How to find final kinetic energy for a charged particle

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SUMMARY

The discussion focuses on calculating the final kinetic energy of an alpha particle subjected to a uniform electric field of 12.5 kV/m over a distance of 2.5 cm. The participant attempted to use the potential energy formula U = 1/2 QV, where Q is the charge of the alpha particle (2e). The calculation yielded an incorrect result, prompting a request for clarification on converting electric field strength (V/m) to energy (Joules). The correct approach involves understanding the relationship between electric field, charge, and potential energy without assuming a capacitor model.

PREREQUISITES
  • Understanding of electric fields and their units (V/m)
  • Knowledge of charge quantization (specifically the charge of an electron)
  • Familiarity with potential energy equations in electrostatics
  • Basic principles of kinetic energy and its relation to work done by electric fields
NEXT STEPS
  • Study the relationship between electric fields and potential energy in charged particles
  • Learn about the work-energy theorem as it applies to charged particles in electric fields
  • Explore the concept of electric potential and its calculation from electric fields
  • Review examples of kinetic energy calculations for charged particles in uniform electric fields
USEFUL FOR

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

Dana Fishel
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Homework Statement


An alpha particle (helium nucleus, charge +2e) starts from rest and travels a distance of 2.5 cm under the influence of a uniform electric field of magnitude 12.5 kV/m. What is the final kinetic energy of the alpha particle?

Homework Equations


U = 1/2 QV = 1/2 C(V^2) = (Q^2)/2C
Q = CV
C = (Eo A)/d

where U = potential energy, Q = charge, C= capacitance, A= area, d = distance, and V = potential difference

The Attempt at a Solution


I tried using U = 1/2 QV by calculating that the charge is 2 times the constant charge of an electron (1.602x10^-19) and used 12.5 kv/m as V (but changed it back to just volts.
U = 0.5 (3.204x10^-19) (12500 v/m)
U = 2.0025x10^-15
But this is wrong. Also, in a practice version of the problem it said that given magnitude = 18.5 kv/m and distance = 1.6 cm the answer would be 9.48x10^-17.
Any help with this problem would be greatly appreciated.
 
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How can you get from the E-field (V/m) to an Energy (Joule = CV)?
This is not a Capacitor ... the general Potential Energy formula is U = Q_total V_average ...
(this gives a ½ for Energy STORED in a Capacitor, but at maximum charge Q and maximum voltage V.)
 

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