What Is the x Component of the Electric Field at a Point on the x-Axis?

In summary, we have a charged rod of length 1.00 m with a linear charge density of λ = ax, where a = −90.0 μC/m, centered on the x-axis. We need to find the x component of the electric field at a point on the x-axis located at a distance of 2.00 m from the end of the rod. Using Coulomb's law and integrating from -L/2 to L/2, we can calculate the electric field at this point to be 4.05 x 10^5 N/C.
  • #1
Broem
9
0

Homework Statement


A charged rod of length L = 1.00 m lies centered on the x-axis as shown. The rod has a linear charge density which varies according to λ = ax where a = −90.0 μC/m.
What is the x component of the electric field at a point on the x-axis a distance of D = 2.00 m from the end of the rod?

Homework Equations


E=kQ/r^2
charge density = Q/A

The Attempt at a Solution


I really do not understand...Where do I get the r from? Since its not uniformly distributed how can there be any field outside?
Would my r = (D-(half of L))?
I'm assuming lambda is just regular charge density at a location of x?
 

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  • #2
Hi Broem! :smile:
Broem said:
Would my r = (D-(half of L))?
I'm assuming lambda is just regular charge density at a location of x?

Yes. :smile:

(and of course it's a linear charge density, ie coulombs per metre, not per metre3 :wink:)
Since its not uniformly distributed how can there be any field outside?

Not following you. :confused:
 
  • #3
Thanks for the quick response!
Ok so here's where I am:

I now know that lambda = Q/L so Q = L * (lambda)

So I'm left with
9e9(-9e-6)/(2-.5)^2

This however is not the correct answer...Where can I go from here?
 
  • #4
oh, no, you'll have to integrate

use coulomb's law to find the electric field from a tiny section [x,x+dx] (whose charge will be λdx, and which you can assume is concentrated entirely at x), and integrate from -L/2 to L/2 :wink:
 
  • #5
I got it!
Thank you so much for your help!
 

What is an electric field at a point?

An electric field at a point is a measure of the force exerted on a positive test charge placed at that point. It is a vector quantity, meaning it has both magnitude and direction.

How is the electric field at a point calculated?

The electric field at a point is calculated by dividing the force exerted on the test charge by the magnitude of the test charge. It is also influenced by the distance from the source of the electric field and the permittivity of the medium.

What is the unit of measurement for electric field at a point?

The unit of measurement for electric field at a point is Newtons per Coulomb (N/C), or Volts per meter (V/m) in terms of potential difference.

How does the direction of the electric field at a point relate to the direction of the force on a test charge?

The direction of the electric field at a point is the direction in which a positive test charge would move if placed at that point. This direction is also the same as the direction of the force exerted on the test charge.

What are some real-life applications of understanding the electric field at a point?

Understanding the electric field at a point is crucial in fields such as electrical engineering, as it helps in designing circuits and predicting the behavior of electric charges. It is also important in the study of electromagnetism and the functioning of electronic devices.

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