# Paradox in calculating potential of a sphere

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1. May 7, 2017

### Buffu

1. The problem statement, all variables and given/known data
Find electrostatic potential of a solid sphere with reference point of 0 statvolt at infinite.

2. Relevant equations

3. The attempt at a solution

Potential energy of a solid sphere is $\dfrac35 \dfrac{Q^2}{r}$.

And I know $\displaystyle U = {1\over 2} \int \rho \phi dv$.

So $2\dfrac {dU}{dv} = \dfrac{dU}{dr}\cdot\dfrac{dr}{dv} = \rho \phi$

Differentiating the sides and then plugging for $\displaystyle \rho = {3Q \over 4\pi r^3 }$

I got $\phi = \dfrac{2Q}{5r}$.

But electrostatic potential is $\displaystyle Q \over r$ for a solid sphere.

Why is this paradox ? what I am missing ?

2. May 7, 2017

### Staff: Mentor

How is the charge distributed in your sphere?
The potential of what, where, with which type of charge distribution?

Start with the volume outside the sphere. It is way easier than the calculation you did.

3. May 7, 2017

### Daniel Gallimore

I do not agree with this. For a conducting metal sphere, the potential energy (with respect to $\infty$) is $$U=-W=-\int_\infty^a\vec{F}\cdot d\vec{\ell}$$ where $\vec{F}$ is the Coulomb force (the Coulomb force is the electrostatic force due to a single point charge at the origin or a sphere centered at the origin outside the sphere). From this, you could get voltage.

However, there is an easier way. We know that the electric field of a solid conducting sphere outside of the sphere (i.e., for $r>a$) is $$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2} \, \hat{r}$$ where $Q$ is the charge enclosed by the sphere and $r$ is the radial distance away from the center of the sphere. Voltage is $$V=-\int_\infty^a\vec{E}\cdot d\vec{\ell}$$

Edit: this applies to uniformly distributed charge. The comment @mfb made about the distribution of the charge is relevant.

4. May 7, 2017

### Buffu

I don't think the sphere is conducting.

Uniformly distributed.

I am also a bit confused as to what and where.

Original statement :-
Find the potential of a solid sphere as a function of $a$. For $a < r$ and $a > r$.

There was image showing what is $a$, that I can't post :(. But I can explain what $a$ is.

It is the radius vector from the centre of the sphere to a location in space, something like what is "R" in spherical coordinates.

Sorry.

5. May 7, 2017

### Daniel Gallimore

As long as the charge is uniformly distributed on the surface of the insulating sphere and there is no external electric field, my explanation will still apply outside of the sphere.

Inside the sphere, you need to apply Gauss' law to get the electric field. Recall that the charge enclosed is going to depend on the radius of your Gaussian surface. Once you have that electric field, use $$V_{in}=-\int_\infty^R\vec{E}_{out}\cdot d\vec{\ell}-\int_R^0\vec{E}_{in}\cdot d\vec{\ell}$$ This is the voltage from infinity to the outside of the sphere using the electric field outside the sphere plus the voltage from the outside of the sphere to the inside of the sphere using the electric field inside the sphere. I'm using $R$ for the radius of the sphere rather than $a$ since you seem to be using $a$ for the radial coordinate.

6. May 7, 2017

### Buffu

I get what you are saying but why I am wrong is the main question.

I think I am wrong in saying $3 Q^2/(5R)$ as you said in your first post but see http://farside.ph.utexas.edu/teaching/em/lectures/node56.html (line 601). It agrees with my proposition.

7. May 7, 2017

### Staff: Mentor

Be careful which formula you take from where. What the author is calculating there is the energy needed to form such a sphere from an initial state where the charge has a large separation. Your problem asks for something completely different - the electrostatic potential as function of the position.

8. May 7, 2017

### Daniel Gallimore

To clarify, I really should have said $$V_{in}=-\int_\infty^R\vec{E}_{out}\cdot d\vec{\ell}-\int_R^r\vec{E}_{in}\cdot d\vec{\ell}$$ where $r<R$ is some radius inside the sphere. Letting $r=R$ gives you the self energy like @mfb said, which isn't really what you want.

9. May 7, 2017

### Buffu

But right hand side of $\displaystyle U = {1\over 2} \int \rho \phi dv$ has self energy term, so I thought I just need to put my value in it.

When I am allowed to use this equation to get $\phi$ by differentiating ? or I am not allowed to do so.

10. May 8, 2017

### Staff: Mentor

=> don't use it, unless you want to calculate the self-energy.

11. May 8, 2017

### Buffu

Why ? It looks like a normal integral to me.

12. May 8, 2017

### Staff: Mentor

Yes, but the integral has nothing to do with the question you try to answer here.

13. May 9, 2017

### Buffu

$\displaystyle \int \rho \phi dv$ has a potential term. what does it mean here ??

14. May 9, 2017

### Staff: Mentor

mgh is a potential term as well. Not every potential that exists in physics is relevant for your problem.

The problem is way easier than your approaches here.

15. May 9, 2017

### Buffu

No I did not pick a "potential integral" from anywhere. $\phi$ here is Electrostatic potential not any potential in physics.

I want to know why this approach is wrong more than how to solve this.

Last edited: May 9, 2017
16. May 9, 2017

### ehild

Read the link you submitted in Post #6. Φ(r) is the potential on the surface of a sphere of radius r with homogeneous charge distribution. The sphere is surrounded with vacuum. You use this Φ to calculate the work needed to "fill" a sphere of radius R with charge. But you have to determine the potential inside and outside of a homogeneously charged sphere of radius R. They are different.

17. May 10, 2017

### Buffu

So If I were to determine potential on the surface of sphere, then I can differentiate and get the potential, like I did in original post.

18. May 10, 2017

### ehild

No. The potential in the integral is not the same as the potential of the final charged sphere.
The other thing, it is a volume integral, for a closed volume. You can not differentiate with respect to the volume.

19. May 10, 2017

### Buffu

You are correct I can't differentiate wrt volume.