# Potential that has had me stuck for hours

by Destrio
Tags: hours, potential, stuck
 P: 213 A total amount of positive charge Q is spread onto a nonconducting, flat, circular annulus of inner radius a and outer radius b. The charge is distributed so that the charge density (charge per unit area) is giver by o = k/r^3, where r is the distance from the centre of the annulus to any point on it. Show that (with V = 0 at infinity) the potential at the centre of the annulus is given by: V = (Q/8pie)((a+b)/ab) so I have dV = (1/4pie)(dQ/r) dQ = odA = o2pirdr = 2piQdr/r^2 I'm really confused about the k in o = k/r^3 I'm thinking it must either be the constant k = 1/4pie, but my professor never rights that as k, so I'm thinking that k=Q because otherwise I don't see a way to have Q in the expression. So I tried taking the integral of that expression from a to b with plugging dA and o in, but I'm not getting what I need. Any help is much appreciated! Thanks
P: 2
To find k: You are given the total charge (Q) and the charge density. Integrate the charge density over the annulus to get an expression for the total charge. Equate this with Q and solve for k.

 Quote by Destrio A total amount of positive charge Q is spread onto a nonconducting, flat, circular annulus of inner radius a and outer radius b. The charge is distributed so that the charge density (charge per unit area) is giver by o = k/r^3, where r is the distance from the centre of the annulus to any point on it. Show that (with V = 0 at infinity) the potential at the centre of the annulus is given by: V = (Q/8pie)((a+b)/ab) so I have dV = (1/4pie)(dQ/r) dQ = odA = o2pirdr = 2piQdr/r^2 I'm really confused about the k in o = k/r^3 I'm thinking it must either be the constant k = 1/4pie, but my professor never rights that as k, so I'm thinking that k=Q because otherwise I don't see a way to have Q in the expression. So I tried taking the integral of that expression from a to b with plugging dA and o in, but I'm not getting what I need. Any help is much appreciated! Thanks
 P: 213 so if i integrate the charge density from a to b i get -2k/r^2 Q = -2k/r^2 k = -qr^2/2 is this correct? how will i get rid of this negative?
P: 213

## Potential that has had me stuck for hours

or do i do:
odA = dQ
Q = -4pik((1/b)-(1/a))
Q = 4pik((b-a)/ab)
k = Qab/4pi(b-a)
 P: 740 Your second post is good, except you should check your work. Bringing the negative into the parentheses should change b-a to a-b, and from where did you get the 4?
 P: 213 i used the - to make it 1/a - 1/b then changed it to b-a/ab so i was back at b-a the 4 i got from dA = 2pirdr then *-2 from the integral i used o=k/r^3 and odA = Q should i just use o=k/r^3 to get Q = -2k(1/b - 1/a) k = (Q/2)(ab/b-a) once i get k, what do i do with it?
 P: 213 thats not right, without the dA i end up with Q = -2k(1/b^2 - 1/a^2)
 P: 740 Sorry, that was my mistake. Your b-a is correct, but your k still isn't completely right. You are missing a $\pi$ in there. After you find k, you should then integrate to find the total potential. Remember, that $$V = \int \frac{k \cdot dq}{r}$$ EDIT: How are you finding Q? It should be the case that $$Q = \int dq = \int_a^b 2\pi r \sigma \cdot dr$$
 P: 213 i figure i must be integrating dQ = odA from a to b dq = int k/r^3 dA but i know dA = 2pirdr is that correct? i dont know how to integrate that
 P: 740 Yes, that is correct, What you basically have is $$dq = \sigma dA = \frac{2\pi k \cdot dr}{r^2}$$ then $$Q = \int_a^b \frac{2\pi k \cdot dr}{r^2}$$ which you can do, because it is just using the power rule (take out the constants first). You almost had it right the first time.
 P: 213 or will it be Q = 2pik * int dr/r^2
 P: 213 that was what i did back when i got: k = Qab/4pi(b-a)
 P: 740 Yes, you are right, but try integrating $$\int \frac{dr}{r^2}$$ again. If you do it correctly, then solve for k, you should have a 2, not a 4.
 P: 213 ah yes, terrible mistake... brought over a -2 instead of a -1 so when i intergrate the final part dV = kdQ/r do i want to have dQ = odA integrating that will give me what i got for Q again and what do i do with the r?
 P: 740 Not quite, because you this time you are integrating dQ/(4$\pi\epsilon r$), not just dQ, as you did when finding what k equals in terms of Q. Furthermore, you can replace the k in $\sigma$ with whatever you found k to be in terms of Q, a, and b.
 P: 213 i can take out the 4pie as a constant so im left with integrating dQ/r which im not sure how to integrate.. do I want to change dQ to odA then everything is in terms of r dV = odA/r4pie = k2pirdr/(r^4)(4pie) V = int dV = (Q/4pie)(ab/b-a) * int dr/r^3 = (Q/4pie)(ab/b-a)(-2)(1/b^2 -1/a^2) is this correct so far?
 P: 213 this leaves me with the correct answer, except with 2pie on the bottom instead of 8pie (I had a factor of 2 in the numerator instead of denominator)
 P: 740 You're right, but you need a -1/2 instead of a -2 coming out of the integral.

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