Calculate Electric Potential from Non-uniform Linear Charge

[DapperDan]

Homework Statement

"A rod of length L lies along the x-axis with its left end at the origin. It has a non-uniform charge density λ=αx where α is a positive constant. a) What are the units of α? b) Calculate the electric potential at A.

Homework Equations

Linear charge density: λ = Q/L where Q is charge in Coulombs and L is length in meters.

Electric potential due to a point charge Q: V = KQ/r where K is the Coulomb constant 8.99*10⁹, Q is charge in Coulombs, and r is the radius in meters separating the charge and the point in question.

The Attempt at a Solution

I've been trying to work this problem with guidance from Chegg, but to limited success. Pictured is as far as I can get even with the assistance of having the solved problem before me. I think that my core issue with this problem stems from a lack of understanding as to how to apply calculus concepts to physical phenomena.

For instance, when V becomes dV in step 2, Q logically also becomes dq. But why does r not become dr? The radius would vary just like the charge would, yet it is left as a constant r. In the next step, I subbed in (d+x) for r as it was done in the Chegg solution. I understand that d + x represents the distance from point A to the origin and then the addition of a given x value, but I still don't entirely understand how it is implemented here. Can we sub this in because it is a function of x and we will be integrating with respect to x? Did we leave r as r rather than changing it to dr to allow for this substitution to be made?

Between steps 6 and 7, the solution I was reading made a mathematical jump that I didn't understand when they rewrote x/(d+x) as 1 - d/(d+x). How did this happen?

Many thanks for any help you can provide!

 

[haruspex]

a) What are the units of α?

Poor question.
It is impossible to say what the units of α are since we are not told what the units of λ and x are. All we can say is what dimension it has.

when V becomes dV in step 2, Q logically also becomes dq. But why does r not become dr? The radius would vary just like the charge would, yet it is left as a constant r.

Think through how an integral formula is developed. We start by considering a small segment of the wire. It starts at d+x and finishes at d+x+dx from the origin.
(The choice of 'd' is unfortunate.)
Because it is short, we can take the charge density as near enough constant along it, so its charge is λdx=αxdx. We can also call this dq if you like, but there's no need.
Likewise, the whole segment is, near enough, all the same distance from A.
The potential it creates at A is $k\frac{\lambda dx}{d+x}$. Integration is the process of adding up all these small contributions, while letting the lengths dx become arbitrarily small:
$\int_{x=0}^{x=L}k\frac{\lambda dx}{d+x}$.
So the r does not become dr because the value (d+x) applies to the whole segment length dx.

 

[DapperDan]

I'm still struggling with setting up this integral. So we have our original equation $V = \frac {KQ} r$. We can rewrite this as $V = \frac {KλL} r$ which becomes $V = \frac {Kλx} r$. r for our point at A would be described by d + x so we write $V = \frac {Kλx} {d + x}$. We know, from our given information that λ = αx so we sub into produce $V = \frac {K(αx)x} {d + x}$. So everything is now described in terms of x, so we would now want to consider how it would be described if x became arbitrarily small. So our x's become dx's and we get: $V = \frac {K(αdx)dx} {d + dx}$. I know that this expression is nonsense, but I don't know how to logically piece it together in a different way.

Also if we're treating charge density as nearly constant, and the segment as nearly the same distance from A, why integrate them? Would we not want to remove constant terms from our integral so that we could focus on summing up whatever terms are going to be in flux over the course of the integration?

Forgive me for my misunderstanding, and thank you again for any further assistance you can provide!

 

[rude man]

I'm afraid you're way off base from the get-go.
You will need to integrate. You would need to integrate even if λ was not a function of x but constant over o<x<L.
So consider an element of charge dq = λ dx. How far away from A is it? So what is its potential at A?

BTW it's unfortunate that you were given the distance as d. d should be reserved for the differential.

Sorry about all the re-edits.

