Force on a charge at the tip of a hollow, charged cone

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Homework Statement:

1.3 Force from a cone ** (a) 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
σ. The slant height of the cone is L, and the half-angle at the
vertex is θ. What can you say about the force on the charge q
due to the cone?
(b) If the top half of the cone is removed and thrown away , what is the force on the charge q due to the remaining part of the cone? For what angle θ is this force maximum?

Relevant Equations:

$$\vec{F}=\frac{1}{4\pi \varepsilon _0}\sum_{j=1}^{N}\frac{q_0q_j\hat{r}_{0j}}{r^{2}_{0j}}=\frac{1}{4\pi \varepsilon _0}\sum_{j=1}^{N}\frac{q_0q_j\vec{r}_{0j}}{r^{2}_{0j}}\frac{1}{\left \|\vec{r}_{0j} \right \|}$$
This is the diagram I drew for my calculations:
IMG_7884.jpg

I wanted to see if my work for part (a) makes sense.

If there is a variable ##l## that runs along the slant of total length ##L##, a ring around the cone can have an infinitesimal thickness ##dl##.
By Coulomb's law,
$$\vec{F}=\frac{1}{4\pi \varepsilon _0}\sum_{j=1}^{N}\frac{q_0q_j\hat{r}_{0j}}{r^{2}_{0j}}=\frac{1}{4\pi \varepsilon _0}\sum_{j=1}^{N}\frac{q_0q_j\vec{r}_{0j}}{r^{2}_{0j}}\frac{1}{\left \|\vec{r}_{0j} \right \|}$$
The symmetry of the cone can be used to show that the horizontal components of forces along the ##\hat{x}## direction cancel by considering two charges on the ring ##dQ## on opposite sides of one another. The sum of the forces of these charges on the charge q at the tip of the cone is
$$\vec{F}=\frac{1}{4\pi \varepsilon _0}(\frac{dQq\left \langle lsin\theta ,lcos\theta \right \rangle}{((lsin\theta )^2+(lcos\theta )^2)^{\frac{3}{2}}}+\frac{dQq\left \langle -lsin\theta ,lcos\theta \right \rangle}{((lsin\theta )^2+(lcos\theta )^2)^{\frac{3}{2}}}$$
$$\Rightarrow \vec{F}=\frac{2dQqlcos\theta }{4\pi \varepsilon _0l^{3}}\hat{j}$$
For the total force on q caused by the ring, I wrote
$$\vec{F}_{ring}=\frac{1}{4\pi \varepsilon _0}\sum_{i=1}^{N}\frac{dQ_iqlcos\theta }{4\pi \varepsilon _0l^3}\hat{j}$$
where N is the number of charges dQ on the ring.
Since there is a surface charge density σ, I wrote that the surface area can be written as
$$2\pi (lsin\theta )dl\Rightarrow q_{ring}=\sigma 2\pi (lsin\theta )dl$$
The total force is the sum of the forces of all the rings of infinitesimal thickness ##dl## that make up the cone:
$$F_{cone}=\int_{0}^{L}\frac{q\sigma 2\pi (lsin\theta )lcos\theta }{4\pi \varepsilon _0l^3}\hat{j}dl$$

I assume part (b) could be done by just integrating from 0 to L/2, but I wanted to make sure this part made sense.
 

Answers and Replies

  • #2
Just providing my thoughts, I am not sure if I am correct. You could probably take the result of standard solution of "Electric field on the axis of a ring of charge" and then integrating for the cone which will make things easy.
 
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haruspex
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As @Physicslearner500039 implies, it would be more usual to integrate around the ring. But in effect, that is what you did.
You do have a typo in one equation, where you duplicated the ##4\pi\epsilon_0## denominator.
Result looks right, but you could cancel a little.
 
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  • #4
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As @Physicslearner500039 implies, it would be more usual to integrate around the ring. But in effect, that is what you did.
You do have a typo in one equation, where you duplicated the ##4\pi\epsilon_0## denominator.
Result looks right, but you could cancel a little.
I'm almost afraid to just edit that typo because on my computer the TeX acts weird and won't adjust correctly after I try to edit. I fixed the duplication of that term on my paper though.
For part (b), I wrote that the force on charge q would be
$$F_{cone}=\int_{\frac{L}{2}}^{L}\frac{q\sigma 2\pi (lsin\theta )lcos\theta }{4\pi \varepsilon _0l^3}\hat{j}dl$$
And then I have to find the maximum of the term ##sin\theta cos\theta##. My trig identities suck, but what I did was:

$$\frac{\mathrm{d} }{\mathrm{d} \theta }(sin\theta cos\theta )=cos^2\theta -sin^2\theta =cos(2\theta )$$
$$cos(2\theta )=0\Rightarrow \frac{cos^{-1}(0)}{2}=\theta =45^{\circ}$$
$$\frac{\mathrm{d} ^2}{\mathrm{d} \theta ^2}(sin\theta cos\theta )=\frac{\mathrm{d} }{\mathrm{d} \theta }(cos(2\theta ))=-2sin(2\theta )$$
##-2sin(2\theta )=-2sin(90^{\circ})=-2\Rightarrow \theta =45^{\circ}## is a maximum for sin(θ)cos(θ).
 
  • #5
haruspex
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Aren’t supposed to perform the integrals? After the cancellation you can do, they're very easy.

Wrt maximising wrt angle, it is not entirely clear whether the charge density is to be taken as constant or the cone's total charge, but you have probably guessed right.
The answer makes sense in that the force will obviously be zero at 0 and 90.

Wrt the trig identities, it would have been slightly easier with the first step ##\sin(\theta)\cos(\theta)=\frac 12\sin(2\theta)##.
 
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  • #6
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Aren’t supposed to perform the integrals? After the cancellation you can do, they're very easy.

Wrt maximising wrt angle, it is not entirely clear whether the charge density is to be taken as constant or the cone's total charge, but you have probably guessed right.
The answer makes sense in that the force will obviously be zero at 0 and 90.

Wrt the trig identities, it would have been slightly easier with the first step ##\sin(\theta)\cos(\theta)=\frac 12\sin(2\theta)##.
I probably should evaluate it. It’s good practice. I’ll do it when I get back home.
After I solved it I saw that it the solutions used the trig identity you just gave. I really need to work on my trig tbh, I bought a book called Advanced Trigonometry by Durell and Robson, and I will work through it at some point. Another place I need to improve is my vector calculus. I’m a bit more than halfway through MIT OpenCourseWare course on multivariable calculus. It’s excellent.

Off topic, but I was registered for a course in applied linear algebra, I had to drop it because I couldn’t handle that along with everything else, but I was working through a proof based linear algebra book along with the “applied” book for the class, which I found eye opening, especially since the “non math major” definition of a vector that they give us is just a magnitude and a direction, while in reality there’s more to it than just that.

Thanks for the help.
 

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