How Close Will Two High-Speed Charged Particles Get?

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In summary, the problem involves calculating the minimum separation of two particles with an initial velocity of 3.0 X 10^6 m/s and a mass of 6.6 X 10^-27 kg, assuming no deflection from their original path. The equation used is Fe = kq1q2/r^2 and the phrase "separated by an enormous distance" may affect a term in the energy equation. Further calculations are needed to determine the final answer.
  • #1
Bradsteeves
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Homework Statement


Two particles, separated by an enormous distance, approach each other. Each has an initial speed of 3.0 X 10^6 m/s. Calculate their minimum separation, assuming no deflection from their original path. The mass of a particle is 6.6 X 10^-27 kg.

Particle charge = 3.2 X 10^-19 C

Given:
V = 3.0 X 10^6m/s
mass of particle = 6.6 X 10^-27 kg
r = ?




Homework Equations


E = Eprime
Fe = kq1q2/r^2



The Attempt at a Solution


EE + EK = EEprime + EKprime
Kq1q2/r + 1/2 mv^2 = Kq1q2/r + 1/2m(vprime)^2

Dont know where to go from there
 
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  • #2
How does the phrase "separated by an enormous distance" affect a term in the energy equation?

When you plug in numbers to your equation, what do you get? (I noticed you combined the kinetic energy of both particles into one term on each side, so you'll have to adjust the mass to deal with that.)
 
  • #3



Hello, based on the information provided, we can use the equation for electrostatic force (Fe = kq1q2/r^2) to solve for the minimum separation between the two particles. However, since we are assuming no deflection from their original path, we can also use the conservation of energy equation (EE + EK = EEprime + EKprime) to solve for the minimum separation.

To do this, we first need to find the kinetic energy (EK) of the particles. We can use the formula EK = 1/2mv^2, where m is the mass of the particle and v is its initial speed. Plugging in the given values, we get EK = 1/2(6.6 X 10^-27 kg)(3.0 X 10^6 m/s)^2 = 2.97 X 10^-11 J.

Next, we can use the equation for electrostatic force (Fe = kq1q2/r^2) to solve for the minimum separation (r). We can rearrange the equation to get r = √(kq1q2/Fe), where k is the Coulomb's constant (9 X 10^9 Nm^2/C^2), q1 and q2 are the charges of the particles, and Fe is the electrostatic force between them.

Plugging in the given values, we get r = √[(9 X 10^9 Nm^2/C^2)(3.2 X 10^-19 C)^2/(2.97 X 10^-11 J)] = 3.06 X 10^-12 m.

Therefore, the minimum separation between the two particles is approximately 3.06 X 10^-12 m. I hope this helps! Let me know if you have any further questions.
 

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