Potential difference/cone

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In summary, the conversation discusses finding the potential difference between two points on a conical surface with a uniform surface charge. Two methods are suggested, one involving integrating over the cone's surface and the other building up from rings with charge density. It is advised to consider the charge and distance from the points when setting up the integrals.
  • #1
bigplanet401
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Hi,

A conical surface (an empty ice-cream cone) carries a uniform surface charge [tex] \sigma[/tex]. The height of the cone is h, and the radius of the top is R. Find the potential difference between points a (the vertex) and b (the center of the top.)

I've tried integrating over the conical surface (zenith [tex]\phi[/tex] fixed):

[tex]
V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \int d\mathbf{a} \frac{\sigma(\mathbf{r}^\prime)}{|\mathbf{r} - \mathbf{r}^\prime|} \quad \rightarrow \quad
\frac{1}{4\pi\epsilon_0} \int r^{\prime 2} dr^\prime \, d\theta^\prime
\frac{\sigma}{\sqrt{1 - r^{\prime 2} \cos^2 \phi}} \, ,
[/tex]

but I think that's wrong. Next I tried building up from a series of rings with charge density [tex]\lambda[/tex]:

[tex]
V_{\text{ring}} = \frac{\lambda}{2 \epsilon_0} \frac{R}{\sqrt{R^2 + z^2}} \, ;
[/tex]

unfortunately, I don't know how to set up the integration for this. Any help is appreciated,hopefully sooner than later--my written qualifier is ~3 weeks away!
 
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  • #2
There shouldn't be any angular dependence in the integrand if you choose to use the first method, so the integral as you have it set up now is wrong. For each area element rdrd[itex]\phi[/itex] in the xy plane, think about the corresponding portion of the surface of the cone. Find the amount of charge in this surface element (this will involve the slope of the cone), and the distance from it to the point you want to calculate the potential. Then integrate over the disk in the xy plane corresponding to the cone.

If you want to use the second method, you'll need to think of rings on the cone corresponding to some dr or dz and then integrate over the height or radius of the cone. For example, if you choose to use dz, find the charge between z and z+dz as a function of z, then find the potential due to this ring, which will be proportional to dz, and then integrate over z.
 
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  • #3


Hello,

Thank you for reaching out for help with this problem. I understand the importance of finding accurate solutions and I am happy to assist you with this question.

Firstly, your approach of integrating over the conical surface is correct. However, the integration limits you have chosen are not appropriate. Instead of integrating from 0 to h, you should integrate from 0 to R, since the potential difference between points a and b is only affected by the charge on the surface of the cone, not inside the cone.

Also, the expression for the potential you have written is for a point charge, not a surface charge. For a surface charge, the potential is given by:

V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0} \int d\mathbf{a} \frac{\sigma(\mathbf{r}^\prime)}{|\mathbf{r} - \mathbf{r}^\prime|} \quad \rightarrow \quad
\frac{1}{4\pi\epsilon_0} \int d\phi^\prime \int_0^R r^{\prime} dr^\prime
\frac{\sigma}{\sqrt{r^{\prime 2} + z^2}} \, ,

where z is the distance from the cone's vertex to the point of interest. This expression takes into account the fact that the potential at a point is affected by all the charge elements on the cone's surface.

As for your second approach, using rings, it can also work. You just need to set up the integration properly. The potential at point b due to a ring of charge at a distance z from the vertex is given by:

V_{\text{ring}} = \frac{\lambda}{4\pi\epsilon_0} \frac{R}{\sqrt{R^2 + z^2}} \, ,

where \lambda is the linear charge density of the ring. Then you can integrate over all the rings from 0 to R to get the total potential at point b.

I hope this helps and I wish you all the best for your qualifier in 3 weeks. Keep up the good work!
 

1. What is potential difference?

Potential difference, also known as voltage, is the difference in electric potential energy between two points in an electric field. It is measured in volts (V) and is a key concept in understanding electricity and circuits.

2. How is potential difference related to current?

Potential difference is directly related to current, as stated by Ohm's Law. The higher the potential difference, the greater the current. This is because a larger potential difference creates a stronger electric field, which pushes electric charges through a circuit at a faster rate.

3. What is the difference between potential difference and electric potential?

Potential difference refers to the difference in electric potential energy between two points, while electric potential refers to the amount of potential energy per unit charge at a specific point. In other words, potential difference is a measure of the difference in energy, while electric potential is a measure of the energy itself.

4. How does potential difference affect the flow of electrons?

Potential difference is what drives the flow of electrons in a circuit. When there is a difference in potential between two points, electrons will flow from the higher potential point to the lower potential point. This flow of electrons is what creates an electric current.

5. What is the role of potential difference in a circuit?

Potential difference plays a crucial role in a circuit by providing the energy needed to move electrons through the circuit. It is responsible for creating the electric field that pushes electrons through the circuit and allows devices to receive the necessary energy to function.

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