Print ViewElectric Potential Energy versus Electric Potential

[Badger]

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 \vec{F}, electric field \vec{E}, 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 \vec{E} = -\vec{\nabla}V, where \vec{\nabla} is the gradient operator:

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

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

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

\vec{E} =

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 \vec{F}(z), the electric force on the charged particle at height z.
Express \vec{F}(z) in terms of q, E, z, and \hat{z}.

\vec{F}(z) =-qE\hat{z}

From here I don't know where to go. Suggestions?

 

[Mindscrape]

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

\nabla V(z) = \frac{dV}{dz} \hat{z} = E \hat{z}

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.

 

[Badger]

Sounds good. Thank you.