The electric field due to a line of charge?

In summary, In this conversation, the speaker is asking for help understanding how to find the electric field at a distance of r<3a along the x-axis due to a line charge with a charge density of \lambda in two separate regions. They suggest splitting the charge into infinitesimal elements and integrating, but are having trouble with the distances involved. The person helping them suggests using different variables for the position of the test charge and the distance between it and the infinitesimal piece of wire to clarify the equation.
  • #1
jeebs
325
4
Hi,
I am so stuck with this and I have the feeling that its a simple thing I'm just not realizing.

I have a line charge stretching along the x-axis from 0 to a, and from 2a to 3a. with a charge density of [tex]\lambda[/tex] in each region. I have to find what the electric field is at a distance of r<3a along the x axis. I have to show that

E(r) = [[tex]\lambda[/tex]a/4[tex]\pi[/tex][tex]\epsilon[/tex]].( 1/[r(r-a)] + 1/[(r-3a)(r-2a)] )

{sorry about the greek letters being too high, i don't mean 4 to the power pi.epsilon}.

Attempt at solution:
What I thought might be the way to think of is to split the charge up into infinitesimal elements and integrate along from 0 to a, adding up the field at the distance r due to each charge element, then doing the same from 2a to 3a.

But, i just cannot seem to get started. My problem seems to be with the distances (the r's and the a's).

I've started off by saying [tex]\lambda[/tex] = dQ/dr

and dE = dQ/(4[tex]\pi[/tex].[tex]\epsilon[/tex].r2) = [tex]\lambda[/tex]dr/(4[tex]\pi[/tex].[tex]\epsilon[/tex].r2)

But then i am just confused, this has been doing my head in all night now.

Can anyone help? very much appreciated.
thanks.

PS. if you think i haven't given enough detail or a good enough attempt at solution please say so if you aren't going to reply because of it. I only need a nudge in the right direction anyway I think. cheers.

DIAGRAM: the x's are to show where the charge is spread over, the full stops are empty space.

xxxxxxxxxx.....xxxxxxxxxx...
|---------|---------|---------|----->x
0...a....2a...3a
 
Last edited:
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  • #2
You've got the idea pretty correct so far. To make things a little clearer, trying calling the position of the test charge (the "r" in E(r), that is) "r", and the distance from this test charge to an infinitesimal piece of the wire "[itex]l[/itex]".

Now if you do your equation for dE, you should have a way to tell the difference between the position of the test charge and the position of the tiny piece of wire. The replacement should be [itex]dQ = \lambda~dl[/itex], while the r[itex]^2[/itex] in the denominator should be the square of the distance between the test charge (at r) and the infinitesimal piece of wire.
 

1. What is an electric field?

An electric field is a physical quantity that describes the influence a charged object has on other charged objects in its vicinity. It is a vector field, meaning it has both magnitude and direction.

2. How is the electric field due to a line of charge calculated?

The electric field due to a line of charge is calculated using the formula E = k*q/r, where k is the Coulomb's constant, q is the charge of the line, and r is the distance from the line to the point where the electric field is being measured.

3. Is the electric field due to a line of charge uniform?

No, the electric field due to a line of charge is not uniform. It decreases as the distance from the line increases, following an inverse-square relationship.

4. What is the direction of the electric field due to a line of charge?

The direction of the electric field due to a line of charge is always perpendicular to the line and points away from it for a positively charged line, and towards it for a negatively charged line.

5. Can the electric field due to a line of charge be negative?

Yes, the electric field due to a line of charge can be negative if the line is negatively charged. In this case, the electric field will point towards the line instead of away from it.

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