# Homework Help: Electromagnetism: Finding Electric Field of non Uniform Charged Rod

1. Jun 26, 2011

### sky.flower123

1. The problem statement, all variables and given/known data

Here is the question:

Thin rod AB has length l=100 cm and total charge q=37 nC that is distributed in such a way that its line density ? is proportional to the square of the distance from the end A, i.e. ?(x) =kx^2. Determine electric field E at the end A of the rod.

? stands for Lambda, sorry I do not understand TeX.

I know that I need to integrate SOMETHING. How I do that I am not sure. Here are a couple of questions:

2. Relevant equations

- Can I assume that at 1m: (37nC)/1m=k(1^2) to find k? (or is k just 8.97E9?)
- Do I use the formula E=lambda/((2)(pi)(epsilson-nought)(x)) because this is a cylinder?

3. The attempt at a solution

This is what I have so far:

solve for k at 1m and see that it equals 37 nC (should I just convert that to C?)

input lambda into the equation and do an integral so you get: k(x^2)/((2)(pi)(epsilon-nought)(x)) from A (0) to B (1 m). From this I got that the integral is k(x^2)/((4)(pi)(epsilon-nought)). This is where would end the problem and put in my values.

I really just want to understand the problem, obviously there is something I am not getting.

PS I know there is this exact question (I did my research on this problem before I came here). I didn't understand the explanation there so please help!

2. Jun 27, 2011

### sky.flower123

I did the math and I got that the electric field equals 9.99E2 (this is converting nC to C)

what would my units be? C/m?

3. Jun 27, 2011

### I like Serena

Welcome to PF, olivia.belli3!

(Have a λ )

Let's start here and try to find k.

Your line density λ defines how much charge is between x and x+dx measured from point A, where dx is a very small distance.
That is, the charge between x and x+dx is:
$$\lambda(x)dx = k x^2 dx$$

To find the total charge you need to calculate the integral of these charges over the length of the rod.

4. Jun 27, 2011

### sky.flower123

I set up my integral that q= the integral of kx^2 dx to get that q=(kx^3)/3
from A=0 to B=1.

this means that k=3Q or k=111 nC (Do I convert to C?)

5. Jun 27, 2011

### I like Serena

Right!

No need to convert from nC to C unless you really want to. ;)
However, your unit is not quite right, since you have that the charge q (coulomb) equals your k times x3 divided by 3. So what should the unit of k be?

Now for the electric field at point A.
I'm not sure where you got your formula, but you should use the formula from Coulomb's law (http://en.wikipedia.org/wiki/Coulomb%27s_law" [Broken]).

That is:
$$dE = \frac 1 {4 \pi \epsilon_0} \frac {dQ} {r^2}$$
where dE is the small contribution of a small charge $dQ=\lambda(x)dx$ at a distance r.

Can you apply this to your problem?
And integrate it over the length of the rod?

Last edited by a moderator: May 5, 2017
6. Jun 27, 2011

### sky.flower123

well in that case my K= 111 nC/m^3.

So I have Q, of course I can integrate over the length of the rod using dr from 0 to 1, but I'm stuck. Where does k come in? I'm a little lost . I'm assuming that you meant to put QQ on the top instead of dQ? In that case I would taket he charge from A (which is zero) then take the charge from B which would lead me to to the answer, but how do you do that?

7. Jun 27, 2011

### I like Serena

$$E_A = \int_0^{AB} dE = \int_0^{AB} \frac 1 {4 \pi \epsilon_0} \frac {\lambda(x)dx} {x^2}=\ ...$$

Last edited: Jun 27, 2011
8. Jun 27, 2011

### sky.flower123

Okay, so I integrated that and put in that lambda=kx^2 (since we know K I am assuming this is okay). I'm assuming that AB=1m?

I got that the integral equals k(x)/((4)(pi)(epsilon-nought)) from 0 to AB which I would think would equal K/((4)(pi)(epsilon-nought)).

9. Jun 27, 2011

### I like Serena

Good!

10. Jun 27, 2011

### sky.flower123

so would my final answer be:

(111nC/m^3)(1m)/((4)(pi)(epsilon-nought)) to yield the electric field as: 9.99E10^11?

11. Jun 27, 2011

### sky.flower123

sorry 9.98E10^11

12. Jun 27, 2011

### sky.flower123

units would be nC/m^2

13. Jun 27, 2011

### I like Serena

Yep!

(Now would be a good time to convert nC to C. )

Uhh, no.
Did you take the unit of $\epsilon_0$ into account?

Btw, do you know how the electric field is related to the force it would exert on a charged object?

Last edited: Jun 27, 2011
14. Jun 27, 2011

### sky.flower123

Eek sorry! forgot about epsilon nought. The fnal units is N/C!

Thank you SO much!
I really appreciate you!

15. Jun 27, 2011

Any time!