- #1
Chansu
Homework Statement
Positive charge Q is distributed uniformly along the x-axis from x = 0 to x = a. Determine
a. The electric field produced by the charge distribution Q at points on the positive x-axis where x > a.
b. A point charge q is then placed at x = a + r. Determine force on q due to Q
c. If r >> a, show that the force approaches that given by two point charges.
With the origin being to the left of the point, it is throwing me off quite a bit. I'm not sure if my answers are correct, but I feel like I'm headed in the right direction.
I've attached the diagram as well as mine to this thread. The video below also is extremely relevant.
Homework Equations
E = kq / r2 = k dq / r2
λ = Q / L = dQ / dL
F = qE
x = a + r
The Attempt at a Solution
[/B]
a. dE = (k dQ) / L2 i^
using λ and solving for dQ,
dE = ∫[(k λ dL) / L2] i^ (from r to x)
taking the integral,
E = [k λ (-L-1)] i^ (from r to x)
E = [k λ(-x-1 + r-1)] i^
E = [k λ(-x-1 + (x-a)-1)] i^ N/C, as long as x > a
b. F = q [k λ(-x-1 + [x-a]-1)] i^
if x = a + r,
F = q [k λ(-[a+r]-1 + [a+r-a]-1)] i^
F = q [k λ(-[a+r]-1 + r-1)] i^ N
c. if r >> a,
F = q k λ [-(a+r)-1 + r-1]
F = q k λ [-r-1 + r-1]
F = q k λ