Potential Energy of an electron in a parallel plate capacitor

Click For Summary
SUMMARY

The potential energy of an electron in a parallel plate capacitor is determined using the equation PE = qV, where q is the charge of the electron and V is the voltage across the plates. The discussion clarifies that the equation 1/2QV represents the energy stored in the capacitor, not the potential energy of the electron. The force acting on the electron due to the electric field is given by F = qE, leading to work done W = F*d = eV as the electron moves between the plates. Therefore, for calculating the potential energy of the electron, the correct approach is to use the qV formula.

PREREQUISITES
  • Understanding of electric fields and forces (E = V/d)
  • Familiarity with the concept of potential energy (PE = qV)
  • Knowledge of capacitor energy storage (1/2QV)
  • Basic physics of motion in electric fields
NEXT STEPS
  • Study the derivation of the potential energy formula for charged particles in electric fields.
  • Learn about the relationship between electric field strength and voltage in capacitors.
  • Explore the concept of energy storage in capacitors and its applications in circuits.
  • Investigate the dynamics of charged particles in varying electric fields.
USEFUL FOR

Physics students, electrical engineering students, and anyone interested in understanding the behavior of charged particles in electric fields and capacitors.

Seydlitz
Messages
262
Reaction score
4

Homework Statement



What is the potential energy of an electron in a parallel plate capacitor if kept on the negative plate before being released?

Homework Equations



1/2mv^2 = qV = PE
Capacitor Energy = 1/2QV

The Attempt at a Solution


1/2mv^2 = qV = PE

However if I'm not mistaken there's also 1/2QV equation that states the energy stored in a capacitor? Which one then should I choose for this electron problem? The normal qV or the 1/2qV?
 
Physics news on Phys.org
The force on the electron due to the E field of the capacitor is F = qE = eV/d.
As it falls the distance between plates, the work done by this force is W = F*d = eV.
So your have it right. The 1/2QV must for something else, perhaps the electrical energy stored in the capacitor.
 

Similar threads

Replies
5
Views
2K
Replies
3
Views
9K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 14 ·
Replies
14
Views
1K
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 28 ·
Replies
28
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
12
Views
3K
Replies
11
Views
3K