The Effect of Electric Field on Proton Motion

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SUMMARY

The discussion centers on the motion of a proton in an electric field, specifically addressing the relationship between electric potential energy and electric potential. The correct answer to the question posed is option d: as the proton moves in the direction of the electric field, both its potential energy and electric potential decrease. This conclusion is supported by the principle that positive charges accelerate toward regions of lower potential, analogous to a ball falling in a gravitational field. The relevant equations discussed include electric potential energy U = k(q1*q2)/r and electric potential V = k(q_not proton)/r.

PREREQUISITES
  • Understanding of electric potential energy and electric potential
  • Familiarity with Coulomb's law and its application
  • Knowledge of the behavior of charged particles in electric fields
  • Basic grasp of physics concepts related to potential energy and kinetic energy
NEXT STEPS
  • Study the relationship between electric potential and electric potential energy in detail
  • Learn about the implications of Coulomb's law on force and motion of charged particles
  • Explore the concept of electric fields and their effects on different charges
  • Investigate the analogy between electric fields and gravitational fields in terms of potential energy
USEFUL FOR

Students of physics, educators teaching electromagnetism, and anyone interested in the dynamics of charged particles in electric fields will benefit from this discussion.

Soaring Crane
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For a proton moving in the direction of the electric field

a. its potential energy increases and its electric potential decreases
b. its potential energy increases and its electric potential increases
c. its potential energy decreases and it electric potential increases
d. its potential energy dcreases and its electric potential decreases


Originally, I thought the answer was b, but this is wrong. Exactly why is b incorrect and what exactly is the correct choice?

Thanks.
 
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electric potential energy U = k \frac{q_{1} q_{2}}{r}
electric potential V = k \frac{q_{not proton}}{r}

since the proton moves in line with the field, it is moving + to - or vice versa??
Now that you knwo the direction you can see what happens in the initial and final conditions
calculate the electric potential energy and electric potential of the proton when it close to the positive end and the negative end, assuming that there are TWO charges of opposite sign creating this field. You know the direction which is bigger than which??
if the values went down then it obviously decreased and if the values went up then increase
 
Hey stunner. A couple questions:
What's "k"?
And what does the charge in the electric potential equation stand for if its not that of the p+?
 
Soaring Crane said:
Originally, I thought the answer was b, but this is wrong. Exactly why is b incorrect and what exactly is the correct choice?
Here's a hint: Positive charges accelerate toward regions of lower potential. (Assuming there are no other forces acting on them.) How do potential and potential energy relate?

Compare this to a ball falling in a gravitational field. Which way does the field point? Where is the PE greater?
 
DB said:
What's "k"?

k = \frac{1} {4 \pi \epsilon_{0}}
 
Ahh, so electrical potential energy is the same as coulombs law?
 
No, Coulomb's law is for the force between two charges.
 
Well, since that hint is true, is it d where both decrease?
 
Soaring Crane said:
Well, since that hint is true, is it d where both decrease?
That's right. The field points toward the region of lower potential. The proton accelerates as it moves through the field, losing potential energy and gaining kinetic energy. (Assuming no other forces act.)
 

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