Find electric potential at a center of charged rod

[gruba]

Homework Statement

Find electric potential at a center of charged rod with charge 'Q' and length '2a' if referent point is at infinity.

Homework Equations

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

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.

 

[SammyS]

Homework Statement

Find electric potential at a center of charged rod with charge 'Q' and length '2a' if referent point is at infinity.

Homework Equations

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

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.

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.

 

[gruba]

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.

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

 

[SammyS]

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

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

 

[gruba]

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

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

 

[SammyS]

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

So, it has no thickness?

 

[gruba]

So, it has no thickness?

No.

 

[SammyS]

No.

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

 

[gruba]

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

It has no thickness.

 

[SammyS]

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.

Your method will not work.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.

 

[gruba]

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.

Your method will not work.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.

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?

 

[SammyS]

I hate to be repetitious, but ...

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.

Your method will not work.

Please answer.

 

[gruba]

I hate to be repetitious, but ...

Please answer.

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.

 

[SammyS]

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.

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

 

[gruba]

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

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

 

[SammyS]

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

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 ?

 

[gruba]

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 ?

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?

 

[SammyS]

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?

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.

 

[gruba]

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.

Do you have theoretical explanation?

 

[SammyS]

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

 

[gruba]

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

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?

 

[SammyS]

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 infinitely many 'dQ' elements which needs to be summed. The same process goes with the other half of a rod (multiplication by 2).

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

Could you check this?

(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.

 

[gruba]

(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.

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

Is this correct?

 

[SammyS]

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

Is this correct?

What are you doing to eliminate ln(0) ?

 

[gruba]

What are you doing to eliminate ln(0) ?

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

 

[SammyS]

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

Then you're finding the potential at some point along the rod that is a distance of $\ a \$ away from the near end of the rod. That's hardly at the center.

 

[gruba]

Then you're finding the potential at some point along the rod that is a distance of $\ a \$ away from the near end of the rod. That's hardly at the center.

Do you know the solution?

 

[SammyS]

Do you know the solution?

Sure. But it goes to ∞ as the point gets close to the rod.

Find the potential a distance, b, from the rod at the rod's center. In other words, at a location a distance, b, in a direction perpendicular to the rod.

$\displaystyle\ V(b)=2\frac{Q}{2a}\int_{0}^{a}\frac{1}{\sqrt{x^2+b^2\,}}dx \ .$

 

[gruba]

Sure. But it goes to ∞ as the point gets close to the rod.

Find the potential a distance, b, from the rod at the rod's center. In other words, at a location a distance, b, in a direction perpendicular to the rod.

$\displaystyle\ V(b)=2\frac{Q}{2a}\int_{0}^{a}\frac{1}{\sqrt{x^2+b^2\,}}dx \ .$

Could you show your full calculations for this problem if you have done it?