# Electric field of bent non conducting rod

## 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?

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∫dEx=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) Im 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!

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kuruman
Homework Helper
Gold Member
Consider subdividing the 120o 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 $\alpha$?

• Consider subdividing the 120o 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!

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?

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