- #1
Jazita
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Homework Statement
The problem is given in the attachment below
Homework 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
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.