What Is the Force on a Point Charge at the Tip of a Hollow Cone?

[Hercuflea]

Homework Statement

The question is 1.3 from Purcell E&M 3rd edition. It is asking you what the force on a point charge q is at the tip of a hollow cone (no bottom surface).

Homework Equations

F = (1/4pie0)*(qQ / r^2)

The Attempt at a Solution

I have worked on this on and off for days and I don't know what I'm doing. Apparently you have to play with the differential area...but that ends up giving me a double integral with only a single differential! It doesn't make sense!

See attached solution...

 

[BvU]

I have Jackson, second edition. But perhaps I can still help, although letting potential helpers guess at the problem formulation is a nono in PF.
My guess is that the cone is uniformly charged (d\sigma A = \sigma d A) and you want the electric field at the tip. Right ?
My question is what dA = d(z r \phi) stands for. Where does it come from ?

 

[Hercuflea]

I apologize for not assigning my variables.

The problem states: A charge q is located at the tip of a hollow cone (such as an ice cream cone without the ice cream) with surface charge density \sigma. The slant height of the cone is L, and the half-angle at the vertex is \theta. What can you say about the force on the charge q at the top of the cone?

Let me upload a picture which is a little more clear, I was kind of rushed when I first posted.

In my formulation,
Q is the total charge on the sphere which is equal to σ*A, A the total area of the cone (there is no bottom).

dq is the differential cone charge, which is equal to d(σ*A). I approximated σ*A with \sigmarl\varphi, which may or may not be the true differential area (I am not familiar with conical coordinates), where r is the differential radius, l is the differential side length, \varphi is the angle around the central azimuth.

The horizontal components of E field cancel out, so I got the integral in the second step for the force.

Little l is the differential length along the side of the cone, so the differential change in radius with side length is lsin\theta.

As you can see, after integrating and canceling some terms out, I got that integral at the bottom which is a double integral with onle one differential! I don't know what I'm doing wrong and there is no solutions in this book :(.

Help?

 

[BvU]

OK, I am somewhat with you. There's a Purcell on the net, but it's second edition. But now I understand why you consider 'Purcell' sufficient reference. Berkeley physics course would have done it for me, but I only own Wichmann (1^st edition...).

d\sigma A = \sigma d A I can understand, but dA = d(z r \phi) I can not. What is z r \phi ?
However, you write ∫ over the cone as a double integral where apparently dA = d(l^2 \sin\theta \phi). It's correct, but can you explain what that is ? Because in the next line I get visions of banana peelings...

My naive approach: a cone with slant height $l$ has area $\pi r l$ where $r = l \sin\theta$ in our case, so $\pi l^2 \sin\theta$. Slant height $l+d l$ gives $\pi (l+dl)^2 \sin\theta$. Thereby $dA=2\pi\sin\theta\enspace dl$.

Another way to get there: $dA= dl\enspace l\sin\theta\enspace d\phi$ (you can picture this as a little rectangle on the cone with height $dl$ and width $l\sin\theta\enspace d\phi$ ) and integrate from $0$ to $2\pi$ as your first double integral already suggests! Point is, if $dA$ is still two dimensional, you want 2 $d$'s, not one! So dA = d(l^2 \sin\theta \phi) is still correct. $\sin\theta$ is a constant, and what you should write imho is dA = 2l \sin\theta d l \enspace d\phi

The other stuff was just fine; after this small correction, the $\int d\phi$ is trivial and the $\int dl$ is almost trivial. Bingo!

 

[TSny]

For reference, here is the question. Note that for (a) you are asked, "what can you say about the force"? The wording implies that you might not be able to actually find the force for this part.

 

[BvU]

Ah, sneaky. My hope was the divergence wouldn't be bothersome because the area and thus the charge would go to 0 fast enough. Not so. Stating that it's infinite and pointing straight up is unsatisfactory!

No wonder this exercise wasn't in the 2nd edition: in those good old days everything was better, even the exercises ;-)

Fortunately the work hasn't been in vain: there is a b part where it comes in most handy.

