HELP Potential energy of proton in capacitor

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SUMMARY

The potential energy of a proton at the midpoint of a capacitor can be calculated using the formula U = qV, where q is the charge of the proton (1.6 x 10^-19 C) and V is the voltage across the capacitor (150V). The correct calculation yields U = 1.6 x 10^-19 C * 150V = 2.4 x 10^-17 J. While an alternative method involving electric field calculations using the integral of the electric field (E) can be applied, the direct formula is the most straightforward approach for this problem.

PREREQUISITES
  • Understanding of electric potential and potential energy
  • Familiarity with the concept of capacitors and their voltage
  • Basic knowledge of electric charge (specifically the charge of a proton)
  • Ability to perform calculations involving scientific notation
NEXT STEPS
  • Study the relationship between electric fields and potential energy in capacitors
  • Learn about the derivation of the electric field between capacitor plates
  • Explore advanced topics in electrostatics, such as Gauss's Law
  • Review calculus applications in physics, particularly integrals related to electric fields
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of electrostatics and capacitor behavior.

csimon1
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Homework Statement



I have attached the image. The question is --

What is the potential energy of a proton at the midpoint of the capacitor?

I know this is an extremely easy problem, but I can't seem to get the formula right...

Homework Equations



I tried U = qV, using 1.6*10^-19 * 150V (because it is in the middle). What am I doing wrong? Is this not the right formula?


The Attempt at a Solution

 

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Hi and welcome to PF!
I know a hard way to solve the question. I'm not sure it is of help in your case, but in case you've taken calculus 3 then it's ok.
[tex]\phi (r)= \int _{\infty}^r \vec E \cdot d \vec l[/tex]. Thus the almost only difficulty would be to calculate the E field between the plates of the capacitor.

There might be of course a much faster way to solve it. So if anyone can help, welcome!
 

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