 

[haruspex]

We can rewrite this as $V = \frac {KλL} r$

Can you say exactly what you mean by V here? The potential due to what?
You should be considering the potential due to the charge on an element of the rod:
$dV = \frac {K\lambda(x).dx} r$

which becomes $V = \frac {Kλx} r$

You forgot to change λ to α when substituting λ(x)=αx.

to produce $V = \frac {K(αx)x} {d + x}$

Because you forgot to replace λ with α earlier, you now have an extra x in the numerator.

how it would be described if x became arbitrarily small. So our x's become dx's

This is another place you go wrong. It is not a matter of replacing all x's with dx, as I tried to explain in post #2. You have to start with the d's in the right places.
$dV = \frac {K\lambda(x).dx} r = \frac {K\alpha x.dx} {d+x}$ etc.

 

[DapperDan]

rude man, no worry about the re-edits! I agree, the use of d as a variable was a terrible choice, especially considering they already opted to use a capital L for the other distance variable. But they're getting paid to write physics books and I am not, so who I am to question their variable selections? lol

If our element of charge was dq, equal to λdx, it would be the distance d plus an additional arbitrarily small distance dx away from A, correct? And it would produce some small change in electric potential at A equal to dV. Which would look like $dV = \frac {Kλdx} {d + dx}$?

I know that the above expression isn't correct, but I just can't see how we can have (d + x) in the denominator without changing to (d + dx) when we begin talking about changes in charge that would correspond to small changes in x values.

 

[haruspex]

it would be the distance d plus an additional arbitrarily small distance dx away from A, correct?

No. As I wrote in post #2, you have an element length dx along the interval (x, x+dx). The end of it at coordinate x is distance d+x from A; the other end is at d+x+dx from A. But since we are going to let dx become arbitrarily small we can take the whole element as being at distance d+x from A.

 

[rude man]

if our element of charge was dq, equal to λdx, it would be the distance d plus an additional arbitrarily small distance dx away from A, correct?

Look at your drawing! Let's try again: what is the distance between A and some element of charge dq along x? In other words, starting with V = kq/r, now it's dV = kdq/r. You have dq right but what about r?

 

[DapperDan]

No. As I wrote in post #2, you have an element length dx along the interval (x, x+dx). The end of it at coordinate x is distance d+x from A; the other end is at d+x+dx from A. But since we are going to let dx become arbitrarily small we can take the whole element as being at distance d+x from A.

Okay, so our small changes in length are occurring at the starting point d + x, and they are so negligible that we can essentially treat the whole thing as the static value d + x rather than as a changing value? That being said can't we just consider it a constant and factor it out in front of our would-be integral? Also, thank you for the continued assistance.

 

[haruspex]

they are so negligible that we can essentially treat the whole thing as the static value d + x

Only for that small element. Other elements are at different distances (different x) from A.

 

[DapperDan]

Look at your drawing! Let's try again: what is the distance between A and some element of charge dq along x? In other words, starting with V = kq/r, now it's dV = kdq/r. You have dq right but what about r?

So if our charge element was positioned at the leftmost end of our rod, it would be the distance (d + x) away from A. So we have $V = \frac {KQ} {d + x}$. When we then think about dV, a small change in electric potential, it corresponds to a ratio of a small change in charge, dq, to a small change in length, which would be some change to (d + x), wouldn't it?

 

[haruspex]

if our charge element was positioned at the leftmost end of our rod, it would be the distance (d + x) away from A

If the element is at the left end of the rod then x is zero for that element.

So we have $V = \frac {KQ} {d + x}$

That equation is correct for a point charge Q. But since we are dealing with small elements, which in the limit will become infinitesimal, it is usual style to start with element notation. Write the charge of the element as dq, say, and its contribution to the potential at A as dV: $dV = \frac {K.dq} {d + x}$.

dV, a small change in electric potential, it corresponds to a ratio of a small change in charge, dq, to a small change in length

No. The whole element of length dx, charge dq, is at d+x (near enough), not at d+dx. Refer to your diagram.

