Potential that has had me stuck for hours

In summary, we are given a total amount of positive charge Q spread onto a nonconducting, flat, circular annulus with inner radius a and outer radius b. The charge is distributed with a charge density o=k/r^3, where r is the distance from the center of the annulus to any point on it. We are asked to show that, with V=0 at infinity, the potential at the center of the annulus is given by:V = (Q/8pie)((a+b)/ab)To find k, we can integrate the charge density over the annulus and equate it with the given total charge Q. Solving for k, we get k = Q
  • #1
Destrio
212
0
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
 
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  • #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.


Destrio said:
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
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
or do i do:
odA = dQ
Q = -4pik((1/b)-(1/a))
Q = 4pik((b-a)/ab)
k = Qab/4pi(b-a)
 
  • #5
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
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
thats not right, without the dA i end up with
Q = -2k(1/b^2 - 1/a^2)
 
  • #8
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
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 don't know how to integrate that
 
  • #10
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
or will it be Q = 2pik * int dr/r^2
 
  • #12
that was what i did back when i got: k = Qab/4pi(b-a)
 
  • #13
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
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
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
i can take out the 4pie as a constant so I am left with integrating dQ/r which I am 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
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
You're right, but you need a -1/2 instead of a -2 coming out of the integral.
 
  • #19
ah yes, i need to brush up on those integrals, thanks!
i wouldn't have been able to solve this on my own
 
  • #20
wait, why would i get -1/2 from integrating 1/r^3
shouldnt it be -2?
 
  • #21
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.
 

What is potential?

Potential refers to the ability or capacity for something to develop, succeed, or become something greater.

What are the types of potential?

There are several types of potential, including kinetic potential (energy of movement), chemical potential (energy stored in chemical bonds), and gravitational potential (energy based on an object's position in a gravitational field).

How do you calculate potential?

The calculation of potential depends on the type of potential being measured. For example, gravitational potential energy can be calculated using the equation PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object.

What factors affect potential?

The factors that affect potential can vary depending on the type of potential being considered. For example, factors such as mass, velocity, and position can affect kinetic potential, while factors such as bond strength and distance can affect chemical potential.

Why is understanding potential important in science?

Understanding potential is crucial in science because it allows us to predict and explain the behavior of systems and objects. It also helps us to understand the concept of energy and how it is transferred and transformed within different systems.

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