Total Electric Charge on an annulus

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SUMMARY

The discussion focuses on calculating the total electric charge on an annulus, defined by inner radius R_1 and outer radius R_2, with a uniform surface charge density, sigma. The total charge, Q, can be derived from the surface area of the annulus using the formula Q = sigma * (2 * pi * (R_2^2 - R_1^2)). Participants clarify that the electric field does not need to be integrated for this calculation, emphasizing the relationship between surface charge density and total charge. The final formula accounts for the double-sided nature of the annulus.

PREREQUISITES
  • Understanding of surface charge density and its relation to total charge.
  • Familiarity with the geometry of an annulus.
  • Basic knowledge of electric fields and their calculations.
  • Ability to manipulate algebraic expressions involving areas and charge densities.
NEXT STEPS
  • Study the derivation of electric fields for various geometries, including annuli.
  • Learn about the relationship between linear charge density and total charge for different shapes.
  • Explore the concept of oscillation frequency in electric fields, particularly for charged particles.
  • Investigate the implications of uniform charge distributions in electrostatics.
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Students and professionals in physics, particularly those focusing on electrostatics, electrical engineering, and anyone involved in problems related to charge distributions and electric fields.

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, E = \frac{Q}{4 \pi \epsilon_0} \frac{1}{r^2} 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 \lambda, charge Q, and length L. 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), \lambda = Q/L.
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? \pi\cdot r^{2} gives the surface area of a disk without any problems. \pi\cdot(r_{o}^{2} - r_{i}^{2}) 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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