Nonuniformly Charged Semicircle

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Homework Help Overview

The problem involves a thin glass rod shaped as a semicircle with a nonuniform charge distribution along its length. The linear charge density is defined as \(\lambda = \lambda_{0} \cdot \sin(\theta)\), where \(\lambda_{0}\) is a constant. The objective is to determine the acceleration of an electron placed at the center of the semicircle.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Newton's 2nd Law and Coulomb's Law to find the electric force acting on the electron. There is an exploration of how to quantify the forces from the charge distribution, with attempts to clarify the correct expression for the linear charge density.

Discussion Status

Participants are actively engaging in clarifying the charge density expression and discussing the method to calculate the electric field components. Some guidance has been provided regarding the use of differential elements to analyze the semicircle's charge distribution.

Contextual Notes

There is a mention of potential confusion regarding the charge density formula and the need to consider components in polar coordinates. Participants are also navigating the implications of the charge distribution on the resulting electric field.

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Homework Statement


A thin glass rod is a semicircle of radius R, see the figure. A charge is nonuniformly distributed along the rod with a linear charge density given by \lambda=\lambda_{0} + sin(\theta), where \lambda_{0} is a positive constant. Point P is at the center of the semicircle.

GIANCOLI.ch21.p50.jpg


Determine the acceleration (magnitude and direction) of an electron placed at point P, assuming R = 1.9 and \lambda_{0} = 2.0 \mu C/m.

Homework Equations


Coulomb's Law F_{e}=kq_{1}q_{2}/r^{2}
Electric Field Equation E=F_{e}/q
Newton's 2nd Law

The Attempt at a Solution


I attempted to use Newton's 2nd Law to solve this problem. I know the mass of an electron, and since gravity can be ignored, I knew that all I needed to do was to find the electric force on the electron to find the acceleration. I know the the force points straight up do to the positive charge above and the negative charge below, but I'm having trouble quanitfying the forces. I attempted to use Coulomb's Law by plugging in the charges for just one electron and proton, but I know that's not right. Please help!
(Also, let me know if the image disappears, it should work out fine though...)
 
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Welcome to PF!

Hi saxyliz ! Welcome to PF! :smile:

(have a theta: θ and a lambda: λ and a mu: µ :wink:)
saxyliz said:
A thin glass rod is a semicircle of radius R, see the figure. A charge is nonuniformly distributed along the rod with a linear charge density given by \lambda=\lambda_{0} + sin(\theta), where \lambda_{0} is a positive constant. Point P is at the center of the semicircle.

Determine the acceleration (magnitude and direction) of an electron placed at point P, assuming R = 1.9 and \lambda_{0} = 2.0 \mu C/m.

Are you sure it isn't λ = λ0 times sinθ (that seems more consistent with the image)? :confused:
 


tiny-tim said:
Hi saxyliz ! Welcome to PF! :smile:

(have a theta: θ and a lambda: λ and a mu: µ :wink:)


Are you sure it isn't λ = λ0 times sinθ (that seems more consistent with the image)? :confused:


Oh yes sorry! It's times! Silly me!
 
saxyliz said:
Oh yes sorry! It's times! Silly me!

:rolleyes: :smile:
saxyliz said:
I attempted to use Newton's 2nd Law to solve this problem. I know the mass of an electron, and since gravity can be ignored, I knew that all I needed to do was to find the electric force on the electron to find the acceleration. I know the the force points straight up do to the positive charge above and the negative charge below, but I'm having trouble quanitfying the forces. I attempted to use Coulomb's Law by plugging in the charges for just one electron and proton, but I know that's not right.

ok, then … yes, the force will point up, so you only need the up-components.

But you must slice the semicircle into bits of length dθ, and plug in the charge for that. :smile:
 
So, dq = \lambda d \theta? Then I plug that into Coulomb's Law to solve for dE, but how do I solve for the components?

To be more specific, what I get is:

dE=k\lambda d \theta /R^{2}

Which is the same as

dE=k\lambda_{0} sin(\theta) d \theta /R^{2}

Do I use a sector for the components? Like polar coordinates?
 
Last edited:
Hi saxyliz! :smile:

(what happened to that θ and λ i gave you? :confused:)
saxyliz said:
So, dq = \lambda d \theta? Then I plug that into Coulomb's Law to solve for dE, but how do I solve for the components?

No, dq = λ sin θ dθ.

For components, you use cos of the angle as usual … what is worrying you about that? :confused:
 
Sorry! I'm totally new to this and didn't get what you meant in the first place (about the θ, etc.)

Why isn't it λ_0 ? Wouldn't it be sin^2 then?

I think I get what you mean about the components, give me a second.

EDIT:
I got it! Thanks so much for your help!
 
Last edited:

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