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 E

_{z}(z), andF(z) = qE

_{z}(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

dE

_{z}= dE cosθ = [k dQ / (r^{2}+ z^{2})] cosθcosθ = z / (r

^{2}+ z^{2})^{1/2}dE

_{z}= kz dQ / (r^{2}+ z^{2})^{3/2}dQ = σdA = σ (rdφ)dr = σrdrdφ [φ is in the x-y plane]

dE

_{z}= kz σrdrdφ / (r^{2}+ z^{2})^{3/2}E(z) =∫dE

_{z}(z) = kzσ ∫_{0}^{2π}dφ ∫_{a}^{b}rdr / (r^{2}+ z^{2})^{3/2}E(z) = 2πkzσ ∫

_{a}^{b}rdr / (r^{2}+ z^{2})^{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.