So r = (.43 - x), as x varies from 0 to .14. This seems like it would be correct because at x = 0, (the left end of the rod) r = .43, and at x = .14,(the right end of the rod), r would equal .29, which was established earlier as the distance between the right end of the rod and the point p. Most...
So if r varies with x would it be r = (.29 + x)?
EDIT: Actually, that doesn't sound quite right.
as x increases, the distance between the charge and the point decreases, so the value of r would have to decrease as x increases.
I'm not quite sure I follow.
If I visualize the point as (.07m + .36m) = .43m from the origin, I would obtain a distance Δx = .29m.
So would I just place this as r in the formula E = ∫keq/r^2, change the q into λdx, and solve accordingly?
So I can choose any location, so long as I integrate over the entire length of the rod. That would mean that what I was doing previously where I was integrating from the end of the rod was incorrect because I simply integrated from the space between the rod and the point.
So I should set the...
Thank you. :biggrin:
I tried to place the origin at the end of the rod, though I'm not too sure I did that successfully.
Though I wonder now, would the numbers work out if I placed my origin in the center of the rod, and instead integrated over the interval of .07m to .36m?
Homework Statement
A rod 14.0 cm long is uniformly charged and has a total charge of -22.0μC. Determing (a) the magnitude and (b) the direction of the electric field along the axis of the rod at a point 36.0 cm from its center.
d (the distance between the center of the rod and the point)...