Understanding Electric Potential and Kinetic Energy in Particle Movement

In summary: I think the author is just trying to illustrate an important point. In summary, the book discusses general physics topics such as electric fields, optics, waves, and more, but does not go into detail on the equations or solving sample problems. The book focuses more on history than on explaining equations or sample problems. The author notes that the relativistic formula is more useful for solving problems than the Newtonian formula, but the answer given in the book for the speed of a proton is incorrect.
  • #1
physicsuser
82
1
I am reading a book about general physics kinematics, optics, waves, etc.
It focuses more on history than on explaining equations or sample problems.

One of the sample problems in Electric Fields section asks to find energy and speed when a positron moves from one plate that has electric potential of 1e6 V to another that has zero electric potential.

As far as the energy is concerned it is the potential which equals
U=1e6 x 1.6e-19 = 1.6e-13 which is what the answer is at the back of the book states.

But for the speed I get a different answer.
I assume that potential is turned into kinetic as positron reached the other plate so:
Ek=0.5mvv

then v=sqr(2Ek/m)

m=9.1e-31 Kg
Ek=1.6e-13

so from above v=5.9e8

but in the book the answer is 1.4e7m/s


What am I missing or not understanding?
 
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  • #2
physicsuser said:
But for the speed I get a different answer.
I assume that potential is turned into kinetic as positron reached the other plate so:
Ek=0.5mvv

then v=sqr(2Ek/m)

m=9.1e-31 Kg
Ek=1.6e-13

so from above v=5.9e8

but in the book the answer is 1.4e7m/s


What am I missing or not understanding?

This one had me scratching my head for a bit, until I noticed that 5.9e8 m/s is about twice the speed of light.

So what you need to do is to use the realtivistic formula for energy, not the Newtonian one.

The relativistic formula will be

mec^2 / sqrt(1-(v/c)^2) = U + mec^2

It's simpler if you remember (or look up) the fact that mec^2 is .51 Mev (million electron volts)

But I still don't get the answer the book quoted, so maybe something else is wrong here.
 
  • #3
Ok sorry I found what is wrong... the problem started by saying that the positron is moving but ended up asking the speed and energy of a proton ...

the author must be laughing at how stupid some people are...

Sorry about this.
 

1. What is an electric field?

An electric field is a physical field that is created by electrically charged particles and affects other charged particles in its vicinity. It is a vector field, meaning it has both magnitude and direction.

2. How is an electric field calculated?

The magnitude of an electric field at a certain point is calculated by dividing the force experienced by a test charge at that point by the magnitude of the test charge. The direction of the electric field is always in the direction of the force experienced by the test charge.

3. What are some common problems involving electric fields?

Some common problems involving electric fields include finding the electric field strength at a specific point, determining the force experienced by a charge in an electric field, and calculating the potential difference between two points in an electric field.

4. How do you represent an electric field graphically?

An electric field can be represented graphically using field lines, which show the direction and strength of the field at different points. The closer the field lines are to each other, the stronger the electric field at that point.

5. What are some real-world applications of electric fields?

Electric fields have many practical applications, such as in electrical circuits and devices, electrostatic precipitators used in air pollution control, and particle accelerators used in scientific research. They also play a crucial role in the functioning of our nervous system and muscles.

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