- #1

Sourabh N

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- Thread starter Sourabh N
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In summary: But WHO exactly uses this meaningless question so often?" In summary, the potential energy of a uniformly charged spherical shell is given by the formula U = Q^2/8*pi*epsilon_0*r. This potential energy is independent of the radius of the shell. However, it can only be partially extracted, as some of the charge must remain within the shell. The concept of "self-energy" is a convention and can be set to any level. The general formula for energy for a charge distribution is ½ [Integral over volume rho*V*(dxdydz)], and for a spherical shell, it is W= ½ Integral over surface( s*V*da).

- #1

Sourabh N

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Physics news on Phys.org

- #2

pam

- 458

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The PE is QV/2.

- #3

rohanprabhu

- 414

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How is the potential energy independent of radius?? I'm getting something like:

[tex]

U = \frac{\sigma^2 \pi r^5}{9\epsilon_0}

[/tex]

where, [itex]\sigma[/itex] is the charge density of the shell.

OR:

[tex]

U = \frac{Q^2}{8\pi \epsilon_0 r}

[/tex]

[tex]

U = \frac{\sigma^2 \pi r^5}{9\epsilon_0}

[/tex]

where, [itex]\sigma[/itex] is the charge density of the shell.

OR:

[tex]

U = \frac{Q^2}{8\pi \epsilon_0 r}

[/tex]

Last edited:

- #4

pam

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V=Q/4piepsilonzero r

- #5

Ulysees

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How low can the potential go = how much energy we can extract from the sphere.

Can we squash it?

Can we stretch it?

Can we squash it?

Can we stretch it?

Last edited:

- #6

Shooting Star

Homework Helper

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Ulysees said:

Can we squash it?

Can we stretch it?

??

The higher the potential energy, the more we can extract (if possible). If we squash it, we have to put in energy. By allowing it to stretch, we can get energy out of it.

- #7

Ulysees

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If it is insulating and rigid so charges cannot move, you can get or give 0 energy.

If it is conducting but rigid, charges can only move inside the conductor. Therefore you can only GIVE some energy to the charges by forcing them to redistribute under external force.

So what is the "self-energy" of the shell then? By convention it's the energy that the charges lose if they are allowed to repel each other to infinity. Only by convention. Potential energy can be set to whatever level you want, it's only the changes that matter, what you get or what you give.

- #8

Ulysees

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Ill-stated question:

How much gravitational potential energy does the universe

- #9

Lojzek

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The second formula is right, the first one is wrong. The question is about a sphere, not a ball, so one must insert Q=4*Pi*sigma*r*r, sorohanprabhu said:

[tex]

U = \frac{\sigma^2 \pi r^5}{9\epsilon_0}

[/tex]

where, [itex]\sigma[/itex] is the charge density of the shell.

OR:

[tex]

U = \frac{Q^2}{8\pi \epsilon_0 r}

[/tex]

the energy is:

E=2*Pi*r^3*sigma^2/epsilon

- #10

Shooting Star

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Ulysees said:Whether you can get energy out or give energy to the system depends on whether the charges can move.

The total PE, calculated with the potential set to zero at infinity, can never be extracted, because we can never send all of the charge to infinity, but

It can be broken up and the KE of the mutually repellent flying pieces may be collected to get some energy. Also, how did you get the charges in there in the first place? If that had required work to be done, perhaps the process can be reversed and the energy collected in a neater manner.If it is insulating and rigid so charges cannot move, you can get or give 0 energy.

If it is conducting but rigid, charges can only move inside the conductor. Therefore you can only GIVE some energy to the charges by forcing them to redistribute under external force.

Just connect a wire to it and ground it, and collect part of the energy as the current flows.

Who's saying anything to the contrary?So what is the "self-energy" of the shell then? By convention it's the energy that the charges lose if they are allowed to repel each other to infinity. Only by convention. Potential energy can be set to whatever level you want, it's only the changes that matter, what you get or what you give.

Ulysees said:hasthis much energy" which is not strictly correct. Here's why.

Ill-stated question:

How much gravitational potential energy does the universehave?

I agree with you. But

- #11

Sourabh N

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Well, thanks for the answers and comments, but was expecting the method to derive the equation.

- #12

Ulysees

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You know as well as I know that current means the charge has to go somewhere. It can't go to infinity in all directions. So you can't ever extract the full energy that we have (by convention) associated with the shell.

>> Ill-stated question: how much gravitational potential energy does the universe

> I agree with you. But

You must have missed the point. It's the bit in bold ("

- #13

Shooting Star

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Sourabh N said:Well, thanks for the answers and comments, but was expecting the method to derive the equation.

You never asked for the method...

The general formula for energy for a charge distribution is ½ [Integral over volume rho*V*(dxdydz)], where V is the potential at a point. To modify that for a spherical shell, we have to find W= ½ Integral over surface( s*V*da), where s is the surface charge density. The total charge is q and the radius is R., e = epsilon_nought.

V at the surface of the sphere = k*q/R, where k = 1/(4*pi*e). Then,

W = ½ kq/R Integral s*da = k/2(q^2/R) = [1/(8*pi*e)]*q^2/R.

- #14

Shooting Star

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Ulysees said:You know as well as I know that current means the charge has to go somewhere. It can't go to infinity in all directions. So you can't ever extract the full energy that we have (by convention) associated with the shell.

In post #10, for that very reason, I have written

You must have missed the point. It's the bit in bold ("have") Not whether the whole question profoundly concerns the whole of humanity or something. "How much potential energy" is like saying "how close together can masses come theoretically".

Hey, I do get the point. I was just kidding a bit by asking

Cheers.

The formula for calculating the potential energy of a charged spherical shell is given by U = Q^2 / (4πε0R), where Q is the charge of the shell, ε0 is the permittivity of free space, and R is the radius of the shell.

The potential energy decreases as the distance from the charged spherical shell increases. This is because the electric field strength decreases with distance, resulting in a weaker force and, consequently, a lower potential energy.

Yes, the potential energy of a charged spherical shell is affected by the presence of other charges. This is because the electric field created by the other charges can interact with the electric field of the shell, leading to changes in the potential energy.

Yes, the potential energy of a charged spherical shell can be negative. This can occur when the shell has a negative charge, and it is placed in an external electric field with the opposite direction. In this case, the potential energy will be negative, indicating that work is done by the external electric field to bring the shell to that position.

The potential energy of a charged spherical shell is directly proportional to its electric potential. This means that an increase in the electric potential will result in an increase in the potential energy, and vice versa. The exact relationship between the two is given by U = QV, where V is the electric potential.

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