Finding an electric field from a scalar field

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Homework Help Overview

The discussion revolves around finding a scalar potential field that corresponds to a given electric field expressed in vector form. The subject area involves concepts from electromagnetism, specifically the relationship between electric fields and scalar potentials.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss integrating components of the electric field to find the scalar potential, with some suggesting the inclusion of arbitrary functions to account for missing information. Questions arise regarding the correct interpretation of derivatives and the implications for the scalar field.

Discussion Status

The discussion is ongoing, with participants exploring different methods of integration and clarifying the relationship between the electric field components and the scalar potential. Some guidance has been offered regarding the integration process and the role of arbitrary functions, but clarity on obtaining these functions remains a point of contention.

Contextual Notes

Participants are navigating the complexities of partial derivatives and the implications of integrating vector components, with an emphasis on the need for additional information to fully define the scalar potential. There is an acknowledgment that the scalar potential is not uniquely defined due to the presence of arbitrary functions.

Noone1982
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Say I know an electric field

E = (yz - 2x)x-hat + (xz)y-hat + (xy)z-hat

How do I find the scalar field that would produce that? If I integrate each part I get

Vx = xyz - x^2
Vy = xyz
Vz = xyz

Vt = 3xy - x^2

To find E, I would take E = gradient cross the scalar field, but that would clearly not work. What I am doing wrong?
 
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Pick any component you want and integrate. For instance, if you picked the x-component then your guess "answer" would look like [tex]V = xyz - x^2 + f(y,z),[/tex] where f is an arbitrary function of y and z. This is because you only know that the partial derivative of V with respect to x is equal to yz - 2x. To figure out what your function f is, make use of your information about the other two partial derivatives of V. Of course, V is still undefined up to a constant.
 
Each component of the E-Field is the (-) derivative of the scalar potential with THAT coordinate , yes E is the (-) gradient .

Notice that d(x^2)/dy = 0 , so that E_y cannot give info about purely "x" terms in V, (or x and z terms, either). Similarly, E_x gives no info about purely y or z terms.

Don't add the xyz terms, just make sure they all agree.
 
Last edited:
Im still fuzzy on how to obtain the unknown function. Can you further explain?
 

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