# Simple Coulomb's Law problem, struggling with basic calculus method

• mikey555
In summary, we are trying to find the force on a point charge located a distance d from one end of a uniformly charged rod with total charge Q and length L. We can use Coulomb's Law by considering an infinitesimal length from l to (l + dl) and integrating over l, which ranges from 0 to L. This allows us to convert r, the distance from the point charge to each dQ, in terms of l, and simplifies the integration process.
mikey555

## Homework Statement

A point charge +q is located a distance d from one end of a uniformly charged rod. The rod has total charge +Q and length L. (The rod and the point charge are each held fixed in place.)

What is the force on the point charge due to the rod?​

## Homework Equations

$$F = \frac{k q Q}{r^2}$$ (Coulomb's Law)

## The Attempt at a Solution

This is a really conceptual sort of question that comes up all the time for me--a simple integration problem of getting a common variable and then integrating. It's the setup that I've never understood.

So here's what I think we're doing conceptually: we're applying Coulomb's law an infinite number of times to an infinite number of small charges on this line of charge and adding them together.

If $$F = \frac{k q Q}{r^2}$$,

then $$dF = \frac{k q dQ}{r^2}$$, where $$dQ = Q (\frac{dl}{L}) = \frac {Q}{L} dl = \lambda dl$$

(to find dF, we let dQ be an infinitely small charge and dF will be its infinite contribution. to find dQ, we take the infinite length that this charge possesses, dl, and take the percentage of the entire line that dl takes up. dl / L is like an infinite percentage that we multiply by Q to get dQ.)

So we have:

$$F = \int \frac{k q dQ}{r^2} = \int \frac{k q Q dl}{L r^2}.$$

Here's where I have trouble. r is obviously changing for each charge, so I can't just pull it out of the integral. How to I write r, the distance from the point charge to each dQ, in terms of dl, an infinite length of the line, so that I can integrate?

More generally, how should I go about setting up integrals like this? I'm never sure if I should integrate dl and convert r, or integrate dr and convert dl. Does anyone have any tips on how to think about this?

mikey555 said:
… Here's where I have trouble. r is obviously changing for each charge, so I can't just pull it out of the integral. How to I write r, the distance from the point charge to each dQ, in terms of dl, an infinite length of the line, so that I can integrate?

Hi mikey555!

(btw, I'd always use x rather than l, because it's easier to write and to read )

Your difficulty is that you're using dl, but not l itself.

That's the tail without the dog!

GENERAL RULE:
Your proof (or your thought process) should begin, not with "consider an infinitesimal length dl", but "consider an infinitesimal length from l to (l + dl)".

Then you can immediately see that l goes from 0 to L (which gives you your limits of integration), and r = |d - l|.
More generally, how should I go about setting up integrals like this? I'm never sure if I should integrate dl and convert r, or integrate dr and convert dl. Does anyone have any tips on how to think about this?

If the point charge were at the end of the rod, then r and l would be the same. It isn't, so r is double-valued over the length 2d … you need a single-valued parameter to integrate over, so it needs to be l, not r.

tiny-tim said:
r is double-valued over the length 2d … you need a single-valued parameter to integrate over, so it needs to be l, not r.

Thanks for the fast response.

What does double-valued mean? Where does 2d come from (I thought we were only dealing with the length d!)?

I think what you said could really help but I don't understand it! If you could rephrase what I quoted above, it would be really helpful.

mikey555 said:
What does double-valued mean? Where does 2d come from (I thought we were only dealing with the length d!)?

I meant that r is measured from that point d from the end, so for values of r less than d, there's two sections at distance r, one on the right and one on the left (but for r > d, there's only one section).

So if you integrated over r, you'd have to integrate from r = 0 to r = d, and then start again and integrate over r = 0 to r = L-d (unless you're willing to have negative values of r, which really doesn't appeal to me), but if you integrate over l, measured from the end of the rod, then l goes from 0 to L.

## 1. What is Coulomb's Law?

Coulomb's Law is a fundamental law in physics that describes the electrostatic interaction between two charged particles. It states that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

## 2. How do I solve a simple Coulomb's Law problem?

To solve a simple Coulomb's Law problem, you will need to use the formula F = k(q1q2)/r^2, where F is the force, k is the Coulomb's constant, q1 and q2 are the charges of the particles, and r is the distance between them. Plug in the values and solve for F.

## 3. What is the Coulomb's constant?

The Coulomb's constant, denoted as k, is a proportionality constant that appears in Coulomb's Law. Its value is approximately 8.99 x 10^9 Nm^2/C^2.

## 4. Can I use Coulomb's Law to calculate the force between more than two charges?

Yes, Coulomb's Law can be extended to calculate the force between multiple charges. The total force is the vector sum of the individual forces between each pair of charges.

## 5. How does calculus come into play when solving a Coulomb's Law problem?

Calculus is used to find the derivative of the Coulomb's Law equation, which gives the electric field strength at a particular point. This can be useful in more complex Coulomb's Law problems, where the charges are not point charges and the distance between them is not constant.

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