# The electric field on the axis of a charged disk

1. Nov 10, 2014

### Karol

1. The problem statement, all variables and given/known data
A disk of radius 10 [cm] is charged with 5E-9 [Coulon]. where on it's axis is the greatest electric field.

2. Relevant equations
The field on the axis of a charged disk:
$E=\frac{1}{4\pi\varepsilon_0}\frac{q\cdot \cos\alpha}{s^2}$

3. The attempt at a solution
I express $\cos\alpha$ in terms of the radius R and the diagonal distance s:
$\cos\alpha=\frac{\sqrt{s^2-R^2}}{s}$
I insert that into part of the equation of the field:
$\frac{\cos\alpha}{s^2}=\frac{\sqrt{s^2-R^2}}{s^3}$
The derivative of this expression is (maybe i have a mistake here):
$\left( \frac{\sqrt{s^2-R^2}}{s^3}\right)'=\frac{3R^2-2s^2}{s^4\sqrt{s^2-R^2}}$
I equal the numerator to 0 and get s=1.22R=12.2[cm]
The horizontal distance on the axis that corresponds to this diagonal is:
$\cos\alpha=\frac{1.22R}{R}\rightarrow\alpha=34.95^0$
To this angle the horizontal distance is 14.3[cm] while it should be around 8[cm]

#### Attached Files:

• ###### Electric Field of a Ring.jpg
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2. Nov 10, 2014

### rude man

I don't think that's right!
Think of the disc as an infinite number of infinitely small annuli ranging in radius from 0 to R
where R = radius of disc.
You probably have the formula for the E field along the axis for a ring (thin annulus) somewhere.

Extra thought: what is the E field very close to a charged plane? I.e. for you that would imply s = R.
Hint: don't try setting a derivative to zero to find the max E field.

3. Nov 10, 2014

### rude man

Extra credit: what if there is a hole in the center of the disc, of radius R2? What then would be the point of maximum E field along the axis?
(Warning: don't try this at home without good math software!)

4. Nov 11, 2014

### Karol

The question is about a thin annulus

5. Nov 11, 2014

### haruspex

So there are two radii specified? Or is it so thin it's effectively a ring, not a disc?

6. Nov 11, 2014

### Karol

yes, a ring, i didn't know how to say in english a flat disk. and what is an annulus? also a ring?

7. Nov 11, 2014

### rude man

@haruspex, the OP's problem clearly states that it's a disc, not an annulus (ring). Even a thin disc is not the same thing as an annulus. A disc has area = pi R^2 while an annulus has an area = pi(R2^2 - R1^2).

I added the "extra credit" as a new, far more challenging problem, mathematically speaking. There is only one radius in the OP's problem.

My "extra credit" was a NEW problem, and in fact a tease since the math, which WOULD involve taking a derivative and setting it to zero, would probably blow your mind if not your computer!
[/quote]

No it's not. An annulus is a ring and this problem is a solid disc.

8. Nov 11, 2014

### rude man

A ring and an annulus are somewhat the same thing, although typically a ring is said to have zero dimensions so charge is in terms of linear (coulombs/meter) rather than surface density (coulombs/sq. meter). An annulus can have an arbitrarily large width so it can be a disc with a small hole in it and still qualify as an annulus.

A disc is a solid circle of one radius only, and with charge density typically defined in terms of surface charge density.

9. Nov 11, 2014

### Karol

I am almost sure the question is about a ring, look at the drawing, the book explained about that ring and in the explanation gave also only one radius R.

10. Nov 11, 2014

### haruspex

OK. In the OP, your differentiation looks wrong.
You will find the algebra much easier if you take an angle as the independent variable instead of a distance.

11. Nov 12, 2014

### rude man

I agree, the drawing looks like a ring. But a ring is not a disc, and the wording says disc.

I will let others take over here ...

12. Nov 13, 2014

### Karol

I express s in terms of $\alpha$:
$\frac{\cos\alpha}{s^2}=\frac{\cos^2\alpha}{R^2}$
$\left( \frac{\cos^2\alpha}{R^2}\right)'=\frac{1}{R^2}(\cos^2\alpha)'=\frac{-2}{R^2}\cos\alpha\cdot\sin\alpha$
And if i equal this to 0 then sin and cos may become 0 and i get 00 and 900

13. Nov 13, 2014

### Karol

$\frac{\cos\alpha}{s^2}=\frac{\cos\alpha\cdot\sin^2\alpha}{R^2}$
$\left(\cos\alpha\cdot\sin^2\alpha\right)'=2\sin\alpha\cos^2\alpha-\sin\alpha\cdot\sin^2\alpha$
$\tan^2\alpha=2\rightarrow\alpha=54.7^0\rightarrow x=7.1[cm]$
Better but still not true, it should be between 8 and 10

14. Nov 13, 2014

### haruspex

I agree, but no need to calculate the angle. The distance you want to calculate is $R/\tan\alpha = 5\sqrt 2cm$
How do you know?

15. Nov 13, 2014

### Karol

I was told to calculate the field E at distances 0, 5, 8, 10 and 15 [cm] from the origin.
$E=\frac{1}{4\pi\varepsilon_0}\frac{q\cdot \cos\alpha}{s^2}$
I calculate only:
$\frac{q\cdot \cos\alpha}{s^2}$
At 0 [cm] it's 0.001
At 5 [cm] it's 0.00356
At 8 [cm] it's 0.00381
At 10 [cm] it's 0.0035
At 15 [cm] it's 0.00256
The peak is either between 5 and 8 or 8 and 10, but i guess closer to 8. i will draw a graph and attach

#### Attached Files:

• ###### 2-14.JPG
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16. Nov 13, 2014

### my2cts

r=R/sqrt(2)

17. Nov 14, 2014

### haruspex

Exactly. If you want to find out which, by the same method, you need to calculate some more points. Why not just trust the caclulus method?

18. Nov 14, 2014

### Karol

what is the calculus method? do you mean just to calculate more points or make a graph?

19. Nov 14, 2014

### haruspex

The differentiation method in post #12.

20. Nov 14, 2014

### Zondrina

The question looks like a ring of radius $r$ and radial width $dr$. In this case the electric field of the "ring" would be given by:

$$\vec E(x) = \frac{kqx}{(x^2 + R^2)^{3/2}}$$

Where you would take the derivative...