Electric potential energy of two charged spheres

In summary, two insulating spheres with radii of 0.3cm and 0.5cm, masses of 0.1kg and 0.7kg, and charges of -2 \muC and 3 \muC are released when they are 1m apart. The question is how fast each sphere will be moving when they collide. The radius of the spheres is used for calculating the potential energy of the system after the collision. The electric potential energy is not equivalent to anything, but the difference in electric potential energy is equal to the total kinetic energy of the system. If the spheres were conducting, the separation of charges would make the problem much harder to solve. The potential energy lost by the system is
  • #1
semc
368
5
Two insulating spheres have radii 0.3cm and 0.5cm, masses 0.1kg and 0.7kg and uniformly distributed charges of -2 [tex]\mu[/tex]C and 3 [tex]\mu[/tex]C. they are released when the distance from their center are 1m apart. How fast will each be moving when they collide.

Firstly, what is the radius of the sphere use for? Secondly, I calculated the electric potential energy of the system but what is this energy equivalent to? Lastly, how can we solve this question if the spheres were conducting spheres?
 
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  • #2
Okay, well, first let me say that I'm not sure how to answer your question, but I would like to think through it a bit myself. The electrical potential energy at the beginning should be equal to the sum of the kinetic energies of the spheres at the end. I think conservation of momentum still holds, so using the energy and momentum equations I think one should be able to calculate both velocities. I'm not sure how the radius of the spheres comes into play either. It would affect the potential energy of each spherical charge distribution, but that doesn't seem like it should be relevant to me since the spheres aren't changing, so we can set our zero of potential energy at the total potential energy inherent within the spheres. If the spheres were conducting, it seems to me that the charge on each sphere should redistribute itself such that field is zero and the spheres don't move. Is this possible?

Sorry for the long, useless response, but I would like an answer to your question as well!

Kelly
 
  • #3
That's what I had in mind at first, so do you think the energy lost by the system is equivalent to the kinetic energy gain by the sphere? So can we treat one sphere as stationary and just calculate the relative velocity of the moving sphere? I think if the spheres are conducting they would polarize each other as they move together?
 
  • #4
semc said:
Firstly, what is the radius of the sphere use for? Secondly, I calculated the electric potential energy of the system but what is this energy equivalent to? Lastly, how can we solve this question if the spheres were conducting spheres?

The radii are used for calculating the potential energy of the system after the collision. The electric potential energy isn't equal to anything, but the difference in electric potential energy is equal to the total kinetic energy of the system, since total energy is conserved.

If the spheres were conducting, the separation of charges would make the problem much harder to solve. I don't think the solution is trivial.
 
  • #5
semc said:
That's what I had in mind at first, so do you think the energy lost by the system is equivalent to the kinetic energy gain by the sphere? So can we treat one sphere as stationary and just calculate the relative velocity of the moving sphere?

The potential energy lost by the system is equal to the kinetic energy gained by TWO spheres. You can't treat one sphere as stationary because in any inertial reference frame, it isn't.

I think if the spheres are conducting they would polarize each other as they move together?

Yes. That would make the problem extremely difficult to solve.
 
  • #6
So if we use [tex]\Delta[/tex]U = [tex]\frac{1}{2}[/tex]m1v12 +[tex]\frac{1}{2}[/tex]m2v22 and conservation of momentum m1v1= -m2v2 simultaneously is able to solve it? I am not able to get the answer?
 
Last edited:
  • #7
Yes, that's how you do it. You should be able to get an answer. How did you try to solve the equations?
 
  • #8
ke [tex]\frac{Qq}{r^2}[/tex]= [tex]\frac{1}{2}[/tex] m1v12 + [tex]\frac{1}{2m2}[/tex] (m1v1)2
The distance r is 1m?

I gotten v1 to be 0.793m/s but its not the answer. So how do we use the radius of the sphere?
 

1. What is electric potential energy?

Electric potential energy is the potential energy that a charged object possesses due to its position in an electric field. It is the energy that is required to move a charged object from one point to another in the presence of an electric field.

2. How is the electric potential energy of two charged spheres calculated?

The electric potential energy of two charged spheres is calculated using the formula U = k(Q1Q2)/r, where k is the Coulomb's constant, Q1 and Q2 are the charges of the two spheres, and r is the distance between them.

3. How does the distance between two charged spheres affect their electric potential energy?

The electric potential energy between two charged spheres is inversely proportional to the distance between them. This means that as the distance between the spheres increases, the electric potential energy decreases, and vice versa.

4. Can the electric potential energy of two charged spheres be negative?

Yes, the electric potential energy of two charged spheres can be negative. This happens when the two spheres have opposite charges, and the potential energy is released as they move towards each other.

5. How does the charge of the spheres affect their electric potential energy?

The electric potential energy between two charged spheres is directly proportional to the product of their charges. This means that as the charges of the spheres increase, the electric potential energy also increases, and vice versa.

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