How Does Linear Charge Density Affect Electric Field Calculation?

[kahlua13]

Homework Statement

This is a wire whose shape is given by y = acos(x/L). This wire has a linear charge density of +λ, and is it desired to determine the electric field at the point (0,y) where y > a.

a) If a=0, determine the amount of charge the wire has.

b)If a > 0, is the total charge on this wire greater than, less than, or equal to the charge when a = 0. Why?

c) If a > 0, is there any statement you can make about the electric field at the POI? Why or why not?

d) a > 0. Locate a point on the wire. What are its coordinates?

e) If I want to move along the wire a small distance, ds,away from this point. Give the x and y-components of ds.

f) Construct, but do not evaluate, the expression for the electric field at the POI. Give enough detail to explain your reasoning.

Homework Equations

linear charge density = charge/length

The Attempt at a Solution

I am able to answer a-d.
For part A, length * linear charge density = charge
For part B, because the wire would be longer if a > 0, the total charge would be greater than if a = 0.
For part C, I believe the answer is that the x-components of the E-field will cancel, allowing us to use only the y-component to calculate the electric field.

However I really don't know where to start with part E! My chosen point is (0,a)

 

[TSny]

Hello. Welcome to PF.

Your answers to (a), (b), and (c) look good.

I suspect for part (d) that you are meant to choose an arbitrary point on the wire rather than the special point (0, a). How would you express the coordinates of the arbitrary point? Keeping in mind your objective in part (f), try to express the coordinates of the point in a way that will be most helpful when you get to (f).

I think that for (e) you are supposed to consider $ds$ as a little displacement that has x and y components. The length $ds$ can then be constructed from these components. You should express the x and y components of $ds$ in a way that will be helpful for part (f).