Uniform Distributed Electric Charge

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Homework Help Overview

The discussion revolves around a problem involving a ring-shaped conductor with a uniform positive charge and the calculation of the electric field and force at a specific point due to this charge distribution. The subject area includes electrostatics and electric fields.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formula for the electric field due to a charge distribution and question the correct application of charge values in calculations. There is an exploration of the geometry involved and the significance of the unit vector in the context of the problem.

Discussion Status

Participants are actively engaging with the problem, identifying potential errors in charge values and discussing the need for a diagram to clarify the geometry. Some guidance has been provided regarding the integral needed for the electric field calculation, and there is recognition of a previous misunderstanding related to charge units.

Contextual Notes

There is mention of a specific charge distribution and the need to clarify the coordinate system used for the calculations. The original poster has encountered issues with the online system's feedback regarding their calculations.

stylez03
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Homework Statement


A ring-shaped conductor with radius a = 2.20 cm has a total positive charge Q = 0.123 nC uniformly distributed around it.

(Question relevant to main problem)
(a)What is the magnitude of the electric field at point P, which is on the positive x-axis at x = 35.0 cm?

[**Actual Question**]
A particle with a charge of - 2.70 mC is placed at the point P described in part (a). What is the magnitude of the force exerted by the particle on the ring?

Homework Equations



F = 1/4*pi*e_o * |q| / r^2 for single point charge

e_o = 8.85 * 10^-12

The Attempt at a Solution



1/(4*pi*8.85 * 10^-12) * ( 0.123*10^-9 * 2.70*10^-9 / .35^2 )

The online system says I'm off by an additive constant,

I also tried

1/(4*pi*8.85 * 10^-12) * ( 0.123*10^-9 * 2.70*10^-9 / .35^2 + 0.0220^2)

Since that equation was also in the book, but still no luck.
 
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The electric fields at a point due a to a charge distribution is given by:

[tex]\mathbf{E} = \frac{1}{4\pi \epsilon_0} \int \frac{\sigma}{r^2} \mathbf{\hat{r}} dl[/tex]

[itex]\sigma[/itex] is the charge density of the ring of charge.
 
Kurdt said:
The electric fields at a point due a to a charge distribution is given by:

[tex]\mathbf{E} = \frac{1}{4\pi \epsilon_0} \int \frac{\sigma}{r^2} \mathbf{\hat{r}} dl[/tex]

[itex]\sigma[/itex] is the charge density of the ring of charge.

Given your equation:

e_o = 8.85 * 10^-12

rho = 0.0220m

r = .35m

I'm not sure what r_hat is in this situation?
 
The r-hat is the unit vector pointing in the direction of the field. You will have to perform the integral which will be different depending on what coordinate system you use.

Do you have a diagram? The geometry is important aswell. I assume the axis of rotation of the circle is the x-axis.
 
Kurdt said:
The r-hat is the unit vector pointing in the direction of the field. You will have to perform the integral which will be different depending on what coordinate system you use.

Do you have a diagram? The geometry is important aswell. I assume the axis of rotation of the circle is the x-axis.

Here is the diagram:

http://www.uploadyourimages.com/view/505148yf.figure.21.21.jpg
http://www.uploadyourimages.com/view/505148yf.figure.21.21.jpg
 
Last edited by a moderator:
Oh hang on problem diagnosed. You've put 2.7 nC in the equation instead of 2.7mC (2.7x10-3). I jumped the gun and started answering the first part.
 
Wait, so what am I suppose to do?
 
Your original method should work you just mistook the units of one of the charges. When I said I jumped the gun I wasn't kidding I had you on the path of deriving the equation you already had. I just didn't recognise it at first because you'd put all the numbers in instead of symbols.
 
  • #10
yea haha, okay thanks. I see what you mean.
 
  • #11
I must be cursed with you. I keep sending you down the wrong path :eek: many apologies.
 
  • #12
it's okay. Thankfully homework isn't worth much in my overall grade, least this will help me prepare for the exam which is worth a lot more.
 

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