Uniform Distributed Electric Charge

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
11 replies · 10K views
stylez03
Messages
139
Reaction score
0

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.
 
Physics news on Phys.org
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:
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.
 
yea haha, okay thanks. I see what you mean.
 
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.