Energy of a Dipole in a Constant E Field

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SUMMARY

The discussion focuses on calculating the mechanical and kinetic energy of an electric dipole in a constant electric field, specifically when oscillating between plus and minus 65 degrees. The potential energy (U) is defined by the equation U = -pEcosθ, where p is the dipole moment and E is the electric field strength. At 180 degrees, the potential energy is given as 2 μJ, which allows for the calculation of pE. The kinetic energy is maximized when the dipole is aligned with the electric field (θ = 0).

PREREQUISITES
  • Understanding of electric dipoles and their behavior in electric fields
  • Familiarity with the equation for potential energy of a dipole (U = -pEcosθ)
  • Knowledge of mechanical energy concepts in physics
  • Basic trigonometry to calculate energy at different angles
NEXT STEPS
  • Calculate the dipole moment (p) using the given potential energy at 180 degrees
  • Explore the relationship between potential energy and kinetic energy in oscillating systems
  • Investigate the effects of varying angles on the potential energy of dipoles
  • Learn about the dynamics of dipoles in non-uniform electric fields
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in the mechanics of electric dipoles in electric fields.

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Homework Statement


The graph shows the potential energy of an electric dipole which is in a constant electric field; only the electric force is acting on the dipole. Consider a dipole that oscillates between plus and minus 65 degrees.

knight_Figure_29_10.jpg


a) What is the dipole's mechanical energy?
b) What is the dipole's kinetic energy when it is aligned with the electric field?

Homework Equations


U = -pEcos\theta


The Attempt at a Solution


I know that when the graph is at the trough it's -pE and at it's peak it's +pE. But I don't know how to factor in the 65 degrees. Please help!
 
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At 180 degrees U = 2 μJ. Substitute in the relevant equation and find pE.
At θ = 0, all the potential energy in converted to K.E. Similarly you can find P.E. at 65 degrees by substituting the value of pE and θ in the equation.
 
rl.bhat said:
At 180 degrees U = 2 μJ. Substitute in the relevant equation and find pE.
At θ = 0, all the potential energy in converted to K.E. Similarly you can find P.E. at 65 degrees by substituting the value of pE and θ in the equation.

Thank you so much! That really cleared it up for me!
 

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