Physics w/Calculus II Electric Potential, Non-Uniform Charge

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1. Mar 20, 2015

ChrisCrary

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

Picture of problem is attached

A rod of length L lies along the positive y axis from 0<=y<=L, with L = 0.400 m. It has a positive nonuniform linear charge density (charge per unit length) λ = cy2, where c is a positive constant and has a numerical value of 8.00 x 10-7. (a) What is the unit of c in terms of meter, Coulomb?
(b) find the electric potential V due to the rod at point A on the y axis at h = 0.450 m from the origin of the xy plane;
(c) find the electric potential V due to the rod at point B on the x axis at d = 0.650 m from the origin of the xy plane;
(d) How much work (in Joules) is required to bring another charge q0 = + 3.50 μC from point A to point B? The Coulomb constant ke = 1/(4πξ0) = 9.00 x 109 N m2/C2, and 1 μC = 10-6 C.

2. Relevant equations

The electric potential due to a single charged particle at a distance r from that particle is V = kq/r, where k is the Coulomb constant.

λ = q / L, or λ = dq / dy

3. The attempt at a solution

My attempt at a solution can be shown in the pictures attached below. I think I got part A and B correct, but please check me as I'm not too sure and I do not have the answers. I worked a bit on part C but got stuck on the integral... I'd rather not do by Trig Substitution. Also, if anyone could manage to explain Part D somewhat, it would be much appreciated. Please tell me if I made any mistakes anywhere!!!

Last edited by a moderator: May 7, 2017
2. Mar 20, 2015

Simon Bridge

Images didn't come out for me ... please describe the reasoning you used when you were completing the problem.

3. Mar 20, 2015

ChrisCrary

Here are the pictures that were supposed to be in the original post: One of the problem, and two of the work.
Can you see them now? My efforts stemmed from working with differential elements of charge, and setting up an integral involving the electric potential equation I gave in the original post, kq/r, but I changed it to k(dq)/r
The trickiest part is setting up the integral. To get it in terms of y to integrate, I replaced dq and r. If you cannot see the pictures, please tell me to elaborate further.

4. Mar 20, 2015

ChrisCrary

I've done a lot of these problems, and I understand them. I just got thrown off since it was a rod on the y-axis.

5. Mar 20, 2015

Simon Bridge

So you just did it the same way as the others with different variables.

It is important to include your reasoning with your working otherwise it's just a bunch of blank maths and you are forcing us to guess.

Pretend you are trying to explain what you are doing to a student just behind you in understanding.
Often, this process builds confidence in what you've just done. Generally, the quickie test to see if you really do understand something is to see if you can explain it to someone else.

i.e. From what I can see, you have not answered (a) - what is the unit of "c" in $\lambda = cy^2$
The argument should be in words or in terms of dimensional analysis, I don't se any of that in your write-up.

I do see $c = C/M^3$ - which does not make mathematical sense.
Perhaps you mean $[c] = QL^{-3}$ or "the SI units for 'c' are 'Coulombs per cubic meter' "?
Possibly you'll get away with that... but you won't always be read by a teacher: see how the way I wrote it is clearer?

For the future: it is also best practice in these forums to write out the working rather than post (large) images of your handwritten work. For this purpose, it is best to learn LaTeX.

Last edited: Mar 20, 2015
6. Mar 20, 2015

ChrisCrary

Okay, so we need to find the potential of this rod. "The electric potential due to a single charged particle at a distance r from that particle is V = kq/r, where k is the Coulomb constant." And because it is not a point, we need to integrate it as a non-uniform charge distribution. This means that we integrate the electric potential formula (V = kq/r).
So we replace the q with an arbitrary element of the rod, dq. We replace r with the h-y, which is the distance from any point on the rod, to the point. This can be shown in one of the pictures uploaded prior. We want to integrate in terms of dy, so we replace dq with λ dy. This is also shown in the same picture. Then since λ (the linear charge distribution) is non-uniform, we replace λ with the given equation (cy2). Now we can pull constants k and c out of the integral, and integrate.

My two questions stem from the limits of the integral and how to integrate the integral shown in part C. The limits of the integral come from where the rod is on the plane. Now usually, I deal with rods on the x-axis. We usually write the limits from left to right, whether they are 0 to L, or D to D+L. These limits cover the rod. Again, usually since this is on the x-axis, you write the limits from left to right. This seems to be because the point is usually to the left of the rod. But for the y-axis (Part B), do I go from bottom to top? 0 to L? Or do I go from the point's distance from the top of the rod (h - L) to h? This is confusing to explain but I am trying the best I can.

Now my final question is how to integrate the integral ∫(y2 dy)/(√(y2 +d2))
This is shown in the picture for Part C.

I know this is a lot, and I am trying my best to explain my thought process. It would be really helpful if you could try to explain to me. Thank you again for your efforts.

7. Mar 20, 2015

SammyS

Staff Emeritus
The units you have for c are correct.

(C/m3)⋅(m2) = C/m , the unit for linear charge density.

8. Mar 20, 2015

ChrisCrary

[Mentors note: This post was moved from another thread]

I tried u-sub, and then replaced y2 and dy for terms of u. I further simplified, but soon after, got stuck on how to integrate in terms of u. Work is shown below. Again, this is from a prior problem, and the work is shown up until the integral I am asking for help on. The picture includes limits of integration, but I do not need this included in the solution.

Last edited by a moderator: Mar 20, 2015
9. Mar 21, 2015

SammyS

Staff Emeritus
That's a difficult integral. Well, at least it's quite messy.

Essentially it's $\displaystyle \int \frac{x^2}{\sqrt{x^2+a^2}}\,dx\$.

Try either trig substitution: x = a⋅tan(θ) or ..

Use a hyperbolic function substitution.\: x = a⋅sinh(t) .

You can try integration by parts first to simplify the integral somewhat.

$\displaystyle v=\sqrt{x^2+a^2}\,,\$ and $\ u = x \$.

But you still arrive at needing either of the above substitutions.

Last edited: Mar 21, 2015
10. Mar 21, 2015

Simon Bridge

Consider: if you rotate the page 90deg, the y axis is now horizontal ... can you see how to do the limits?

It's OK - now I can see how you are thinking... it seems you have got into a bad habit when it comes to the way you think about integrals ... basically, as a series of special rules to be remembered and followed. You should think, instead, of the integration as a summation. You know how to do sums already. I'll illustrate this by going through the start of your problem more formally.

On the diagram - mark off a position inside the rod and label it "y", then another position slightly greater than that and label it "y+dy".
The amount of charge between $y$ and $y+dy$ along the rod is $dq=\lambda\; dy$, the potential at position $y=h$ on the axis, due to this charge is given by $$dV = \frac{k\lambda\; dy}{h-y}$$. Now you just have to add up the all contributions due to the charge inside the interval $0 < y < L$ ... we write that as: $$V=\int_0^L \frac{k\lambda\; dy}{h-y}$$ ... note: the integration symbol is actually a stretched out varient of the letter "S" (for "sum").

The same approach works no matter how the rod is aligned. i.e. if it were aligned 45deg to the x axis.

Does that help clear your thinking?

FWIW: you can always change the axes so the rod is aligned to whatever one you want ... Nature does not care which way up we draw a line.
You are already getting advise on how to complete the actual integration so I left that part alone.