Drawing a diagram really helps :smile:
So I would set dq=λdx
and then dE=dq/4∏ε0r2, and I would substitue dq for λdx and integrate both sides because the change in charge q over the change in distance x is equal to the charge density.
Is this correct direction I'm heading towards? I still do...
Ahhh I see this makes more sense
So the equation you are using is dE=dq/(4∏r2)
You made a cube because you pulled out a (r-r'). Did you pull it out so so it becomes a vector quantity?
And anywhere on the x-axis, the vector field point parallel to the x-axis in the positive direction I believe...
Since this is an infinitely long line of charge, I calculated the electric field by assuming there is a cylinder wrapped around the line of charge because the electric field is uniform on the surface of the cylinder. I chose length L as an arbitrary length for the length of the cylinder. The...
Homework Statement
1. Consider a line of charge (with λ charge per unit length which extends along the x-axis from x=-∞ to x=0
(a) Find all components of the electric field vector at any point along the positive x-axis
(b) Find the electric potential difference between any point on the...