Find the electric force of circular disk

Question

An annulus is a circular disk with a hole through the center. Suppose an annulus with an outer radius of 9 cm and inner radius of 5 cm is placed horizontally and bears a uniform surface charge density of 325 nC/m2. Find the electric force on a -18 mC test charge placed 12 cm directly above the exact center of the annulus. Assume the same permittivity as in the previous problem.

Answer 1

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), and

 

F(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 / (r² + z²)] cosθ

 

cosθ = z / (r² + z²)^1/2

 

dE_z = kz dQ / (r² + z²)^3/2

 

dQ = σdA = σ (rdφ)dr = σrdrdφ  [φ is in the x-y plane]

 

dE_z = kz σrdrdφ / (r² + z²)^3/2

 

E(z) =∫dE_z(z) = kzσ ∫₀^2π dφ ∫_a^b rdr / (r² + z²)^3/2

 

E(z) =  2πkzσ ∫_a^b 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.