Print ViewElectric Potential Energy versus Electric Potential

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SUMMARY

This discussion focuses on the relationship between electric potential (V) and electric potential energy (U), specifically in the context of a uniform electric field. The electric field can be derived from the electric potential using the formula ∇V = -E, where ∇ is the gradient operator. The participants clarify that the electric force F = -qE can be expressed in terms of the charge (q) and the electric field (E). The correct potential to use in this context is V(z) = -Ez, which simplifies the calculation of the electric field as E = -∂V/∂z.

PREREQUISITES
  • Understanding of electric fields and forces
  • Familiarity with calculus, specifically partial derivatives
  • Knowledge of electric potential and electric potential energy concepts
  • Basic understanding of Gauss's Law
NEXT STEPS
  • Study the derivation of electric fields from electric potentials using ∇V
  • Explore the application of Gauss's Law in calculating electric potential
  • Learn about the implications of electric potential energy in different electric fields
  • Investigate the relationship between electric force and electric potential energy in various scenarios
USEFUL FOR

Students and educators in physics, particularly those focusing on electromagnetism, as well as anyone seeking to deepen their understanding of electric potential and electric potential energy relationships.

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Homework Statement


Learning Goal: To understand the relationship and differences between electric potential and electric potential energy.

In this problem we will learn about the relationships between electric force [tex]\vec{F}[/tex], electric field [tex]\vec{E}[/tex], potential energy U, and electric potential V. To understand these concepts, we will first study a system with which you are already familiar: the uniform gravitational field.


Part 9
The electric field can be derived from the electric potential, just as the electrostatic force can be determined from the electric potential energy. The relationship between electric field and electric potential is [tex]\vec{E} = -\vec{\nabla}V[/tex], where [tex]\vec{\nabla}[/tex] is the gradient operator:

[tex]\vec{\nabla}V = \frac{\partial V}{\partial x}\hat{x}+ \frac{\partial V}{\partial y}\hat{y}+ \frac{\partial V}{\partial z}\hat{z}[/tex].
The partial derivative [tex]\frac{\partial V}{\partial x}[/tex] means the derivative of [tex]V[/tex] with respect to [tex]x[/tex], holding all other variables constant.

Consider again the electric potential [tex]V=-Ez[/tex] corresponding to the field [tex]\vec{E}=E\hat{z}[/tex]. This potential depends on the z coordinate only, so [tex]\frac{\partial V}{\partial x}=\frac{\partial V}{\partial y}=0[/tex] and [tex]\frac{\partial V}{\partial z}=\frac{dV}{dz}[/tex].

Find an expression for the electric field [tex]\vec{E}[/tex] in terms of the derivative of [tex]V[/tex].
Express your answer as a vector in terms of the unit vectors [tex]\hat{x}[/tex], [tex]\hat{y}[/tex], and/or [tex]\hat{z}[/tex]. Use [tex]dV/dz[/tex] for the derivative of [tex]V[/tex] with respect to [tex]z[/tex].

[tex]\vec{E}[/tex] =

Homework Equations


Not sure if they are relevant but...
F = qE
U = qV
V = K (q/r)

The Attempt at a Solution


I have no idea. I look at this and just stare.

Part 4 asked:
Find [tex]\vec{F}(z)[/tex], the electric force on the charged particle at height z.
Express [tex]\vec{F}(z)[/tex] in terms of q, E, z, and [tex]\hat{z}[/tex].

[tex]\vec{F}(z) =-qE\hat{z}[/tex]

From here I don't know where to go. Suggestions?
 
Last edited:
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What potential are you supposed to be using? I am confused as well. This is the exact problem description?

You have everything right if you are using the potential V(z) = -Ez because

[tex]\nabla V(z) = \frac{dV}{dz} \hat{z} = E \hat{z}[/tex]

If it is the general potential of a charged particle, however, you would have to find the potential at a distance z with calculus, or most likely Gauss's Law.
 
Sounds good. Thank you.
 

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