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Potential that has had me stuck for hours

  • Thread starter Destrio
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  • #1
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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
 

Answers and Replies

  • #2
2
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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.


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
 
  • #3
212
0
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?
 
  • #4
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or do i do:
odA = dQ
Q = -4pik((1/b)-(1/a))
Q = 4pik((b-a)/ab)
k = Qab/4pi(b-a)
 
  • #5
737
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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?
 
  • #6
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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?
 
  • #7
212
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thats not right, without the dA i end up with
Q = -2k(1/b^2 - 1/a^2)
 
  • #8
737
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Sorry, that was my mistake. Your b-a is correct, but your k still isn't completely right. You are missing a [itex]\pi[/itex] in there.

After you find k, you should then integrate to find the total potential. Remember, that

[tex]V = \int \frac{k \cdot dq}{r}[/tex]

EDIT:

How are you finding Q? It should be the case that

[tex]Q = \int dq = \int_a^b 2\pi r \sigma \cdot dr[/tex]
 
Last edited:
  • #9
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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
 
  • #10
737
0
Yes, that is correct, What you basically have is

[tex]dq = \sigma dA = \frac{2\pi k \cdot dr}{r^2}[/tex]

then

[tex]Q = \int_a^b \frac{2\pi k \cdot dr}{r^2}[/tex]

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.
 
  • #11
212
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or will it be Q = 2pik * int dr/r^2
 
  • #12
212
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that was what i did back when i got: k = Qab/4pi(b-a)
 
  • #13
737
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Yes, you are right, but try integrating

[tex]\int \frac{dr}{r^2}[/tex]

again. If you do it correctly, then solve for k, you should have a 2, not a 4.
 
  • #14
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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?
 
  • #15
737
0
Not quite, because you this time you are integrating dQ/(4[itex]\pi\epsilon r[/itex]), not just dQ, as you did when finding what k equals in terms of Q. Furthermore, you can replace the k in [itex]\sigma[/itex] with whatever you found k to be in terms of Q, a, and b.
 
  • #16
212
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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?
 
  • #17
212
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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)
 
  • #18
737
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You're right, but you need a -1/2 instead of a -2 coming out of the integral.
 
  • #19
212
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ah yes, i need to brush up on those integrals, thanks!
i wouldn't have been able to solve this on my own
 
  • #20
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wait, why would i get -1/2 from integrating 1/r^3
shouldnt it be -2?
 
  • #21
737
0
Well, think about taking the derivative. If you have a -1/2 in front, taking the derivative of 1/r^2 would give you a -2 in front, which would cancel the -1/2, leaving you with 1/r^3, which is what you have as the integrand.
 

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