Physics

[marspotter]

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Hi. y'all.

I have this question EK physics #149 that I have tough time understanding.

Question goes,

A positively charged particle starts at rest 25cm from a second positively charged particle which is held stationary throughout the experiment. The first particle is released and accelerates directly away from the second particle. When the first particle has moved 25cm, it has reached a velocity of 10 m/s. What is the maximum velocity that the first particle will reach?

I would really appreciate a thorough explanation.

 

[mmc48]

The easiest way to look at it is by analyzing the energies present within the system of interest.

Holding two positive charges at any distance near each other (or 2 negative charges for that matter) constitutes a potential energy within the system. The particles wish to move away from each other so as to minimize potential energy (like a ball at the surface of the earth being held at 10 m above the ground has potential energy due to the gravitational force of the earth). The potential energy between two charged particles can be calculated by U=kqq/r. Since the initial radius is 25 cm and then doubles to 50 cm, look at what happens to the equation. Radius increases by 2, and since radius is inversely proportional to U, then U decreases by a factor of 2.

Now a decrease in potential energy U is directly proportional to an increase in kinetic energy (KE). Think about a ball falling towards the earth. It loses potential energy as it falls and gains kinetic energy. Same thing here. As the 1st particle moves away, it gains KE and loses U. So KE increases by a factor of two. The kinetic energy equation is KE=.5mv^2. If KE doubles, then velocity increases by a factor of 1.4 (the square root of 2 is 1.4 and you assume mass stays constant). Multiply by a factor of 10 and you get 14 m/s.

 

[marspotter]

Thank you so so much.

Now that I have overlooked thoroughly with your explanation, it was just the conservation energy type. You made my day!!