Could charges on a sphere cause it to explode?

In summary, the conversation discusses the potential due to a uniformly charged sphere and the distribution of free charges on its surface. The question is raised about the possibility of the sphere exploding due to repelling forces between the charges, but it is explained that the forces are in equilibrium and the net force is the same in all directions. The topic of parallel plate capacitors is also briefly mentioned, with the comparison of finding the force on one plate to finding the force on a uniformly charged sphere. The conversation also touches on the application of this concept in the physics of water droplets and the potential for survival and growth with the presence of a net charge. Finally, the conversation ends with a discussion about the self-capacitance of a sphere and the
  • #1
manofphysics
41
0
Ok, this question may sound stupid to some. But still, I 'll ask.

In electrostatics we frequently discuss, potential due to uniformly charged sphere.where the whole sphere's surface has positive or negative charge.
So, this means, free charges of the same type(+ve or -ve) have been distributed all over the surface.

Now, my question is , won't the sphere explode because of repelling forces between the free charges which are present on the surface of the sphere(as they are all postive or all negative)?
 
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  • #2
Well if the charges are distributed evenly over the surface, then the repulsive force are not radial.

The forces acting on the charges are in equilibrium so everything is balanced hence if the sphere. The equilibrium exists because no matter in which direction you look from a particular charge you see the same number of charges at every distance. So the net force is the same in all directions I believe.
 
  • #3
If charges are distributed uniformly over the surface of a sphere, they repel each other and this manifests in a decrease of the pressure in the sphere (there is net radial outward force proportional to surface area). Try to calculate the force! For this you must find the intensity right at the radius where charges are.
 
  • #4
I would also like to ask, what about the free charges on one plate of a parallel plate capacitor?
 
  • #5
That's the same. You just find the intensity at the first plate (caused by the second plate) and then multiply it by the total charge on the first plate and you get the force. However, the sphere is a bit more interesting. This has an application in the physics of water droplets. When the droplet starts forming in the air, it is very small and thus has very high pressure inside due to capillarity. This leads to high rate of evaporation of molecules from the droplet and it is likely that such a droplet would quickly vaporize. But imagine that our droplet contains some net charge. Then this charge will distribute on its surface and will cause decrease of the pressure inside and therefore to decrease of evaporation rate. Then this droplet can more easily survive and starts to grow. That's probably how Wilson chamber works - charged particles facilitates droplet forming.
 
  • #6
Thanks very much for the answers , Jano L.
 
  • #7
There are two 1 Farad (1 million microfarad), 20 volt aluminum capacitors (3 " diameter) for sale at www.digikey.com. (1/2)C V^2 = 200 joules; not very much energy.
[Edit] According to Wiki, the self capacitance of a sphere of radius R is
C = 4 pi e0 R, so for a sphere of radius 0.5 meters
the capacitance C = 56 picoFarads.
If we took the charge (Q = CV = 20 Coulombs) on the 1 Farad capacitor above, and charged this sphere, the voltage would then be

V = Q/C =20 Coulombs/56 picoFarads = 357 billion volts.

So 20 Coulombs is a lot of charge, especially in a 3" diameter capacitor.
See
http://en.wikipedia.org/wiki/Capacitance#Self-capacitance
 
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What is the concept of free charges on a sphere?

Free charges on a sphere refer to the distribution of electric charges on the surface of a spherical object, where the charges are free to move and redistribute themselves.

What is the equation for calculating the electric field of free charges on a sphere?

The electric field at any point on the surface of a sphere due to free charges can be calculated using the equation E = Q/(4πε0r2), where Q is the total charge on the sphere, ε0 is the permittivity of free space, and r is the distance from the center of the sphere to the point of interest.

How does the distribution of charges on a sphere affect the electric field?

The distribution of charges on a sphere affects the electric field by changing the direction and magnitude of the field at different points on the surface. This is due to the fact that the electric field is stronger near areas with a higher concentration of charges.

What is the difference between free charges and bound charges on a sphere?

Free charges are those that are able to move and redistribute themselves on the surface of a sphere, while bound charges are those that are fixed in place and cannot move. Bound charges are typically found within the material of the sphere, while free charges are located on the surface.

How does the electric potential of a sphere with free charges compare to one with bound charges?

The electric potential of a sphere with free charges is typically higher than that of a sphere with bound charges, as the free charges are able to move and redistribute themselves to create a stronger electric field. This results in a higher potential energy at the surface of the sphere.

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