 

[DapperDan]

Okay tell me if I'm visualizing this correctly. Is it correct to say that since $λ = \frac Q L$ and therefore $λL = Q$ then charge is going to vary with x in the same way as the radius (d + x) does? And that begets $dq = λdx$ where the small change in charge is accompanied by a similar small change in x? But since $λ = αx$ it will still vary with x in the same way that the radius (d + x) does when it becomes $dV = \frac {K(αx)dx} {d+x}$?

I have attached an image demonstrating what I mean if I worded this poorly. I hope it is legible!

 

[rude man]

. Is it correct to say that since $λ = \frac Q L$ and therefore $λL = Q$

Incorrect if by Q you mean all the charge on the line.

then charge is going to vary with x in the same way as the radius (d + x) does? And that begets $dq = λdx$ where the small change in charge is accompanied by a similar small change in x? But since $λ = αx$ it will still vary with x in the same way that the radius (d + x) does when it becomes $dV = \frac {K(αx)dx} {d+x}$?

This part and you concluding statement are correct. So proceed ...

Also: Q = λL would be correct only if λ were constant. Actually, Q = integral of λ(x)dx from x=0 to x=L, giving Q = αL²/2.

 

[DapperDan]

Incorrect if by Q you mean all the charge on the line. This part and you concluding statement are correct. So proceed ...

Also: Q = λL would be correct only if λ were constant. Actually, Q = integral of λ(x)dx from x=0 to x=L, giving Q = αL²/2.

Okay so radius is a function of x. Q is a function of charge density and x. But charge density is a nested function within our charge function, and it varies differently than charge. Charge density varies in tandem with radius, as they are functions of the same input, which is the position along the rod. dV is an arbitrarily small change in charge which is corresponds to an arbitrarily small change in x, but is also dependent on the factors of charge density and radius, which vary at a much grander scale. At any given point where an arbitrarily small change in electric potential/charge/change in x would occur, charge density and radius would be like set values, because to affect them would take a much greater change along the x-axis than the arbitrarily small one.

Is this all still a correct statement? If this is correct, then I think I might finally be getting the concept here.

 

[haruspex]

Q is a function of charge density and x.

If Q is the total charge, I would say it depends on the density function, λ(x), and the total length L. Specifically, Q=∫_x=0^Lλ(x).dx. Note that although x appears in that expression it is as a "dummy variable". You could change it to anything else, y, t, ... and the value of the expression does not change. Hence, it does not make Q a function of x.

dV is an arbitrarily small change in charge

dV would be an arbitrarily small change in, or contribution to, a potential.

At any given point where an arbitrarily small change in electric potential/charge/change in x would occur, charge density and radius would be like set values, because to affect them would take a much greater change along the x-axis than the arbitrarily small one.

Yes, you could think of it that way.

 

[rude man]

Okay so radius is a function of x.

OK but I wouldn't call it "radius". Just call it distance.

dV is an arbitrarily small change in charge

not charge; potential[quote

Is this all still a correct statement? If this is correct, then I think I might finally be getting the concept here.[/QUOTE]You're a bit verbose and I'm not sure what to make of it all, but if you understand that you should integrate x dx/(x+d) then you are pretty much there.

 

[DapperDan]

I think I'm starting to understand. My only other question is in regards to the transition from step 6 to step 7 in my picture. How does $\frac x {d+x}$ become $1 - \frac d {d+x}$?

 

[rude man]

I think I'm starting to understand. My only other question is in regards to the transition from step 6 to step 7 in my picture. How does $\frac x {d+x}$ become $1 - \frac d {d+x}$?

That's an indentity and just a suggestion to make the integration easier for you. I never used it; just look up the integral in step 6 in a table of integrals.

 

[DapperDan]

Excellent! Well with that, I think my questions have been answered. Thank you both very much for your help, you've been very insightful!

 

[rude man]

Excellent! Well with that, I think my questions have been answered. Thank you both very much for your help, you've been very insightful!

Keep at it!