Electric field of bent non conducting rod

[cookiemnstr510510]

Homework Statement

You are given a non conducting rod carrying uniformly distributed charge, -Q, that has been bent into a 120° circular arc of radius, R. The axis of symmetry of the arc lies along the x-axis and the origin is at the center of curvature of the arc.
(a) in terms of Q and R, what is the linear charge density, λ?
(b) in terms of Q and R, what is the magnitude and direction the the resulting electric field at the origin?
(c)If the arc is replaced by a point particle carrying charge, -Q, at x=R, by what factor is the resulting electric field at the origin multiplied?

Homework Equations

λ=Q/L
E=KQ/R²
dq=λds
ds=Rdθ

The Attempt at a Solution

(a) λ=Q/L, λ=Q/(⅓2piR) = 3Q/2piR

(b) Cant seem to figure out how to write an integral with bounds. The bounds on my integral below will be from θ=0° → θ=60°
E=2∫dE_x=2∫kλdθcosθ/R = 2kλ/R ∫cosθdθ (Like previously stated this integral is being evaluated from θ=0° to θ=60°)
E=(2kλ/R)[sin60°-sin0°] = (kλ√3)/R, it wants in terms of R and Q so now plug in our above statement for λ
E=(3kQ√3)/2piR²

(c) I am a little unsure about what this is asking, from my understanding:
to obtain the new electric field from a point particle which is simply kQ/R² we must multiply the previous electric field from bent rod by a factor of 2pi/3√3. Not sure if this is correct or if I'm thinking about the question in the wrong way.
Any help would be much appreciated
Thanks!

 

[kuruman]

Consider subdividing the 120^o arc into infinitesimal elements $ds$.
1. What is the charge on one of these elements?
2. What is the electric field contribution $dE$ from this charge element to the point of interest?
3. Add all such contributions to find the net electric field. Remember that the electric field element $dE$ is a vector so you have to find its components and add them separately.

Part (c) is asking this: Say the field due to the arc is $E_0$, If you replace the arc by a single charge -Q at distance $R$, you will get another field $E_1$. If you write an equation relating the two as $E_1=\alpha E_0$, what is the value of constant $\alpha$?

 

[cookiemnstr510510]

Consider subdividing the 120^o arc into infinitesimal elements $ds$.
1. What is the charge on one of these elements?
2. What is the electric field contribution $dE$ from this charge element to the point of interest?
3. Add all such contributions to find the net electric field. Remember that the electric field element $dE$ is a vector so you have to find its components and add them separately.

Part (c) is asking this: Say the field due to the arc is $E_0$, If you replace the arc by a single charge -Q at distance $R$, you will get another field $E_1$. If you write an equation relating the two as $E_1=\alpha E_0$, what is the value of constant $\alpha$?

Okay, Ill work on this and post my results! Thanks!

 

[Cutter Ketch]

Answering is easy when people make mistakes. It’s much harder when everything seems perfect. I’m too lazy to really check every factor of 2 or pi, so I may be mistaken, but this looks really good to me. What has you worried?

 

[cookiemnstr510510]

Answering is easy when people make mistakes. It’s much harder when everything seems perfect. I’m too lazy to really check every factor of 2 or pi, so I may be mistaken, but this looks really good to me. What has you worried?

Yea just part c, wasn't sure what it was asking but @kuruman helped me out with that one. Just have a test in a few days and want to make sure I'm doing these problems correctly. Always love reassurance! LOL

 

[Cutter Ketch]

You saw the symmetry and ignored the y component. You also used symmetry to cut the integral in half. You integrated in theta and didn’t forget the extra R.

I don’t think you have much to worry about, but I’ll wish you good luck anyway.