# Electric field of bent non conducting rod

In summary, the electric field at the origin due to a point particle with charge -Q is multiplied by a factor of 2pi/3 when the arc is replaced by a single charge.

## 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) 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!

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##?

kuruman said:
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?

Cutter Ketch said:
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

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.

## What is an electric field?

An electric field is a physical quantity that describes the force exerted on a charged particle by other charged particles in its vicinity. It is represented by a vector, with direction pointing towards the direction of the force and magnitude determined by the strength of the force.

## What is a non-conducting rod?

A non-conducting rod is a material that does not allow the flow of electric current through it. This means that the electrons in the material are tightly bound and cannot move freely, unlike in conducting materials.

## How is the electric field of a bent non-conducting rod calculated?

The electric field of a bent non-conducting rod can be calculated by using Coulomb's Law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. This can be applied to each charged particle along the bent rod to determine the overall electric field at any point.

## What factors affect the electric field of a bent non-conducting rod?

The electric field of a bent non-conducting rod is affected by the magnitude and distribution of the charges on the rod, the distance from the rod, and the angle of the bend. The closer the charges are to each other, the stronger the electric field will be. Additionally, the electric field will be stronger at points closer to the rod and weaker at points further away.

## How does the electric field of a bent non-conducting rod differ from that of a straight rod?

The electric field of a bent non-conducting rod differs from that of a straight rod because the distribution of charges is not uniform along the bent rod. This means that the electric field will not be constant along the length of the rod and will vary depending on the position of the charged particles. In a straight rod, the electric field is constant at all points along its length.