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

[Dongorgon]

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!

 

[cosmic dust]

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)

 

[Dongorgon]

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?

 

[MrWarlock616]

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

For D would the answer be along these lines?

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... >>

 

[Nickie]

Your approach for part A is correct, using the equation F=QE. For part B, you are on the right track by using the equation W=Fd, where F is the force calculated in part A and d is the distance between the plates. Your answer of 56.2 aJ is correct.

For part C, the increase in kinetic energy of the particle can be calculated using the equation KE=1/2mv^2, where m is the mass of the particle (which is not given in the question) and v is the final velocity of the particle. Since the particle starts from rest and accelerates to a final velocity between the plates, the increase in kinetic energy will be equal to the work done by the electric field (calculated in part B).

For part D, the change in potential energy of the particle can be calculated using the equation PE=qV, where q is the charge of the particle and V is the potential difference between the plates. In this case, the potential difference can be calculated using the equation V=Ed, where E is the electric field and d is the distance between the plates. So, the change in potential energy will be equal to qEd.

Overall, your approach and calculations are correct. Keep up the good work!