# Uniformly Charged Rod

1. Sep 2, 2014

### jls

1. Problem Statement
A uniformly charged rod of length L and total charge Q lies along the x axis as shown in in the figure below. (Use the following as necessary: Q, L, d, and ke.)

(a) Find the components of the electric field at the point P on the y axis a distance d from the origin.

(b) What are the approximate values of the field components when d >> L?

2. Equations
I have a diagram and understand that E=kQ/r^2, however, I can not figure out how to define each component.

3. Attempt
I know that I must integrate to solve once I have defined the component, however I do not know how to define them.
Would Ex=rsin(θ)[(kQ)/r^2] and Ey=rcos(θ)[(kQ)/r^2] be at all in the right direction?

Last edited: Sep 2, 2014
2. Sep 2, 2014

### BvU

Hello jls, and welcome to PF.
Can't find a picture... or is the figure below in your book only ? In that case:
Is the center of the rod at x=0 ? (would make things easier...)

3. Sep 2, 2014

### vela

Staff Emeritus
$E = \frac{kQ}{r^2}$ only applies for a point charge. There's probably an example done in your textbook that you might find very helpful.

4. Sep 3, 2014

5. Sep 4, 2014

### BvU

Yup. Now we set up the integral (which you already expected to be needed). We take a little chunk of rod from x to x+dx and write down the x and y components of $\vec E$ at point $\vec P = (0, y_P)$. Is one way.

Your Ex=rsin(θ)[(kQ)/r^2] and Ey=rcos(θ)[(kQ)/r^2] looks like an integration over $\theta$; is fine too.

Both cases you need to express the things that vary in terms of the integrand: r($\theta$), Q($\theta$) -- or rather the dQ from $\theta$ to $\theta + d\theta$. Or express them in x and dx and let x run from 0 to L.

6. Sep 4, 2014

### jls

How do you turn that chunk (x to x+dx) into the components? I think I could figure it out if I knew what that meant..

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