Electric Field Due to a Charged Disk

[jrz5036]

Homework Statement

Suppose you design an apparatus in which a uniformly charged disk of radius R is to produce an electric field. The field magnitude is most important along the central perpendicular axis of the disk, at a point P at distance 2.60R from the disk. Cost analysis suggests that you switch to a ring of the same outer radius R but with inner radius R/2.00. Assume that the ring will have the same surface charge density as the original disk. If you switch to the ring, by what percentage will you decrease the electric field magnitude at P?

Homework Equations

E field equations for disk and ring

The Attempt at a Solution

please tell me how to do it and the answer is ...Thanks```

 

[AEM]

Homework Statement

Suppose you design an apparatus in which a uniformly charged disk of radius R is to produce an electric field. The field magnitude is most important along the central perpendicular axis of the disk, at a point P at distance 2.60R from the disk. Cost analysis suggests that you switch to a ring of the same outer radius R but with inner radius R/2.00. Assume that the ring will have the same surface charge density as the original disk. If you switch to the ring, by what percentage will you decrease the electric field magnitude at P?

Well, you're going to have to do some integrating. Also, you'll need to make use of symmetry. I'll get you started. It would be a good exercise to separately compute the electric field for a solid disk and then the ring. Draw yourself a picture of a disk of radius R and mark the point P where you want the electric field. You're going to have to sum up the contribution of little charge elements on this disk to the total electric field at point P. Label a little piece of charge on the disk and sketch in the electric field it makes at P.

You can define your bit of charge as

$$dq = \rho da$$ where rho is the charge density per unit area. Now, for each piece of charge on the disk there is another piece diametrically opposite to it. That means that at P there will be a cancellation of components of the electric field vectors such that the resultant electric field is along the axis perpendicular to the disk. That's great because it simplifies your analysis.

By now you should have figured out that you'll need to integrate over the disk to add up the charge . I'll give you the integral:

$$E = \frac{1}{4 \pi \epsilon_o} \int \frac{dq}{r^2}$$

I gave you dq above, so you'll have to figure out da. You will also have to figure out how to express $$r^2$$ in terms of the geometry. Important note: r in the equation above is the distance from the piece of charge to point P. You will have to represent the radius of the disk as you integrate with some other letter or you will mess up! Your limits of integration will depend on whether you choose to express da in terms of one variable , or two. Hint: if you choose two integrals, one of them will have limits from 0 to 2 pi.

I think this will get you started. Once you understand how to calculate the electric field for the whole disk, the ring will be a piece of cake.

 

[thomate1]

To solve this problem, we can use the equation for the electric field due to a charged disk, which is:

E = (σ/2ε₀) * (1 - (z/√(z² + R²)))

Where:
E = Electric field magnitude
σ = Surface charge density
ε₀ = Permittivity of free space
z = Distance from the center of the disk
R = Radius of the disk

We can also use the equation for the electric field due to a charged ring, which is:

E = (σ/2ε₀) * (R²/(z² + R²))

Where:
E = Electric field magnitude
σ = Surface charge density
ε₀ = Permittivity of free space
z = Distance from the center of the ring
R = Radius of the ring

Now, to find the percentage decrease in the electric field magnitude at point P, we can set the two equations equal to each other and solve for z:

(σ/2ε₀) * (1 - (2.60R/√(2.60R)² + R²)) = (σ/2ε₀) * (R²/(2.60R)² + R²))

Simplifying this equation, we get:

1 - (2.60R/√(2.60R)² + R²) = R²/(2.60R)² + R²

Solving for R, we get:

R = 0.72R

This means that the distance z for the ring would be 0.72 times the distance z for the disk. Therefore, the electric field magnitude at point P would decrease by approximately 28% if we switch from the disk to the ring. This is because the electric field decreases as the distance from the charged object increases.

In conclusion, switching to the ring would result in a 28% decrease in the electric field magnitude at point P. This decrease may be beneficial in terms of cost analysis, but it is important to also consider the overall effectiveness and efficiency of the apparatus in producing the desired electric field. Further analysis and testing may be needed to make a final decision on whether to switch to the ring or not.