A very long line of charge with linear charge density lies along the x-axis, ste
ID: 1701485 • Letter: A
Question
A very long line of charge with linear charge density lies along the x-axis, steching from x=-L to x=L . For a given point on th z-axis find the magnitude of E (E is a vector) using:
1. Gauss' Law in the limit L => Hint: your gaussian surface is a cylinder, but there is no flux through the ends
2. Integrate over the charge distribution. Show that your answer matches 1 in the limit L => Hint even though you'll be integrating over x, you'll find the z component, so your integral will have both x and z
3. The electric potential. First integrate to find the potential Voltage. Then from Voltage find E show that the answer matched 1 in the limit L => . Hint potentials are scalars, so they don't have components.
Explanation / Answer
Due to symmetry you know that there is only E in the positive z direction. In other words since it is an infinite line of charge, for every small segment of charge on the left of the midline there is an equal small segment of charge on the right of the midline. These two charges will have equal and opposite x-components of E so the z-components will add. I've done this problem using trig substitutions and integrating but you should be able to position the gaussian cylinder on the midline and evaluate using gauss's law. The final answer should come out to E = lambda/4(pi)Eoz Note: Eo is epsilon naught