a) The cables are sufficiently long as to be considered infinitely long. Calculate the magnitude of the electric field strength between the two cables as a function of $\lambda$ and r(the distance from the positively charged cable. Use epsilon as the permittivity of free space and assume the wires are separated y a Distance D
(b) if the charge density separation between the two cables is 40$\mu$C and the distance between the two cables is 30.0m, calculate the electric field midway between the two cables.



3. The Attempt at a Solution :
well its asking for electric field between two line charges so fine the net vector sum of the electric fields
(a) Using Gauss' Law
E_1 = q/($\epsilon$Acos(0)) because electric field and area vector are parallel
E_2 = q/($\epsilon$Acos(180)) because electric field and area are in diff directions

adding them:
E_1 = $\lambda$l/($\epsilon$2$\pi$rl*1) A=2pi(r)(l)
E_2 = -$\lambda$l/($\epsilon$2$\pi$rl*-1)

E_1+E_2 = $\lambda$/($\epsilon$*2$\pi$r)+$\lambda$/($\epsilon$*2$\pi$r)
= $\lambda$/($\epsilon$$\pi$r)

(b) inputting the values for lambda, r and epsilon
= 40*10^-6/ (8.85*10^-12*$\pi$*(30/2))= 9.59*10^4 N/C

Yet no matter how many times i plug that in, it keeps saying I'm wrong.

 

[Electra]

Pleasee, i urgently need help :(

 

[Ginger14]

The equation is a bit more complicated than the equation to answer the second part of the problem. The second part of the problem is simple Gauss' Law. Just remember that r=D/2

 

[rude man]

1. I see nothing wrong with the way you did this, and I get the same answer. Who says it's the wrong answer?

2. Same deal except you use two cylindrical surfaces of different radii. r1 + r2 = D