Electric Potential/Potential Energy problem

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SUMMARY

The discussion focuses on solving two physics problems involving electric potential and potential energy. In problem (a), a particle with a mass of 6 grams and charge of 54 μC is fired at 25 m/s towards another stationary particle with a mass of 13 grams and charge of 22 μC, requiring the calculation of the distance of closest approach. Problem (b) involves four identical particles, each with a charge of 4 μC and mass of 2 grams, released from the vertices of a square with a side length of 42 cm, and requires determining their speed when their distance from the center triples. Key equations include conservation of energy (ΔU + ΔK = 0) and the potential energy formula (U = kQ1Q2/r).

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Homework Statement



a) From a large distance away, a particle of mass 6 grams and charge 54 μc is fired at 25 m/s to the right straight toward a second particle, originally stationary but free to move, with mass 13 grams and charge 22 μc. Find the distance of closest approach between the charges.

b) Four identical particles each have charge of 4 μC and mass 2 grams. They are released from rest at the vertices of a square of side 42 cm. How fast is each particle moving when their distance from the center of the square triples?

Homework Equations


a) ΔU + ΔK = 0
b) U=(kQ1Q2)/r , ΔU + ΔK = 0, K = (1/2)mv2, V = (KQ)/r

The Attempt at a Solution


a) not sure how to do this problem other than knowing that Vi = 25 m/s for the 6 gram particle and Vi = 0 for the 13 gram particle.
 
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jawhnay said:

Homework Statement



a) From a large distance away, a particle of mass 6 grams and charge 54 μc is fired at 25 m/s to the right straight toward a second particle, originally stationary but free to move, with mass 13 grams and charge 22 μc. Find the distance of closest approach between the charges.

b) Four identical particles each have charge of 4 μC and mass 2 grams. They are released from rest at the vertices of a square of side 42 cm. How fast is each particle moving when their distance from the center of the square triples?


Homework Equations


a) ΔU + ΔK = 0
b) U=(kQ1Q2)/r , ΔU + ΔK = 0, K = (1/2)mv2, V = (KQ)/r


The Attempt at a Solution


a) not sure how to do this problem other than knowing that Vi = 25 m/s for the 6 gram particle and Vi = 0 for the 13 gram particle.
For (a), how can you describe the energy at their closest approach?

(b) can be solved in a similar way.
 
I think (a) also requires conservation of momentum.
 
mfb said:
For (a), how can you describe the energy at their closest approach?

(b) can be solved in a similar way.

i'm not sure... my knowledge of energy is kind of rusty since i haven't taken mechanics in a while...
 
voko said:
I think (a) also requires conservation of momentum.
Right
jawhnay said:
i'm not sure... my knowledge of energy is kind of rusty since i haven't taken mechanics in a while...
Then you should try to fix that. While it is possible to solve it without the concept of energy, this would require solving differential equations, which is NOT easier.
 

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