Electric potential, potential difference, and potential energy

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SUMMARY

This discussion focuses on calculating potential differences and speeds of charged particles, specifically electrons and protons, using energy conservation principles. The first problem successfully determined that an electron must be accelerated through a potential difference of 135.159V to reach 2.3% of the speed of light. The second problem, involving an electron's speed reduction from 2x10^6 m/s to 500,000 m/s over 1 cm, incorrectly calculated the potential difference as -10.66054142V. The third problem attempted to find the speed of a proton accelerated through 69V, but resulted in an imaginary number due to incorrect application of the energy conservation formula.

PREREQUISITES
  • Understanding of electric potential and potential difference
  • Familiarity with the conservation of energy principle in physics
  • Knowledge of the mass and charge of fundamental particles (electron and proton)
  • Ability to manipulate equations involving kinetic and potential energy
NEXT STEPS
  • Study the derivation and application of the kinetic energy formula for charged particles
  • Learn about the relationship between potential difference and electric field strength
  • Explore the concept of relativistic effects on particle speeds at high velocities
  • Investigate the implications of negative potential differences in particle physics
USEFUL FOR

Students and professionals in physics, particularly those focusing on electromagnetism and particle dynamics, as well as educators seeking to clarify concepts related to electric potential and energy conservation.

btaylor
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1. Here is a problem that I know how to solve
Through what potential difference would an electron need to be accelerated for it to achieve a speed of 2.3% of the speed of light (2.99792x10^8 m/s), starting from rest? Answer in units of V.

For this problem I used:
Code:
deltaK + deltaU = 0
(1/2)mv^2 - 0 = -qdeltaV

It works out to be 135.159V.

2. Now here is a similar problem that I can't seem to solve
An electron moving parallel to the x-axis has an initial speed of 2x10^6 m/s at the origin. Its speed is reduced to 500000 m/s at the point p, 1cm away from the origin. The mass of the electron is 9.10939x10^-31 kg and the charge of the electron is -1.60218x10^-19 C. Calculate the magnitude of the potential difference between this point and the origin. Answer in units of V.

I tried to use the same approach for this problem:
Code:
(1/2)m2v2^2 - (1/2)m1v1^2 = -qdeltaV
m1 and m2 are the same, so the equation becomes:
Code:
(1/2)m(v2^2 - v1^2) / -q = deltaV
(1/2)(9.10939x10^-31)(500000^2 - (2x10^6)^2) / 1.60218x10^-19 = deltaV
-10.66054142V = deltaV

This is not the right answer, and I don't know what I could be doing wrong.

3. Here is something else that I can't seem to solve
Calculate the speed of a proton that is accelerated from rest through a potential difference of 69V. Answer in units of m/s.

I attempt to use the same formula:
Code:
(1/2)mv^2 - 0 = -qdeltaV
(1/2)(1.67262158x10^-27)v^2 = (-1.60218x10^-19)(69)

However, this yields an imaginary number.


Any hint as to what concepts I'm missing here would be greatly appreciated. :)
 
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I figured it out.
 
how do u do it?
 

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