Nonuniform Rod Electric Field

[mattbeatlefreak]

Homework Statement

A thin rod of length 2ℓ has a linear charge density that is λ₀ at the left end but decreases linearly with distance going from left to right in such a way that the charge on the entire rod is zero.

What is the magnitude of the electric field along the rod's axis at a position P that is a distance d>ℓ to the right of the center of the rod? Use the notation l for the variable ℓ.
Express your answer in terms of the variables l, λ₀, d, and Coulomb's constant k.

Homework Equations

Electric field = kq/d²
λ=q/l

The Attempt at a Solution

I am completely lost on this one, although I have some parts down I think.
I know that I will need to integrate and should be using a small section of rod dl for charge dq.
One end of the rod has charge density of λ₀, so the other end of the rod must have charge -λ₀ in order for the entire rod to have a zero net charge. I also think that you would integrate from 0 to 2l. I have tried several different setups but do not feel confident on any of them and could not find any guidance from a similar problem on the Internet or my textbook.
Because the charge decreases linearly, I was thinking that dq = λ₀ - ax (if one end of the rod is placed at the origin), but that just added another variable which confused me further.
At first I had tried to consider the rod as a dipole, since one end was positive and one was negative, but it seems wrong that way since there is charge throughout and not just at both ends.
Any help with this is greatly appreciated, thanks.

 

[gneill]

Try placing the center of the rod at the origin. You should be able to come up with a simple expression for the charge density that uses the sign of x to set the sign of the charges. Then you can integrate from -L to +L.

Starting with a symmetrical layout often simplifies the math.

 

[mattbeatlefreak]

Try placing the center of the rod at the origin. You should be able to come up with a simple expression for the charge density that uses the sign of x to set the sign of the charges. Then you can integrate from -L to +L.

Starting with a symmetrical layout often simplifies the math.

Thanks for the help.
So when I put the center at x=0, then you can express any dq = λ₀⋅x?
When I set up my integral, I get
E=∫kdq/d² (integral bounds are -L to L)
I substitute dq with λ₀x and pull out kλ₀/d²
E=kλ/d²∫xdx
Evaluating this integral I reach an answer of E=kλ₀L²/d²
Is this the correct set up or am I completely misguided..

 

[gneill]

Thanks for the help.
So when I put the center at x=0, then you can express any dq = λ₀⋅x?

$λ_o$ is a charge density. You want the density to be $λ_o$ at x = -L, which would make a dq there equal to $λ_o dx$. So when you scale the density the scaling factor should be signed but unitless, and it should be +1 at either end of the rod. How might you accomplish that?

When I set up my integral, I get
E=∫kdq/d² (integral bounds are -L to L)
I substitute dq with λ₀x and pull out kλ₀/d²
E=kλ/d²∫xdx
Evaluating this integral I reach an answer of E=kλ₀L²/d²
Is this the correct set up or am I completely misguided..

You want to integrate over x from -L to +L, but the distance that you use in Coulomb's law must be the distance between the point in question (d) and the charge element dq. That distance varies from d + L to d - L. So your integral is going to be a bit more complicated because you can't pull the $d^2$ out.

Here's a diagram:

 

[mattbeatlefreak]

$λ_o$ is a charge density. You want the density to be $λ_o$ at x = -L, which would make a dq there equal to $λ_o dx$. So when you scale the density the scaling factor should be signed but unitless, and it should be +1 at either end of the rod. How might you accomplish that?

Well, λ could just be set equal to x, or it could be λ=ax where a is some constant value... since as x increases, the charge density decreases.

You want to integrate over x from -L to +L, but the distance that you use in Coulomb's law must be the distance between the point in question (d) and the charge element dq. That distance varies from d + L to d - L. So your integral is going to be a bit more complicated because you can't pull the $d^2$ out.

Here's a diagram:
View attachment 94924

Okay I'll try this again. I think I'm slowly grasping what I should be doing. So you set dq=λdx. λ=x. the distance you can use in the integral will be d+x?

E=k∫λdx/(d+x)² (integrating from -L to L)
E=k∫xdx/(d+x)²
At this point, I'm either still setting this up wrong or I haven't learned enough about integration to solve this.
Also, your diagram was beautiful and extremely helpful!

 

[gneill]

Take a look at the diagram and the line representing the charge density in particular. What is the slope of that line? What's the equation of the line then? $λ(x) = ?$

If both x and d are measured from the origin, what is the distance between them? (It's labelled "R" in the diagram). Note that Coulomb's Law for the electric field uses $R^2$. If it was electric potential you'd use $R$.

If the integral is more complicated than what you've dealt with up to now then you might try a table of integrals or the Wolfram online integrator. Or, see if you can cast it into a generic form and take it to the Calculus and Beyond homework forum to work on it. You'll need to take a stab at it first though; It looks like it might be one of those integrals that requires variable substitution and maybe integration by parts.

 

[mattbeatlefreak]

Take a look at the diagram and the line representing the charge density in particular. What is the slope of that line? What's the equation of the line then? $λ(x) = ?$

If both x and d are measured from the origin, what is the distance between them? (It's labelled "R" in the diagram). Note that Coulomb's Law for the electric field uses $R^2$. If it was electric potential you'd use $R$.

If the integral is more complicated than what you've dealt with up to now then you might try a table of integrals or the Wolfram online integrator. Or, see if you can cast it into a generic form and take it to the Calculus and Beyond homework forum to work on it. You'll need to take a stab at it first though; It looks like it might be one of those integrals that requires variable substitution and maybe integration by parts.

The slope is y/x, so in this case it would be -λ/L.
If I use R² in the integral, won't my answer be in terms of R? I can only use l and d for distance measures in my answer, that is why I was trying to avoid using the R. Thank you for the advice with the integral, I'll give it a try first if I can ever figure out how to set it up

 

[gneill]

The slope is y/x, so in this case it would be -λ/L.

Well, it would be $-λ_o/L$ for that particular line, right? So $λ(x) = (-λ_o/L) x$. Multiply that by dx and you get a charge element dq at location x.

If I use R² in the integral, won't my answer be in terms of R? I can only use l and d for distance measures in my answer, that is why I was trying to avoid using the R.

The R is just a label in the diagram designating the distance that would be the required "R" in Coulomb's Law. Examine the diagram and write its equivalent in terms of x and d.

Thank you for the advice with the integral, I'll give it a try first if I can ever figure out how to set it up

Okay. Good luck!