 

[Hercuflea]

I don't understand? Sorry, this is my first "real" physics course. You can't use Gauss' Law / Divergence theorem here because there is no symmetry, right?

And by the way, when I originally wrote zr\varphi, I meant to say l * r * \varphi. That was my approximation for the differential area even though I'm not sure if it's right.

 

[BvU]

1I don't understand? Sorry, this is my first "real" physics course. 2You can't use Gauss' Law / Divergence theorem here because there is no symmetry, right?

3And by the way, when I originally wrote zr\varphi, I meant to say l * r * \varphi. That was my approximation for the differential area even though I'm not sure if it's right.

1) No reason to apologize at all. You are clearly trying hard. we try to help.

2) Gauss doesn't help here at the very tip because of the singularity. Also, the theorem doesn't depend on symmetry, it's just that we often can make good use of symmetries present. There is rotational symmetry (phi) here...

3) Still, is it clear now that $dA$ is not $d( l r \phi)$ ?

 

[Hercuflea]

3) Still, is it clear now that $dA$ is not $d( l r \phi)$ ?

See attached response (my attempt at using LATEX was an utter epic failure.)

 

[BvU]

Yes. To the question "because it's so small". No. To the suggestion that something was removed.

Derivatives have to do with limits. Physicists have casual ways of dealing with those, but mathematicians are generally backing them up, especially when there is no pathological behaviour (discontinuities or discontinuous derivatives).
You know that e.g. $\frac{d}{dl}(l^2)=2l$ is a limit: $\frac{d}{dl}(l^2)\equiv\displaystyle\lim_{dl\downarrow 0}\frac{(l+dl)^2-l^2}{dl}$. Here you don't remove anything, you just take a limit. You did that when you learned about differentiating.

The casual way is to treat the buggers as quotients (Δ instead of $d$, if you want to be explicit) and not to worry too much about taking limits until it matters.

We write $d(l^2)=2l\enspace dl$ without thinking, in the certainty that a more formal treatment makes no difference.

--

It's weekend, so I'll also expand on

(my attempt at using LATEX was an utter epic failure.)

I learned $\TeX$ thirty years ago and still benefit from the investment. Knuth deserves something equivalent to the Nobel prize for TeX in itself (an equivalent, because supporting science and even making science possible for some reason isn't counted as science). Either that or because mathematics doesn't have a Nobel prize (but how about physics? they benefit!).
Life was tougher in those days. Nowadays with cutting and pasting you get a leg-up that makes it three to four times easier. The future will be even better. Failure is no reason to not try again (and fail better...).
Depending on where you are heading, the investment is well worth it. MS equation editor IS TeX under the hood (and it sucks).
Learning things under pressure is no good. So if you have to hand in your HW tomorrow, don't bother. But recreationally learning TeX is fun, I assure you. Sniff How to Type Mathematical Equations or some TeX turorial and discover a world of beauty!

 

[BvU]

Yes. So I understand that A= \pirl = \pil^{2}sin(\theta)

Just to get you started: once you are in TeX, you don't need the itex thingy any more: just removing those gets you
A= \pi rl = \pi l^2sin(\theta)
Which doesn't look like a failure at all. It gets perfect when you use \sin $(\sin)$ instead of sin $(sin)$ Not only not italic, but also because the spacing gets adapted: $A= \pi rl = \pi l^2\sin(\theta)$ (very subtly in this case).

The 'quick symbols' in advanced mode are smart enough that they can be used within and without TeX (probably a font thing).

And I must say that this quoting button in PF that gets you to see the source is a great help!
(The source above comes from the notification mail; it shows what you had before you changed it to a picture and an expression of despair. Not good PF practice to quote from it anyway, but: just trying to help...)

Nobody is perfect: I can't get $[$tex$]$ to appear with normal backets in the TeXt...

 

[Subhash]

Choose a differential ring which is at a slant distance 'l' from the vertex and of thickness 'dl'.
Now try and find out the force exerted by the point charge on the ring, not the other way round (this is simpler to calculate).
Proceed along these lines and integrate over the cone. The final integration will turn out simple.