The Electric Field Due to a Charged Disk

[brcole]

Homework Statement

A circular plastic disk with radius R = 1.80 cm has a uniformly distributed charge of Q = + 2.15 106 e on one face. A circular ring of width 30 µm is centered on that face, with the center of the ring at radius r = 0.50 cm. What charge is contained within the width of the ring?

Homework Equations

E=σ/2ε(1-z/√(z^2+R^2 ))

The Attempt at a Solution

e=8.85e-12
o= 2.15e6
z=.5 cm
r= 1.8
Ans=3.15 e10

 

[Delphi51]

I don't see it as an electric field question. The question asks "what charge ...".

 

[thomate1]

I would like to point out that the equation used in the attempt at a solution is not the correct one for determining the electric field due to a charged disk. The correct equation is given by:

E = (σ/2ε) * [1 - (z/√(z^2 + R^2))]

where σ is the surface charge density, ε is the permittivity of free space, z is the distance from the center of the disk, and R is the radius of the disk.

Using this equation, we can calculate the electric field at the center of the disk (z = 0) as:

E = (2.15 * 10^6 C/m^2) / (2 * 8.85 * 10^-12 C^2/Nm^2) = 1.22 * 10^17 N/C

This value is independent of the radius of the disk, so it does not matter that the ring is centered at a different radius.

To determine the charge contained within the width of the ring, we can use the equation for the electric field at a distance z from the center of the disk:

E = (σ/2ε) * [1 - (z/√(z^2 + R^2))]

Solving for σ, we get:

σ = 2ε * E * [1 - (z/√(z^2 + R^2))]

Plugging in the values for z = 0.50 cm and R = 1.80 cm, we get:

σ = 2 * 8.85 * 10^-12 C^2/Nm^2 * 1.22 * 10^17 N/C * [1 - (0.50 cm/√(0.50^2 + 1.80^2) cm)] = 2.12 * 10^-5 C/m^2

To determine the charge within the width of the ring, we can use the surface charge density formula:

Q = σ * A = (2.12 * 10^-5 C/m^2) * (30 * 10^-6 m^2) = 6.36 * 10^-11 C

Therefore, the charge contained within the width of the ring is 6.36 * 10^-11 C. This is a much smaller value than the attempt at