# Homework Help: Find electric potential at a center of charged rod

1. Jun 23, 2015

### gruba

1. The problem statement, all variables and given/known data
Find electric potential at a center of charged rod with charge 'Q' and length '2a' if referent point is at infinity.

2. Relevant equations
Electric potential of a point charge: V=kQ/r
Electric potential via electric field: integral (Edl)

3. The attempt at a solution
I used relation for potential of point charge:
V=kQ/(1/r_A-1/r_B), where r_A is a distance between charge and point, and r_B is the distance of infinity. From there, I think that r_A=a because that is the maximum distance between point and charge (point lies on the rod). This gives
V=kQ/a

Is it right that if we calculate potential of this rod in a given point, where referent point is at infinity, we can look at the rod as a point charge?
Could someone show the steps if this isn't correct?

Thanks for replies.

2. Jun 23, 2015

### SammyS

Staff Emeritus
Is this the complete problem statement as it was given to you?

It seems to me, that unless you have some finite (rather than infinitesimal) radius for the rod, the potential at any point on the rod will be infinitely large.

3. Jun 23, 2015

### gruba

The rod is a line of length '2a'. There are no calculations (values).

4. Jun 23, 2015

### SammyS

Staff Emeritus
So, it doesn't have a radius of R, or diameter of D, for instance?

5. Jun 23, 2015

### gruba

No, it is a straight line (finite rod).

6. Jun 23, 2015

### SammyS

Staff Emeritus
So, it has no thickness?

7. Jun 23, 2015

### gruba

No.

8. Jun 23, 2015

### SammyS

Staff Emeritus
or .. is that, Yes, it has no thickness?

9. Jun 23, 2015

### gruba

It has no thickness.

10. Jun 23, 2015

### SammyS

Staff Emeritus
What you are trying to do here is not entirely clear.

If this is an assigned problem, then please state the entire problem. If this is a problem of your own creation, then there is a flaw in it.

If you want to find the electric potential (due to the rod) at a point sufficiently far from the finite rod, then, yes, you can treat the rod as a point charge. However, it appears that you want to find the electric potential at a point directly on the rod, namely at its center. You will be integrating 1/r as r → 0 . That's a problem.

11. Jun 23, 2015

### gruba

Okay, I thought we can look at the rod as point charge if finding potential ON the rod (center). What am I doing wrong? Can you show your calculations?

12. Jun 23, 2015

### SammyS

Staff Emeritus
I hate to be repetitious, but ...

13. Jun 23, 2015

### gruba

I am sorry, but I don't understand what is unclear about the problem. Finite straight rod of length '2a' is charged with 'Q'. Find electric potential at the center of a rod (point lies on a rod) if referent point is at infinity.

14. Jun 23, 2015

### SammyS

Staff Emeritus
Is this actually a legitimate exercise from a book, or given to you by an instructor? Is it the COMPLETE statement of the problem?

15. Jun 23, 2015

### gruba

Yes, it is complete statement. I had this problem in my exam paper.

16. Jun 23, 2015

### SammyS

Staff Emeritus
Then the electric potential at any point directly on the rod is ± ∞, with the sign corresponding to the sign of charge Q. This is relative to the electric potential being zero at an infinite distance from the rod.

Do you know how to find either the electric field or electric potential by integration ?

17. Jun 23, 2015

### gruba

I know to use superposition method when point is not directly on a charge. I don't get how did get potential on center of a rod is equal to infinity? Can you show?

18. Jun 23, 2015

### SammyS

Staff Emeritus
Well, that's the problem. The center of the rod is a point directly on a charge and in this case that can't be overcome.

19. Jun 23, 2015

### gruba

Do you have theoretical explanation?

20. Jun 23, 2015

### SammyS

Staff Emeritus
Do you know how to find either the electric field or electric potential by integration ?

21. Jun 23, 2015

### gruba

I have calculated the new result for potential at the center of rod:

V=Q*ln(a)/(4pie_0*a)

I used superposition. If one half of a rod has length 'a', then from point at center to 'a' is infitinely many 'dQ' elements which needs to be summed. The same process goes with the other half of a rod (muliplication by 2).

V=2*k*integral(0-a)(Q'dl/r)=Q*ln(a)/(4pie_0*a)

Could you check this?

22. Jun 23, 2015

### SammyS

Staff Emeritus
(The overall solution is more complicated than this), but the main problem is in evaluating that integral which does show how the potential behaves near the rod.

$\displaystyle\ \int_{0}^{a}\frac{Q}{r}dr=Q\left(\ln(r) \right|_0^a \$

$\ \ln(0)\$ is undefined.

23. Jun 23, 2015

### gruba

V=(Q/(8pi*e_0*a))*ln2

Is this correct?

24. Jun 23, 2015

### SammyS

Staff Emeritus
What are you doing to eliminate ln(0) ?

25. Jun 23, 2015

### gruba

Instead of using limits of integration (0-a), I used (2a-4a). That is because length of the rod is '2a'