Uniform elecric fields- force and kinetic energy

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SUMMARY

A particle with a mass of 3 x 10^-12 kg and a charge of +4.0 x 10^-14 C is released between two parallel conducting plates in a vacuum. The net force on the particle is calculated using the formula F = VQ/d, where V is the potential difference and d is the separation of the plates. As the separation of the plates increases, the electric field strength decreases, leading to a reduction in electric force. However, the distance the particle travels also increases, which compensates for the decrease in force, resulting in a constant kinetic energy gain as long as the potential difference remains unchanged.

PREREQUISITES
  • Understanding of electric fields and forces
  • Familiarity with the concepts of kinetic energy and work
  • Knowledge of the relationship between electric potential difference and electric force
  • Basic physics of charged particles in electric fields
NEXT STEPS
  • Study the relationship between electric field strength and plate separation in parallel plate capacitors
  • Learn about the work-energy theorem in the context of charged particles
  • Explore the effects of varying potential differences on electric forces
  • Investigate the motion of charged particles in non-uniform electric fields
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in the dynamics of charged particles in electric fields.

chanderjeet
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Homework Statement


A particle of mass 3x 10^-12kg carrying a charge of +4.0x 10^-14C is released at the upper plate of a pair of horizontal parallel conducting plates placed in a vacuum.


Homework Equations


(1) Calculate the net force on the particle: I added the electric force VQ/d and the weight W=mg
(2) Calculate the KE of the particle when it reaches the lower plate: work done =gain in KE= Fd
(3) Explain the effect on force and the KE if the separation of the plates is increased. (this I'm not sure about)

The Attempt at a Solution



I'm thinking that since F= VQ/d if the distance increases then the force would decrease. I could be wrong.

Since, the gain in KE is equivalent to the work done =Fd and since F decreases (I assume) then the KE would also decrease. Am I wrong?
 
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chanderjeet said:
I'm thinking that since F= VQ/d if the distance increases then the force would decrease. I could be wrong.
Yup, you are right. Assuming that the potential difference between the plates remains the same, the larger the plate separation, the lower the electric field strength (and electric force consequently)
chanderjeet said:
Since, the gain in KE is equivalent to the work done =Fd and since F decreases (I assume) then the KE would also decrease. Am I wrong?
You forgot that the distance traveled by the particle, which is the distance between the two plates, also increases.
 
ohh, i did miss that. Thanks very much
 

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