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
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.
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?
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?
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?
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]
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
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.
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.
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?
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.
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?
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)
ah yes, i need to brush up on those integrals, thanks! i wouldn't have been able to solve this on my own