- #1

cookiemnstr510510

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## 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/R

^{2}

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)/2piR

^{2}

(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/R

^{2}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!