 #1
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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 halfangle 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:
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.
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.