# Potential of thin ring of charge

Tags:
1. May 11, 2017

### Buffu

1. The problem statement, all variables and given/known data

A thin disc, radius 3 cm, has a circular hole of radius 1 cm in the middle. There is a surface charge of -4 esu/$cm^2$ on the disc.

a) what is the potential at the centre of the hole ?

b) An electron starting from rest at centre of the hole moves out along the axis. What is the ultimate velocity it gains ?

2. Relevant equations

3. The attempt at a solution

For (a) :

$R = \sqrt{y^2 + s^2}$ and $dq = 2\pi s\sigma ds$

I did not drew $\theta$ but it is the angle between radius vector $r$ and $y$ axis.

Then,
$\displaystyle \phi(0,y,0) = \int^3_1 {dq \over R} = \int^3_1 {2\pi \sigma s ds \over \sqrt{y^2 + s^2}} =2\pi \sigma (\sqrt{y^2 + 9} - \sqrt{y^2 + 1})$

For centre, $\phi = 2\pi\sigma$

For (b):-

Since $\cos \theta = \dfrac{y}{\sqrt{y^2 + s^2}}$

$\displaystyle E = \int^3_1 {dq \over R^2} \cos \theta = \int^3_1 {2\pi \sigma s y \over (y^2 + s^2)^{3/2}} ds =\sigma\left( \frac1{\sqrt{y^2 + 1}} - \frac1{\sqrt{y^2 + 9}}\right)$

$eE = ma = m y^{\prime\prime} = m v\dfrac{dv}{ dy}$

$\displaystyle {e\sigma \over m} \int_0^{\infty} \frac1{\sqrt{y^2 + 1}} - \frac1{\sqrt{y^2 + 9}} dy = \frac{v^2}{2}$

Hence $v = \sqrt{\dfrac{2e \sigma \ln 3}{m}}$

Is this correct ? I think the second answer is not correct as I get $2.16 \times 10^{7} m/s$ as the answer, which is very huge.

2. May 11, 2017

### vela

Staff Emeritus
You dropped a factor of 2, it seems, but it otherwise looks right.

I'm not sure what this means. What $\phi$ supposed to represent?

You seem to be making this part unnecessarily complicated. Use conservation of energy. What's the potential energy of a charge at the center in terms of the electric potential at the center?

3. May 11, 2017

### Buffu

$\displaystyle \int {s\over \sqrt(y^2 + s^2)} = \sqrt{s^2 + y^2}$ ?

I am sorry, this was vague.

$\phi(0,y,0)$ is electrostatic potential which is $2\pi \sigma (\sqrt{y^2 + 9} - \sqrt{y^2 + 1})$.

The question to find the potential at the centre of the disc which is at $y =0$

So, $\phi(0,0,0) = 2\pi \sigma (3 - 1) = 4 \pi\sigma$

.

Let me get first part correct because I can use that to find energy.

4. May 11, 2017

### vela

Staff Emeritus

Sorry, you were right. I messed up the factor of 2.

5. May 11, 2017

### Buffu

Is everything else correct ?

6. May 12, 2017

### haruspex

I think there are some errors in the field calculations.
(It would be easier to follow if you had left the radii as variables instead of plugging in the values 1 and 3. That would have allowed the use of dimensional analysis to sanity-check the equations.)
E should be obtainable by differentiating φ wrt y. If I do that there seems to be a missing factor 2πy.
I certainly don't think you should be getting ln() terms in there when they do not turn up using the energy approach.

As for the magnitude of the answer, I think you are in the right ballpark, but maybe that means you need to use relativistic equations.

7. May 12, 2017

### Buffu

Yes you are correct.

Then $\displaystyle E = {2\pi y \sigma}\left({1\over \sqrt{1+y^2}} - {1\over \sqrt{9 + y^2}} \right)$

and
is
$\displaystyle {e\sigma \over m} \int_0^{\infty} {2\pi y }\left({1\over \sqrt{1+y^2}} - {1\over \sqrt{9 + y^2}}\right) dy = \frac{v^2}{2}$

I got the improper integral as $2$,

So $\displaystyle v = \sqrt{8\pi e \sigma \over m}$ which is $7.38 \times 10^7 m/s$

8. May 12, 2017

### haruspex

That's what I got by considering energy. But that is a quarter light speed, no? Doesn't that mean you need to use the relativistic formula for KE?

9. May 12, 2017

### Buffu

Yes I guess you are correct but I don't know special relativity ;).

10. May 12, 2017

### haruspex

11. May 12, 2017

### Buffu

12. May 12, 2017

### haruspex

I get $\frac{v^2}{c^2}=1-\left(\frac{mc^2}{E+mc^2}\right)^2$.
If v' is the velocity you calculated by Newtonian principles, E=mv'2/2.
But looks like it's not quite fast enough to worry about. If e.g. v' is c/4 then v≈ 0.238c

Last edited: May 12, 2017