# The Electric field due to a charged rod

1. Sep 17, 2012

### Unbounded

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

A rod 14.0 cm long is uniformly charged and has a total charge of -22.0μC. Determing (a) the magnitude and (b) the direction of the electric field along the axis of the rod at a point 36.0 cm from its center.

d (the distance between the center of the rod and the point) = .29m
l = length of the rod

The answer to part (a) is supposed to be 1.59x106C

2. Relevant equations

λ(the linear charge density) = q/l

The electric field E at a point due to one charge element carrying charge Δq =

(keΔq)/r2

3. The attempt at a solution

At first I noticed that the distance between the end of the rod and the point should be .29m, as pointed out above. I assumed I could simply take the formula for the electric field and integrate it from 0 m to .29m, but when I did that it, well, didn't turn out too well.

Then I tried again, and instead integrated the formula over the distance of .14m to .43m. The process looked something like this:

E = ∫(keλdx)/x2

Noting the constants, I moved them outside of the integral, and had:

E = keλ∫dx/x2

Integrating, I got:

E = ke(Q/l) [ -1/x] evaluated from the lower limit .14m and the upper limit .43m.

E = ke(Q/l)[1/.14 - 1/.43]

Using all of that, however, gave me an answer of 6.8 x 106, which is far off of the answer the book gave me, 1.59x 106, and I'm not too sure my units would cancel out properly either.

Where exactly did I mess this thing up?

Thanks in advance for any help.

2. Sep 17, 2012

### TSny

Hi, Unbounded. Welcome to PF.

In this type of problem, it is very important to be clear on your choice of the location of the origin of your x-axis. There are several possible locations of the origin that could be chosen for convenience in setting up the integral. Once you have the origin nailed down, you will be able to logically deduce the limits of integration and the form of the denominator in the integral.

Where did you decide to choose the origin?

3. Sep 17, 2012

### Unbounded

Thank you.

I tried to place the origin at the end of the rod, though I'm not too sure I did that successfully.

Though I wonder now, would the numbers work out if I placed my origin in the center of the rod, and instead integrated over the interval of .07m to .36m?

4. Sep 17, 2012

### TSny

I'm not sure which end of the rod you chose (left or right). You can choose either end, or the middle, or you can even chose the origin at the location of the point where you would like to find E. It won't make much difference in difficulty of setting up and doing the integral. But you do have to make a choice, since the form of the integral will depend on the choice.

Note that you need to integrate over all of the elements of charge of the rod. So, once you choose your origin, the limits of integration will be determined by the x coordinates of the left and right end of the rod.

5. Sep 17, 2012

### Unbounded

So I can choose any location, so long as I integrate over the entire length of the rod. That would mean that what I was doing previously where I was integrating from the end of the rod was incorrect because I simply integrated from the space between the rod and the point.

So I should set the left end of the rod at the origin, and integrate from x = 0 to x = .14m, where x = 0 is the left end of the rod and x = .14 is the right end of the rod.

But then where do the distance between the center of the rod and the point where I have to calculate the electric field come into play?

6. Sep 17, 2012

### TSny

That's right.
That's not a bad choice. Yes, then the limits would be from 0 to .14 m.
Well, that comes in when you decide how to write the integrand. If x is the location of a point of the rod, how would you write an expression for the distance from that point to the point where you want to find E?

7. Sep 17, 2012

### Unbounded

I'm not quite sure I follow.

If I visualize the point as (.07m + .36m) = .43m from the origin, I would obtain a distance Δx = .29m.

So would I just place this as r in the formula E = ∫keq/r^2, change the q into λdx, and solve accordingly?

8. Sep 17, 2012

### TSny

Since x is the variable of integration, you need to express r in terms of x. r is the distance from an element of charge of the rod to the field point. For an element of charge located on the x-axis at position x, how would you express r in terms of x?

9. Sep 17, 2012

### Unbounded

So if r varies with x would it be r = (.29 + x)?

EDIT: Actually, that doesn't sound quite right.

as x increases, the distance between the charge and the point decreases, so the value of r would have to decrease as x increases.

Last edited: Sep 17, 2012
10. Sep 17, 2012

### TSny

If point A is on the x-axis at x = 3 and B is on the x-axis at x = 5, how do you get the distance from A to B?

Likewise, if an element of charge is located at x = x and the field point is located at x = .43 m, how do you get the distance from the element of charge to the field point (r)?

11. Sep 17, 2012

### Unbounded

So r = (.43 - x), as x varies from 0 to .14. This seems like it would be correct because at x = 0, (the left end of the rod) r = .43, and at x = .14,(the right end of the rod), r would equal .29, which was established earlier as the distance between the right end of the rod and the point p. Most importantly, if x =.07, the center of the rod, r will equal .36m, which is the distance between the center and the point p.

So if I understand this, my entire formula would look something like:

E =0.14keλdx/(.43-x)2

Where λ = Q/l?

12. Sep 17, 2012

### TSny

You got it. Good work!

13. Sep 17, 2012

### Unbounded

Thank you! I really appreciate the help!

14. Sep 17, 2012

### TSny

You're very welcome. For practice you might see if you can get the same result by choosing the origin at the center of the rod.