The electric field on the axis of a charged disk

[Karol]

Homework Statement

A disk of radius 10 [cm] is charged with 5E-9 [Coulon]. where on it's axis is the greatest electric field.

Homework Equations

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

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]

 

[rude man]

Homework Statement

A disk of radius 10 [cm] is charged with 5E-9 [Coulon]. where on it's axis is the greatest electric field.

Homework Equations

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

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.

 

[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!)

 

[Karol]

The question is about a thin annulus

 

[haruspex]

The question is about a thin annulus

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

 

[Karol]

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

 

[rude man]

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

@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]

The question is about a thin annulus

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

 

[rude man]

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

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.

 

[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.

 

[haruspex]

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.

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.

 

[rude man]

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.

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 ...

 

[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

 

[Karol]

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

 

[haruspex]

$\tan^2\alpha=2$

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

it should be between 8 and 10

How do you know?

 

[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

 

[my2cts]

r=R/sqrt(2)

 

[haruspex]

The peak is either between 5 and 8 or 8 and 10

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?

 

[Karol]

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

 

[haruspex]

what is the calculus method?

The differentiation method in post #12.

 

[STEMucator]

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...

 

[haruspex]

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...

I think we're past all that. See posts #13 and #14. Further discussion concerned how to determine what range the answer should be in by evaluating the field at sample points.

 

[STEMucator]

I think we're past all that. See posts #13 and #14. Further discussion concerned how to determine what range the answer should be in by evaluating the field at sample points.

I was just tossing in an extra vote for the question being about a ring, and providing an alternative formula.

Excel should probably clean the rest of the problem up nicely.

 

[Karol]

Thanks