EK Physics in book question 149. Electrostatics

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149. A positively charged particle starts at rest 25 cm 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 25 cm, it has
reached a velocity of 10 m/s. What is the maximum
velocity that the first particle will reach?
150.
A. 10 m/s
B. 14 m/s
C. 20 m/s
D. Since the first particle will never escape the electric
field of the second particle, it will never stop accelerating, and will reach an infinite velocity.


I cant seem to understand their explanation, i just guess that it would be B since its unreasonable for it max at 10 and D is definitely wrong.. I kinda get the process for the answer but i need a better explanation....

for those with the book, this question is not part of the 30 min exams but questions in the book along with the content.

 

[Rabolisk]

This is a tough problem that requires understanding of many concepts. Potential or voltage due to a single charge q is kq/r, where r is the distance from that charge. You start off 25 cm, and by the time the particle has moved 25 cm, r has increased from 25 to 50 cm. Now the potential at 50 cm away is half that at 25 cm away. kq/25 = 2 times kq/50. Now remember that potential is just potential energy divided by the charge of the particle you're concerned with. Since potential has decreased to half its initial value, potential energy has decreased to half its initial value. ALL potential energy is converted to kinetic energy since electrostatic forces are conserved. You've "used" up half of your potential energy, and to accelerate away to infinity you can use the other half. The change in kinetic energy between 25 and 50 cm is thus equal to the change in kinetic energy between 50 cm and infinity. Now, kinetic energy is proportional to velocity squared. So velocity would not double to 20 m/s, but rather be multiplied by radical 2 to arrive at 14 m/s.

I'm sorry if that was a very convoluted explanation. I wish I could explain it better.

 

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i get what your talking about but it dont understand that very last part about multiplying by radial 2.. anyone else got some input?

 

[Rabolisk]

All right, so going from 25cm to 50cm takes the same amount of potential energy as going from 50cm to infinity. In another words, the kinetic energy gained as the particle travels from 25 to 50 is the same as that gained from 50 to infinity.

That can be proven by using V = q1q2/r. You can see that voltage at 50 cm is half of that at 25 cm, or to put it another way, it is halfway between the voltage at 25 cm and voltage at infinity (which is 0).

Now we relate voltage to velocity. The change in voltage times the charge of the particle gives the change in work, potential energy, or kinetic energy. (Work-kinetic energy theorem) The question states that the change in velocity was 10 m/s. Relating kinetic energy to velocity we have mv^2/2=KE. Now, the same kinetic energy has to be added onto this particle. Hence the particle at infinity has TWICE the kinetic energy than it did at 50cm. Mass is constant, so velocity squared must double. That means that velocity has to be multipled by sqrt of 2.