# Homework Help: Archived Finding the electric field between two lines of charges using Gauss Law

1. Sep 13, 2011

### Jazita

1. The problem statement, all variables and given/known data
The problem is given in the attachment below

2. Relevant equations

(i)EA= Q/epsilon naught

(ii)Area of cylinder used = 2pi*r*L

(iii)The integral of E*dA =Q/4pi*epsilon naught

(iv) Llamda= Q/L

3. The attempt at a solution
Well I know I needed to choose a suitable Gaussian surface which is a cylinder around the two lines of charges. Using Gauss' Law I get the first equation above using the third equation. I know that I have to take the 1st equation and transpose the Area across. I end up with E=llamda/(2*pi*r*epsilon naught). That is the electric field for one of the lines of charge. I'm not sure how to get the second line of charge involved nor to include the distance, D into the equation. (I'm sure it does play a role in the equation but I'm not sure how)

Oh. I got the answer for the 2nd part of the question using the formula E=llamda/(2*pi*D*epsilon naught) but when I placed it in the equation section, it was wrong.

File size:
50.4 KB
Views:
272
2. Feb 8, 2016

### Alex Duron

This is a Gauss's law problem.

First, let's place the two lines of charge on an x-y coordinate: we'll place the positive line on the x axis and the negative line parallel to the x axis a height D along the y axis.

Second, let's draw the directions of the electric fields of the lines at the point of interest (halfway in between the lines, D/2). The direction of the electric field of the positive line (E+) at this point is up as is the direction of the electric field of the negative line (E-). The electric fields are in the same direction at this point so the net field is their sum.

Third, how can we calculate the electric field of an infinite line? The result can be obtained from any text, and is E = λ/2πε0r, where r = the distance from the line charge to the point where we're trying to find the magnitude of the electric field.

Fourth, let's see how we can derive this equation ourselves. Gauss's law: = EA = Q/ε0. Solving for E gives E = Q/Aε0. The Guassian surface that is symmetrical about an infinite line charge is a cylinder which has a surface area A = 2πrl so E = Q/2πrlε0 but Q/l = λ giving us E = λ/2πε0r. The ends of the cylinder, with area A = πr^2, are not included since their surfaces are parallel to the electric field of the line charge so they have zero electric flux.

Fifth, finding the net electric field. As positioned in our x-y coordinate system both electric fields are up which we generally take to be positive. Let's find the sum of the electric fields. ∑Ey = E+ + E- = λ+/2πε0r + λ-/2πε0r where λ+ = the absolute value of the positive line and λ- = is the absolute value of the negative line. We only use the absolute values because we've already taken into account the directions of the electric fields of the lines in our analysis. One final note, if the distances between the lines and the point of interest are not different then the two electric fields will have different values of r. Thank you.

Alex Duron
foundations-tutoring.com