Electric Potential Energy Question

In summary: The difference in PE is ΔPE=PE2-PE1=kq^2(1/r2-1/r1). As ΔKE+ΔPE=0, then μu22=-kq^2(1/r2-1/r1). Initially r1=5.0*10-12m and finally r2=∞. The distance between the particles is 2r and the reduced mass of the pair of particles is mu=m/2. kq^2/m is the Coulomb constant k=8.99*109 Nm^2/C^2. Combining all that, you get:μu22=-kq^2(1/r2-1/r1)u22
  • #1
MystiqThunder
7
0
1. Homework Statement
Two electrons start at rest with a separation of 5.0*10-12m. Once released, the electrons accelerate away from each other. Calculate the speed of each electron when they are a very large distance apart.

Homework Equations


ΔEK+ΔEE=0

The Attempt at a Solution


Okay so I know that:[/B]
  • The Electric force is in the same direction as the displacement, which means kinetic energy is increasing and electric energy is decreasing (ΔEK=-ΔEE)
  • As they move to a really far distance, the change in electric energy is equal to 0 - EE1 as anything divided by infinity is equal to zero
GIVENS:
mass of electron = 9.109*10-31 Kg
q = -1.60*10-19 C
k = 8.99*109
v1 = 0 m/s
r1 = 5.0*10-12 m

My process:
ΔEK+ΔEE=0
-ΔEE=ΔEK
-[kq1q2(1/r2-1/r1)]=½m(v2-v1)2
√[{-2(kq1q2(1/r2-1/r1)}/m]=v2
√[{-2(8.99*109)(-1.60*10-19)2(0-1/5.0*10-12)/(9.109*10-31)]=v2
1.0*107=v2

Is this correct?

Hopefully that shows it well, if not here is a picture (I write small, probably can't make out much):

EDIT: Looks like pic is a no show

IMG_20141119_213002.jpg



 
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  • #2
Looks to me as if you are moving only one charge to infinity. But the electrons are supposed to move away from each other.
 
  • #3
BvU said:
Looks to me as if you are moving only one charge to infinity. But the electrons are supposed to move away from each other.
You are absolutely correct, as I discovered today walking into class. The conservation of energy comes from the kinetic energy of not just one but both electrons MOVING AWAY from each other, as you just said. Thanks a lot anyways :)
 
  • #4
MystiqThunder said:
1. Homework Statement
Two electrons start at rest with a separation of 5.0*10-12m. Once released, the electrons accelerate away from each other. Calculate the speed of each electron when they are a very large distance apart.

Homework Equations


ΔEK+ΔEE=0

The Attempt at a Solution


Okay so I know that:[/B]
  • The Electric force is in the same direction as the displacement, which means kinetic energy is increasing and electric energy is decreasing (ΔEK=-ΔEE)
  • As they move to a really far distance, the change in electric energy is equal to 0 - EE1 as anything divided by infinity is equal to zero

My process:
ΔEK+ΔEE=0
-ΔEE=ΔEK
-[kq1q2(1/r2-1/r1)]=½m(v2-v1)2

The last equation is wrong. The potential energy of a system of two point charges is kq1q2/d, where d is the separation between them.
The electrons interact and there is no external force. So the centre of mass of the system stays in rest The kinetic energy of the system is the sum of the kinetic energies of both electrons, not the difference between them! Initially, d=5.0*10-12m
 
  • #5
ehild said:
The last equation is wrong. The potential energy of a system of two point charges is kq1q2/d, where d is the separation between them.
The electrons interact and there is no external force. So the centre of mass of the system stays in rest The kinetic energy of the system is the sum of the kinetic energies of both electrons, not the difference between them! Initially, d=5.0*10-12m

If you look at how he is using it, this is not the problem, r1 and r2 correspond to the distance between the electrons at two different times. The problem, as BvU noted, is in the setup and treatment of the kinetic energy.

Mystiq: The expression for the kinetic energy difference is wrong not only because you forgot about one of the electrons. The difference should be proporional to v1^2 - v2^2, not (v1-v2)^2. In this case, one of the velocities was zero so it did not make a difference, but it is something you will generally need to consider.
 
  • #6
Sorry. I see now that r1 and r2 were distances at two different times.
The two-electron system can be replaced by a single particle with reduced mass, μ=m/2 if the particles are identical. The velocity of that "imaginary" particle is equal to the relative velocity u=va-vb of the real particles, while the CoM of the system is stationary. As they are identical va=-vb=v and u=2v. The kinetic energy is then KE=1/2 μ u^2, and the change of the kinetic energy is ΔKE=1/2 μ(u22-u12).
As the particles were in rest initially, u1=0, so the change of the KE is ΔKE=1/2 μu22.
 
Last edited:

Related to Electric Potential Energy Question

1. What is electric potential energy?

Electric potential energy is the energy that an object possesses due to its position in an electric field. It is the potential energy that a charged particle has when placed in an electric field.

2. How is electric potential energy calculated?

Electric potential energy is calculated by multiplying the charge of the object by the electric potential at its location. The formula for electric potential energy is U = qV, where U is the electric potential energy, q is the charge, and V is the electric potential.

3. What is the difference between electric potential and electric potential energy?

Electric potential is the electric potential energy per unit charge, while electric potential energy is the energy that an object has due to its position in an electric field. In other words, electric potential is a measure of the strength of the electric field at a specific point, while electric potential energy is a measure of the amount of work required to move a charged object to a specific point in the electric field.

4. How does electric potential energy change in an electric field?

In an electric field, electric potential energy can either increase or decrease. If the electric field is doing work on a charged object, the electric potential energy will decrease as the object moves towards the lower potential. Conversely, if the charged object is doing work on the electric field, the electric potential energy will increase as the object moves towards the higher potential.

5. What factors affect the electric potential energy?

The electric potential energy is affected by the amount of charge on the object, the distance between the object and the source of the electric field, and the strength of the electric field. The greater the charge and the closer the object is to the source, the higher the electric potential energy will be. Additionally, the stronger the electric field, the greater the change in electric potential energy will be as the object moves through the field.

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