Electric Potential Energy problem

In summary: Two particles each have a mass of 6.1*10-3 kg. One has a charge of +5.0*10-6 C, and the other has a charge of -5.0*10-6 C. They are initially held at rest at a distance of 0.80 m apart. Both are then released and accelerate toward each other.The particle with the charge of -5.0*10-6 C moves faster than the particle with the charge of +5.0*10-6 C.
  • #1
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"Two particles each have a mass of 6.1*10-3 kg. One has a charge of +5.0*10-6 C, and the other has a charge of -5.0*10-6 C. They are initially held at rest at a distance of 0.80 m apart. Both are then released and accelerate toward each other. How fast is each particle moving when the separation between them is one-half its initial value?"

What equations would I need to use in order to solve this problem? I tried using EPE/q=Va-Vb, but the voltage cancels out, so I'm unsure as to where to go from there. Thanks.

~Futron
 
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  • #2
Use kinetic and potential energy relationships. Initially they both have no kinetic energy, but a certain amount of potential energy. When d = d/2 they have a certain about of potential and kinetic energy. Find the potential energy in both cases, any potential energy lost would have been translated into kinetic energy.
 
  • #3
Alright, I tried using EPEi=EPEf+(1/2)mv^2 where EPE=q0*(Va-Vb), but that doesn't seem to work. Any ideas?
 
  • #4
[itex] EPE_f-EPE_i = mv^2/2, and EPE = q(V_b-V_a) [/itex] are the amended versions of your equations. Find the potential with [itex] U = qE\Delta x[/itex]

This will take you in circles though.

The force between the charges is given by coulombs law. Find the difference in potential energy at d = r and d = r/2.
 
  • #5
So if I have [itex] EPE_a-EPE_b=q_0(V_A-V_B) [/itex] where [itex] V=kQ/r [/itex], how would I then find the difference between the two when the distance is halved?
 
  • #6
[itex] U = qE\Delta x[/itex]

The field for point charges is [itex] E = \frac{kq}{r^2} [/itex]

[tex] \Delta x_i = 0.8m, \Delta x_f = 0.4m [/tex]
 
  • #7
Thanks, but I'm still not getting the correct answer. Which charge should I use for [itex] q [/itex] in [itex] EPE=qE\Delta x [/itex], and once I get that, would it simply be a matter of solving [itex] EPE_f-EPE_i=(1/2)mv^2 [/itex]?
 
  • #8
Can you show some work? In your first post you said the voltages cancel out which isn't correct, this would only happen if [itex] V_b = v_a [/itex] which isn't true.

Show me where your messing up, all these techniques should give the same answer, if you prefer voltages we can use that.
 
  • #9
Alright, I tried using EPEf-EPEi=(1/2)mv^2 where EPEf=(5x10^-6)((9*10^9)*(-5*10^-6)/0.4)) and EPEi=(5*10^-6)x((9x10^9)(-5x10^-6)/0.8)), but that gives 5.43m/s, which is not the correct answer.
 
  • #10
I think that the potential energy state is [tex]U=k_{e}q_1q_2/r[/tex], you'll need to find [tex] \Delta U=k_eq_1q_2(1/r_{initial}~-1/r_{final})[/tex], this will equal the negative value of the change in kinetic energy [tex]=-mv^{2}_{final}-0[/tex]
 
  • #11
Thanks! I was jumbling too many equations together at once but that made it much clearer.
 

1. What is electric potential energy?

Electric potential energy is a type of potential energy that is associated with charged particles in an electric field. It is the amount of work needed to move a charged particle from one point to another in an electric field.

2. How is electric potential energy calculated?

Electric potential energy is calculated using the equation U = qV, where U is the electric potential energy, q is the charge of the particle, and V is the electric potential at the location of the particle.

3. What factors affect electric potential energy?

The factors that affect electric potential energy include the distance between the charged particles, the magnitude of the charges, and the medium in which the charges are located.

4. How is electric potential energy related to electric potential?

Electric potential energy is directly related to electric potential. Electric potential is the electric potential energy per unit charge, or V = U/q. This means that as electric potential energy increases, electric potential also increases.

5. What are some real-life applications of electric potential energy?

Electric potential energy has many real-life applications, including in batteries, electric motors, and capacitors. It is also used in power plants to generate electricity and in electrical circuits to power devices. Additionally, electric potential energy is important in understanding the behavior of lightning and static electricity.

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