# The Electric Field Due to a Charged Disk

• brcole
In summary, the formula for calculating the electric field due to a charged disk is E = σ/(2ε<sub>0</sub>), where σ is the surface charge density and ε<sub>0</sub> is the permittivity of free space. The electric field due to a charged disk is different from that of a point charge because the charge is distributed over a larger surface area and decreases more quickly with distance. Factors that affect the strength of the electric field include the magnitude of the charge, radius of the disk, distance from the center, and the permittivity of the medium. The direction of the charge does not affect the electric field, and it can be used in practical applications such as electrostatic precipit
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

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

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

## 1. What is the formula for calculating the electric field due to a charged disk?

The electric field due to a charged disk can be calculated using the formula E = σ/(2ε0), where σ is the surface charge density and ε0 is the permittivity of free space.

## 2. How does the electric field due to a charged disk differ from that of a point charge?

The electric field due to a charged disk is different from that of a point charge because the charge is distributed over a larger surface area, resulting in a more uniform electric field. Additionally, the electric field due to a charged disk decreases more quickly with distance compared to that of a point charge.

## 3. What factors affect the strength of the electric field due to a charged disk?

The strength of the electric field due to a charged disk is affected by the magnitude of the charge on the disk, the radius of the disk, and the distance from the center of the disk. The electric field also depends on the medium in which the disk is located, as the permittivity of the medium affects the electric field strength.

## 4. Is the electric field due to a charged disk affected by the direction of the charge?

No, the electric field due to a charged disk is not affected by the direction of the charge. The electric field lines always point away from a positively charged disk and towards a negatively charged disk, regardless of the direction of the charge.

## 5. How can the electric field due to a charged disk be used in practical applications?

The electric field due to a charged disk can be used in practical applications such as electrostatic precipitators, which use the electric field to remove particulates from air or gas. It can also be used in electrostatic motors, where the electric field is used to produce motion in a charged object.

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