How Do You Calculate the Electric Field from a Charged Cone?

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Homework Help Overview

The discussion revolves around calculating the electric field at a distance 'a' from the end of a solid uniform charge volume shaped like a cone. Participants are exploring the integration of charge distributions and the resulting electric field, with specific focus on the geometry of the cone and the appropriate mathematical setup.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need for a clearer problem description and the importance of understanding the charge distribution's shape. There is mention of using cylindrical coordinates or treating the problem as a stack of disks. Questions arise about deriving expressions for the electric field from a uniformly charged disk and how to set up the integral correctly.

Discussion Status

Some participants have provided guidance on the need for detailed descriptions of variables and integrals. There is acknowledgment of the complexity of the electric field from a charged disk and the necessity to consider the geometry of the cone. Multiple interpretations of the problem are being explored, but no consensus has been reached.

Contextual Notes

Participants note that the problem involves a three-dimensional setup and that the height of the cone and the distance from its end are crucial for the calculations. There are indications of missing information regarding the base width of the cone and the specifics of the charge distribution.

I-aM-Lost
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Homework Statement



Electric field at a distance 'a' from the end of a solid uniform charge volume

dq = Pdv
dv = (pi)(y)(dy)
A = (pi)(y)
dE = f(dq)

Homework Equations



What is the starting integral

The Attempt at a Solution



limits from 0 to b

(k)(p)(pi)(y)(dy)
E = --------------------
(b+a-y)^2
 
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I-aM-Lost said:

Homework Statement



Electric field at a distance 'a' from the end of a solid uniform charge volume

dq = Pdv
dv = (pi)(y)(dy)
A = (pi)(y)
dE = f(dq)

Homework Equations



What is the starting integral

The Attempt at a Solution



limits from 0 to b

(k)(p)(pi)(y)(dy)
E = --------------------
(b+a-y)^2
You'll have to provide a better problem description than that if you want to receive any meaningful help. As it stands I can sort of understand that you want to integrate over a charge distribution and determine the resulting electric field at some point.

But I have no idea what the shape of the distribution is, so I can't check your setup in any way. Also, you didn't actually ask a question, merely presented some work.
 
gneill said:
You'll have to provide a better problem description than that if you want to receive any meaningful help. As it stands I can sort of understand that you want to integrate over a charge distribution and determine the resulting electric field at some point.

But I have no idea what the shape of the distribution is, so I can't check your setup in any way. Also, you didn't actually ask a question, merely presented some work.

Sorry I forgot to add all the question. Find the electric field at a distance 'a' from the end of a solid cone uniform charge volume. Derive the equation. I just am trying to find the integral i can derive the equation from there.

So the picture is a solid cone around the y axis. With the point 'a' a distance above the cone itself. Height of the cone is b
 
So this is a 3D problem? You'll have a triple integral to cover the volume if you want to do it all in one go. maybe choose cylindrical coordinates. Otherwise you might attack it as a stack of disks with a single integral, provided that you know the field from a uniformly charged disk...
 
gneill said:
So this is a 3D problem? You'll have a triple integral to cover the volume if you want to do it all in one go. maybe choose cylindrical coordinates. Otherwise you might attack it as a stack of disks with a single integral, provided that you know the field from a uniformly charged disk...

Correct it is a 3d problem. I was going to try to do the problem as a stack of disks.
 
So where are you at with regards the field from a single disk? Have you derived an expression for the electric field on the axis of a charged disk?
 
gneill said:
So where are you at with regards the field from a single disk? Have you derived an expression for the electric field on the axis of a charged disk?

This all i really have. I have really been challenged by this stuff lately

limits from 0 to b

(k)(p)(pi)(y)(dy)
E = --------------------
(b+a-y)^2
 
It would be better to write out your equations on a single line so that space preservation is not a concern. That or place your text inside code tags " [ code ] ... [ /code ] " (without the spaces inside the square brackets). Best of all learn to use Latex syntax and write your equations using it. For example:

## f(x) = \int_0^\pi \left( 3x^2 + 2x \right) dx ##

As for your problem, you need to describe what it is you're doing in detail. What the variables are, what the integral you're describing is meant to sum. The electric field on the axis of a uniformly charged disk is not as simple as that of a point charge! Take a look at the Hyperphysics website entry.
 
gneill said:
It would be better to write out your equations on a single line so that space preservation is not a concern. That or place your text inside code tags " [ code ] ... [ /code ] " (without the spaces inside the square brackets). Best of all learn to use Latex syntax and write your equations using it. For example:

## f(x) = \int_0^\pi \left( 3x^2 + 2x \right) dx ##

As for your problem, you need to describe what it is you're doing in detail. What the variables are, what the integral you're describing is meant to sum. The electric field on the axis of a uniformly charged disk is not as simple as that of a point charge! Take a look at the Hyperphysics website entry.
## f(x) = \int_0^b\left( (k)(p)(pi-y)(dy) \right ) ## all over (b+a-y)^2

I am trying to derive the electric field for a charge volume of a solid cone.

k = constant
pi-y = From Area of pix^2
p - rho
dy - because the cone is around the y axis
b - total length of cone
a - distance above cone
 
  • #10
Sorry, I'm just not picking out the cone geometry or the field from a disc in that .

Did you take a look at he Hyperphysics link that I placed in my last post? You'll see that the field from a single disk is not simple, depending upon both distance and radius of the disk. You're wanting to sum a stack of these (so over their z variable, but you can change it to y for your setup. What's important is that you get the radius to vary with the height so that the stack forms a cone of the proper dimensions. You haven't mentioned the width of the base of your cone).
 

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