Electric potential reference points confusion

[Shinobii]

Hi gang, I am hoping you can clear something up for me. When evaluating the potential of a solid sphere, I find myself confused about the volumes used such that,

$$\phi = Q \int_{\infty}^R \frac{1}{V_1} r^2 \sin(\theta) \, dr \, d\theta \, d\phi + Q \int_R^r \frac{1}{V_2} r^2 \sin(\theta) \, dr \, d\theta \, d\phi.$$

where $V_1 = 4 \pi r^3/3$ and $V_2 = 4 \pi R^3/3$.

Why is $V_1$ the volume inside and $V_2$ the volume at r = R?

The first volume I understand since we are coming from infinity to the surface of the sphere. But when evaluating the second integral with the second volume, why is it only at r=R.

My current logic is that the volume depends on the initial point of integration (which would be consistent with this problem), but surely this cannot be always true?

I hope someone can shed some light on this matter.

 

[aftershock]

Hi gang, I am hoping you can clear something up for me. When evaluating the potential of a solid sphere, I find myself confused about the volumes used such that,

$$\phi = Q \int_{\infty}^R \frac{1}{V_1} r^2 \sin(\theta) \, dr \, d\theta \, d\phi + Q \int_R^r \frac{1}{V_2} r^2 \sin(\theta) \, dr \, d\theta \, d\phi.$$

where $V_1 = 4 \pi r^3/3$ and $V_2 = 4 \pi R^3/3$.

Why is $V_1$ the volume inside and $V_2$ the volume at r = R?

The first volume I understand since we are coming from infinity to the surface of the sphere. But when evaluating the second integral with the second volume, why is it only at r=R.

My current logic is that the volume depends on the initial point of integration (which would be consistent with this problem), but surely this cannot be always true?

I hope someone can shed some light on this matter.

I don't think I've seen whatever formula you're using, but I think you're attempting to find the potential inside the sphere. You start from zero since that's you're reference point and integrate up to the surface of the sphere. Then you add another integral since the term you're integrating is different inside the sphere. This runs from the surface to whatever point you want to find the potential at, r.

Is your question why you go to r and not 0? That's because that would give you the potential at the center. r gives you the ability to plug in any distance from the center inside the sphere and find the potential.

 

[Shinobii]

Thanks

Yes, you are right about the potential inside a solid sphere; sorry for not being more clear. You answered my question perfectly! I suppose that is also the reason that $V_1$ has r in the denominator, since we are coming in from infinity.

Thanks a bunch.