The work done by an electric field in accelerating a charged particle?

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SUMMARY

The discussion focuses on the work done by an electric field in accelerating a charged particle with a charge of 3.2x10^-19 C in a uniform electric field of 39000 N/C between two plates spaced 4.5x10^-3 m apart. The force on the particle is calculated using F=QE, resulting in 12 fN. The work done by the electric field is determined using the equation WD = FxD, yielding 56.2 aJ, which is also the increase in kinetic energy of the particle. The change in potential energy is equal to the negative of the gain in kinetic energy, confirming the conservation of energy principle.

PREREQUISITES
  • Understanding of electric fields and forces (F=QE)
  • Knowledge of work-energy principles (WD = FxD)
  • Familiarity with kinetic and potential energy concepts
  • Basic grasp of conservation of energy in physics
NEXT STEPS
  • Study the relationship between electric fields and forces on charged particles
  • Learn about the work-energy theorem in greater detail
  • Explore conservation of energy in electric fields and mechanical systems
  • Investigate the implications of potential energy changes in electric fields
USEFUL FOR

Students and educators in physics, particularly those focusing on electromagnetism and energy conservation principles, will benefit from this discussion.

Dongorgon
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Homework Statement


A particle with a charge of 3.2x10^-19 C accelerates from rest between two charged plates. The electric field is uniform and is 39000 N/C. The distance between the plates is 4.5x10^-3 m.

A) Calculate the force on the particle
B) How much work is done by the electric field in accelerating the particle between the two plates?
C) What is the increase in kinetic energy of the particle?
D) What is the change in potential energy of the particle?

Homework Equations



My answers so far:

A) Using F=QE to get an answer of 12fN

B) I'm not sure on the validity of this answer, or infact the right equation to use for this question and parts B and C.

I've used WD = FxD
Therefore, an answer of 56.2 aJ

C) Would the increase in kinetic energy be simply the change in energy experienced for part B)?

Any help would be greatly appreciated!
 
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A) correct!
B) correct!
C) You are right!
D) The change in P.E. will be equall to minus the gain in K.E. (conservation of energy)
 
cosmic dust said:
A) correct!
B) correct!
C) You are right!
D) The change in P.E. will be equall to minus the gain in K.E. (conservation of energy)

So both part B and C have the same answers of 56.2 aJ?

For D would the answer be along these lines?

25iss5w.png
 
Dongorgon said:
So both part B and C have the same answers of 56.2 aJ?

For D would the answer be along these lines?

25iss5w.png

Since the particle acquires a P.E. because of the work done on it and this is converted to K.E., so the answer to (D) is << solution deleted by Mentor -- but use his hint to fix your answer... >>
 
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