# A line charge with variable charge density

1. Jul 3, 2012

### nilesthebrave

1. The problem statement, all variables and given/known data
A line charge in the x directio has a variable charge density given by the equation λ=4 λnot((1-x)/2L), where λnot is a constant. The rod has a length of L.

i)What is the net charge of the rod? Hint-the net charge is calculated by integrating the charge density with respect to x.
ii)Show that the electric field at the origin is given as E=-((2kQ)/L^2) (1-ln2)i

2. Relevant equations

Q=∫λrdx

Ex=(kλdx)/r^2

3. The attempt at a solution

Well I couldn't find much about something like this, so I kind of winged it and have no idea if what I did was even partially right but here we go.

i)Q=∫λrdx
=4λnot∫(1-(x/2L))/L^2 dx
=4λ (x-4L)/4L^2

Q=(λx-4L)/L^2

Again I'm not sure I'm doing that right, but it's what made sense to me.

ii)dEx=((kλdx)/r^2) (x/r)

Ex=kλx∫dx/r^3

Ex=kλx[-1/2r^2]

And now I have no idea what I'm doing...

#### Attached Files:

• ###### variable line charge.png
File size:
2.6 KB
Views:
80
Last edited: Jul 3, 2012
2. Jul 3, 2012

### Yukoel

Hello nilesthebrave
Think again .L is a constant all by itself and your integrated product seems to include the variable.By your picture the rod extends from x=L to x=2L right? Does that give you an idea of the limits of integration? I think Q=∫λ(r)dx is what you meant in your solution.

Think of an element small enough (dx) on the line rod at a distance x from the origin .It acts as a point charge (approximately).What is the charge on this length of element ?(assuming λ remains the same as for the distance x along this element)What is the elemental magnitude(dE) of the field due to such a charge at given distance? Then sum up those contributions by integration.You should be getting an expression for total electric field.The sign on the final expression might give you an idea on the nature of charge?

Does this help?
regards
Yukoel

3. Jul 3, 2012

### nilesthebrave

Ok, let's try this out:

Q=∫(from x=L to x=2L) (4λnot(1-(x/2L))rdx
=4λnot∫(1-(x/2L)rdx
=4λnot[x-1/2x^2]r^2/2)from L to 2L
=4λnot[2L-L^2)-(L-L/2)]
=(3L/2-L^2)4λnot

Is that closer to being correct?

Also, I'm still not quite sure what to do with part ii. Is how I'm starting off at least right?

4. Jul 4, 2012

### Saitama

It would be much better if you use LaTeX to write the equations. Its difficult to follow what you have written. In the question, you have mentioned that $$λ=4λ_o\frac{1-x}{2L}$$ but in the attempt you use:
$$λ=4λ_o(1-\frac{x}{2L})$$
Which one of them is the expression for λ? Also, what is the meaning of the bold "r" you have in your attempt.

As Yukoel said, think of a very very tiny part dx and find its charge. Integrate the expression from L to 2L to obtain the total charge.

EDIT: Seems like that you have to use the latter expression for λ. I calculated the electric field using the second expression and found the electric field as given in the question.

Last edited: Jul 4, 2012
5. Jul 4, 2012

### Yukoel

Hello nilesthebrave,
Why the bold r here?Are you trying to treat charge as a vector?Your integration involves x as a variable so why would you integrate r instead of it ?(ignoring the fact that vectors are not integrated directly in case r was a vector)You need to recheck your integral again.It should be simply ∫λ(x)dx =Q (limits as said earlier)
And as pranav-arora quoted the correct expression for charge is
$$λ=4λ_o(1-\frac{x}{2L})$$ (satisfies dimensions)
For the second part first of all calculate the charge on an elemental length dx using charge density.Treat it as a point charge and find the magnitude of the field created on the origin.In your attempt you have directly shifted λ out of your equation though it is a function of x.The basic formula you use i.e.
dEx=((kλdx)/x^2)
sounds fine to me.(By the looks of your method I think that you have assumed some oblique orientation of your rod in space though your uploaded picture depicts it lying on x axis. )You need to recheck your integral here too.
Hoping this helps.
regards
Yukoel

6. Jul 4, 2012

### nilesthebrave

Alright I think I FINALLY got all it all. Thank you very much for all of your help. I have a habit of overthinking problems and spending far longer on them than I should. So thanks for giving my brain an enema.

Also, how do I use this LaTex, I think it'd be far more convenient for everyone if I started using it :tongue:

Is it just clicking the Sigma up top?
$m_{n}\ =\ 1.67492716(13)\ \times\ 10^{-27}\ kg$

7. Jul 4, 2012

8. Jul 4, 2012

### nilesthebrave

Awesome, thank you very much. It's bookmarked for when I inevitably become stuck on another Physics 2 problem :p