Potential from uniformly charged rod: signs and limits of integration

[uhty]

Homework Statement

http://www.phys.uri.edu/~gerhard/PHY204/tsl330.pdf
full solution here

Homework Equations

The Attempt at a Solution

Here is solution of problem for positively charged rod. What will be the difference if I take negatively charged rod? Why I start integration from d to d+l and not other way around? Should I do the same for negative charge or start on other side (d+L)? From which side I should start integration to get correct sign of potential?

 

[Coto]

That solution does not assume that the rod is positively or negatively charged. You should be able to analyze on your own what happens if Q is positive and when Q is negative. The derivation is the same.

As for the reason you integrate from d to d+L, you need to understand the general equation for calculating the electric potential. When you understand this equation, you will understand why the derivation was done the way it was.

 

[uhty]

I think you are wrong: the picture explicitly shows positively charged rod. In case of negatively charged rod the sign of potential must be negative, right? The integral then should be taken with negative sign, right?
Your explanation that I must understand how equation works is not very helpful, my questions are exactly about that-how it works?
If I integrate from d+L to L the integral will be of opposite sign, but the result should be the same. But which sign is correct for positive or negative charge? How sign in the intergal relates to charge sign?

 

[Coto]

I think you are wrong: the picture explicitly shows positively charged rod. In case of negatively charged rod the sign of potential must be negative, right? The integral then should be taken with negative sign, right?

The picture is just showing you the setup so you understand where all the variables are coming from. If you read the derivation, the charge per unit length can be positive or negative, it would depend on Q. He does not restrict Q to be positive, even though in the picture he draws a positive charge on the rod, the picture would be equally valid had he drawn a negative charge on the rod.

Your explanation that I must understand how equation works is not very helpful, my questions are exactly about that-how it works?
If I integrate from d+L to L the integral will be of opposite sign, but the result should be the same. But which sign is correct for positive or negative charge? How sign in the intergal relates to charge sign?

You mean from d+L to d? You are correct, the integral would have the opposite sign. You need to consider what it means if you integrate from d+L to d. Consider a line of point charges instead of the rod, each of charge q, summing them up their elec. potentials doesn't depend on the order that you sum them, so summing them from d+L to d wouldn't change anything. However, as you know, reversing the direction of integration **does** change something.

As an intuitive example, to calculate the area under the line y = 1, we could integrate y from x = 0 to x = L, what is the area? Now, integrating from x = L to x = 0, what is the area? We know area to be a positive quantity, so which way was the "correct" way? Why did the other way yield a different answer?

These are the same questions that should focus your understanding of why you integrate from the left side to the right side when doing this problem.

Lastly, the sign of the charge has absolutely no bearing on the sign of the integral. Again, the derivation is done for a general charge density and it **does not** assume a positive or negative charge density.

 

[uhty]

Ok, thank you for explanations.
I understand it well in all points except one: why sign of charge has no influence to sign of integral as you suggest? I know that positive charges produce positive potential and negative charges produce negative potential . So, the sign of charge on the rod must be included somewhere to provide sign for final potential. We are talking about absolute potential assuming that potential in infinity is zero, so sign must be present, right?

 

[Coto]

That's correct uhty. It comes down to this, what is your final result? The charge density is in that equation, and so any sign change in the charge density is indeed reflected in a sign change of the final result.

For example, the short rod expression is [itex] V = kQ/d [/itex]. So V will also be negative if Q is negative.

 

[uhty]

If it was some signless density-replacing half of the rod with negative charge would not produce any difference, but it will obviously be different case which needs two separate integrations. So, I think that having negatively charged rod we must include minus sign into integral and with positively charged not. Otherwise I don't get it.

 

[Coto]

:) . But now you have changed the problem. It is implicitly assumed that the charge density is constant along the rod (otherwise you can't pull that [itex]\lambda[/itex] out of the integral).

That derivation no longer holds at all if your charge density is no longer constant along the length of the rod. If for instance you had a situation like your example above, then you are correct, you would have to split the integral into two integrals to solve the problem.

 

[uhty]

what confuse me with integrations -if I integrate from d+L to d-I get different answer, I get
lnd/(d+L) instead of ln(d+L)/d but the result should be the same in both cases.

 

[Coto]

Actually, you get the negative of the same answer:

[tex]\ln{\frac{d}{d+L}} = -\ln{\frac{d+L}{d}}[/tex].

 

[uhty]

that is where my question about signs come from: we have two difefrent signs for the same answer depending on limits order.

 

[Coto]

This goes back to the reasoning about the area in post #4. This same reasoning is applied here. The standard way to integrate is from left to right and this is so mathematics lines up with our notions of reality (e.g. area should be positive)! To understand how this all arises you would need to analyze how the integral results from Riemann sums.