
Arturo O. answered 08/28/17
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I will set this up for you in terms of an integral, and you evaluate the integral (or look it up in a table of integrals).
a = inner radius
b = outer radius
σ = surface charge density
The annulus lies in the x-y plane centered at the origin, so we want the force at z above the x-y plane.
z = distance above annulus along z-axis
q = test charge
It is easier to find E(z), and then find
F(z) = qE(z)
By the symmetry of the problem, the only component of the field that is not zero is the z component, so we only need to find Ez(z), and
F(z) = qEz(z)
θ = angle from the z-axis to a line from (0,0,z) to an infinitesimal charge dQ on the annulus, located a distance r from the z-axis
dEz = dE cosθ = [k dQ / (r2 + z2)] cosθ
cosθ = z / (r2 + z2)1/2
dEz = kz dQ / (r2 + z2)3/2
dQ = σdA = σ (rdφ)dr = σrdrdφ [φ is in the x-y plane]
dEz = kz σrdrdφ / (r2 + z2)3/2
E(z) =∫dEz(z) = kzσ ∫02π dφ ∫ab rdr / (r2 + z2)3/2
E(z) = 2πkzσ ∫ab rdr / (r2 + z2)3/2
Finally,
F(z) = qE(z)
You have everything you need now. Evaluate the integral or look it up, then plug and chug, being careful with the units.