Integrating from R to r: Potential Contribution & Reasoning

[phantomvommand]

Homework Statement

I am trying to calculate the potential in the sphere at a distance r away from the centre, where r < R, R is the radius of the sphere.

Relevant Equations

$V = - \int \vec E \cdot d \vec l$
$V = \frac {1} {4 \pi \varepsilon} \int \frac {\rho} {r} dV$

The potential contribution from R > 0 is simple. My next step is to integrate from R to r. With regards to the integration from R to r, the 2nd method gives a potential contribution that is the negative of the 1st method. What is the reason?

 

[phantomvommand]

Nvm I think the 2nd method is impossible. Each differential element of charge is at a different distance away from the point where radius = r, and furthermore, the denominator should be the distance between the differential element of charge and the point where radius = r. Please check if I'm right on this

 

[vela]

It's not clear what you're doing from your posts. Is the sphere a solid sphere of charge with constant charge density $\rho$, or a spherical shell of charge with surface density $\sigma$?

The potential contribution from R > 0 is simple. My next step is to integrate from R to r. With regards to the integration from R to r, the 2nd method gives a potential contribution that is the negative of the 1st method. What is the reason?

It's a mystery to me what you did in both cases. Did you mean $R>0$ or $r>R$? Are you integrating from infinity?

Nvm I think the 2nd method is impossible. Each differential element of charge is at a different distance away from the point where radius = r, and furthermore, the denominator should be the distance between the differential element of charge and the point where radius = r. Please check if I'm right on this

That's right if I'm thinking what you're thinking, but I'll note the 2nd method is not impossible. Depending on your approach, you may need mathematical techniques you haven't learned yet, though.

 

[phantomvommand]

It's not clear what you're doing from your posts. Is the sphere a solid sphere of charge with constant charge density $\rho$, or a spherical shell of charge with surface density $\sigma$?It's a mystery to me what you did in both cases. Did you mean $R>0$ or $r>R$? Are you integrating from infinity?That's right if I'm thinking what you're thinking, but I'll note the 2nd method is not impossible. Depending on your approach, you may need mathematical techniques you haven't learned yet, though.

To use the 2nd approach, I suppose you would have to add up the potential contribution of all infinitesimal elements in the sphere? So it would be a volume integral, though each elements is at a different distance away.

 

[jbriggs444]

To use the 2nd approach, I suppose you would have to add up the potential contribution of all infinitesimal elements in the sphere? So it would be a volume integral, though each elements is at a different distance away.

Yes. And the contributions are in different directions. So, in the absence of a useful symmetry, you'll be adding vectors.

Edit: Egg on face here as pointed out.

 

[kuruman]

Homework Statement:: I am trying to calculate the potential in the sphere at a distance r away from the centre, where r < R, R is the radius of the sphere.

The potential at fixed point $r$ inside the sphere ($r<R$) is the sum of two terms:
1. The potential at the outer surface of a charged sphere of radius $r$.
2. The potential at the inner surface of a charged shell of inner radius $r$ and outer radius $R$.
The two contributions need to be calculated separately. You do this by considering shells of thickness $dr'$ from $r'= 0$ to $r'=R$ and noting that the electric field is zero when $r'>r$ and goes as $r^{-2}$ for $r'<r$.

 

[phantomvommand]

Yes. And the contributions are in different directions. So, in the absence of a useful symmetry, you'll be adding vectors.

Isnt potential a scaler quantity? i don't think we are adding vectors here.

 

[jbriggs444]

Isnt potential a scaler quantity? i don't think we are adding vectors here.

Right you are. I was thinking of integrating the force to find what would be the gradient of the potential.