Total Electric Charge on an annulus

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Homework Help Overview

The discussion revolves around calculating the total electric charge on an annulus, which is defined as a thin disk with a circular hole at its center. The problem involves parameters such as inner radius R_1, outer radius R_2, and a uniform positive surface charge density, sigma. Participants explore how to approach the problem without directly integrating the electric field.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between surface charge density and total charge, questioning the need for integration and exploring definitions of charge per unit area. They also consider how to calculate the total area of the annulus and relate it to the surface charge density.

Discussion Status

The conversation is active, with participants providing guidance on how to relate surface charge density to total charge. There is an ongoing exploration of how to calculate the total area of the annulus and the implications of uniform charge distribution. Some participants are close to deriving the total charge but are still refining their understanding of the relationships involved.

Contextual Notes

Participants are navigating through the definitions and relationships of charge density and area, with some uncertainty about the correct formulas and their application. There is a mention of the need to account for both surfaces of the annulus in the charge calculation.

Tcat
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I am getting stuck trying to find the total electric charge on the annulus. The Question reads: "A thin disk with a circular hole at its center, called an annulus, has inner radius R_1 and outer radius R_2. The disk has a uniform positive surface charge density, sigma, on its surface. Determine the total electric charge on the annulus." I realize that I need to intergrate dE_mag(r) from R_2 to R_1, but I'm not sure how to do it using the correct variables :frown: .
 
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You don't need to integrate the electric field to solve this problem. There is a much simpler way to obtain the total charge. You don't even know what the electric field is, think about the known quantities ...
 
Well you are given R_1, R_2, sigma, k, Q and based on the formula E = (Q/(4*pi*E_0*r^2))... r^2= (R_1 + R_2)^2 ?
 
Tcat, the formula you quoted, [tex]E = \frac{Q}{4 \pi \epsilon_0} \frac{1}{r^2}[/tex] is only valid for point particles of charge Q (and uniformly charged spheres). The electric field produced by the annulus is actually much more complicated than the above formula. Since the electric field isn't known, all you really have is the radii and the surface charge density. How is the surface charge density defined?
 
surface charge density = E_o*E
 
Try not to think in terms of formulas. Let me give an example. Suppose I give you a wire with uniform linear charge density [tex]\lambda[/tex], charge [tex]Q[/tex], and length [tex]L[/tex]. How are the three related? What does linear charge density mean? It is the charge per unit length, right? So how do I find it? I divide total charge by total length (since the system is uniform), [tex]\lambda = Q/L[/tex].
Now try to ask similar questions about the surface charge density. What does surface charge density mean?
 
surface charge density is the charge per surface area, which is Q/A
 
Right! How can you use this to solve the problem?
 
I'm not sure, do you find the SA of the disk first?
 
  • #10
Surface charge density=Q/A=((Q/pi*((R_2)^2 - ((R_1)/2)^2)
 
  • #11
I mean Q/A=((Q/pi*((R_2)^2 - (R_1)^2) Does that sound right? What do I do from here?
 
  • #12
You are almost there. Just solve for the unknown charge in terms of known quantities.
 
  • #13
Q= surface charge, but I'm not sure what quantites are related with it
 
  • #14
I know the radii and the surface charge density, sigma... does Q=sigma*((R_2)^2 + (R_1)^2)^1/2
 
  • #15
Nope, try again. But you are very close. What you have multiplying sigma is not the area ...
 
  • #16
Q=sigma*pi*((R_2)^2 + (R_1)^2)^1/2
 
  • #17
Why take the square root? [tex]\pi\cdot r^{2}[/tex] gives the surface area of a disk without any problems. [tex]\pi\cdot(r_{o}^{2} - r_{i}^{2})[/tex] works for an annulus.

Also, since the charge is uniformly distributed over the surface of the annulus, don't you need to take into account the bottom surface as well as the top surface?
 
  • #18
So now using this formula how do I get the total electric charge
 
  • #19
You're almost there! You have the charge per unit area, and you can calculate the total area. Combine the two (how?), and voila.
 
  • #20
Well the charge per unit area = sigma*pi*((R_2)^2 - (R_1)^2) so I add that to the total area which is... pi*((R_2)^2 - (R_1)^2) and I get sigma*2(pi*((R_2)^2 - (R_1)^2) for the final answer of total electric charge
 
  • #21
That looks correct. It's the charge per unit surface area times the total surface area of the annulus, where you've taken care to multiply by 2 because it's double-sided.
 
  • #22
Thank-you so much for your help! I have a part 2 of the problem that says "A point particle with mass, m, and negative charge, -q, is free to move along the x-axis (but cannot move off the axis). The particle is originally placed at rest at and released. Find the frequency of oscillation of the particle." I know that E_x =-(sigma/(2*epsilon_0))*(x/((x^2 + R_2^2)^(1/2)) -x /((x^2 + R_1^2)^(1/2))) so using this formula and the formula on frequency I should be able to get the answer. I know that frequency = (1/(2*pi))*(k/m)^1/2. How do I incorporate this formula with the E_x formula to find the overall frequency?
 

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