If I have a hollow cone (surface with no bottom cover ) as the one in the picture. The cone has surface charged density $\sigma$. It rotates around the symmetry axis with an angular velocity $\omega$. I want to find the distribution of current on the surface of the cone.


What I mean with distribution of the current is the following. I can write the current as :

$$I=\frac{dq}{dt}=\frac{dq}{dl}\frac{dl}{dt} \tag{1}$$$

That is useful in other situations like when I have a wire with linear charge density $\lambda$Wire

I can use the eq(1) for finding the current if ther is a current density $\lambda$. As I can write $q=\lambda l$, $\frac{dq}{dl}=\lambda$ and $I=\frac{dq}{dt}=\lambda v$.

The same follows for a charged sheet if width $b$ with current density $\sigma$. As I can write $q=\sigma a=\sigma bl$ , $\frac{dq}{dl}=\sigma b$ and $I=\frac{dq}{dt}=\sigma bv$.


But with a cone what can I do with the cone. I can write $q=\sigma a=\sigma \frac{2\pi r}{2l} $. But as the length $s_n$ of one circular wire of current varies from 0 to $2\pi r$.Different than before that the width was constant. What can I do ?

My attemp the length that varies $s=2\pi r_{change}$ , now $q=\sigma \frac{s}{2l} $. $\frac{dq}{dl}=\sigma \frac{1}{2l}$ and $I=\frac{dq}{dt}=\sigma \frac{v}{2l}$.

It's clear that $v=wr$ since it is circular motion.

enter image description here


Do not open the cone. Think of it in the profile view : You have an isosceles triangle. Now move along the axis of the cone, say a distance $x$ and take an element $\mathrm{d}x$. Somewhat like this :

enter image description here

This small element is similar to the rectangle you described. With length as $2\pi r(x)$ and width $\mathrm{d}x$. You also know the velocity with which the charge moves :

$$v = \omega r(x)$$

$r(x)$ is the radius at that point and omega the angular velocity.

Thus the current would be

$$I = \sigma \omega r(x)\mathrm{d}x$$

for that rectangle.

$r(x)$ can be found by triangle similarity:

$$r(x) = R / H \cdot x$$ where $R$ is the base radius and $H$ the height of the cone

So the current varies with position on the cone and to find the overall current, integrate from $x =0$ to $x= H$.


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