Getting the electric potential function using the electric field components

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Homework Help Overview

The discussion revolves around finding the electric potential function V(x,y,z) from given electric field components Ex, Ey, and Ez. The original poster expresses confusion regarding the integration process and the resulting discrepancies between their calculated electric field components and the original ones.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to integrate the electric field components separately and sum them, questioning the validity of this approach when the results do not match the original components. Some participants suggest that summing the results may not be appropriate.
  • Participants discuss the implications of constants of integration and the need to consider functions of other variables when integrating in multiple dimensions.
  • There are inquiries about the treatment of terms involving products of variables, such as xy, in the integration process.

Discussion Status

Participants are actively engaging with the original poster's confusion, providing insights into the integration process and the nature of partial derivatives. There is a focus on clarifying the relationship between the electric field components and the potential function, with some guidance offered regarding the handling of integration constants.

Contextual Notes

The discussion highlights the complexity of integrating multi-dimensional differential equations and the potential coupling of dimensions, which may affect the integration approach. The original poster's lack of familiarity with partial derivatives is noted as a factor in their understanding of the problem.

ahmeeeeeeeeee
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hello , I want to get the electric potential function with displacement
V(x,y,z)
as

Ex= -8 -24xy
Ey =-12x^2+40y
Ez= zero

when I perform intergration on Ex and Ey separately then sum them , the answer is wrong , why ?? I mean when I do this and then perform partial derivatives again , the Ex in not the same , is this integration wrong ?
 
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ahmeeeeeeeeee said:
hello , I want to get the electric potential function with displacement
V(x,y,z)
as

Ex= -8 -24xy
Ey =-12x^2+40y
Ez= zero

when I perform integration on Ex and Ey separately then sum them , the answer is wrong , why ?? I mean when I do this and then perform partial derivatives again , the Ex in not the same , is this integration wrong ?
Hello ahmeeeeeeeeee. Welcome to PF !

It's wrong, because you shouldn't sum them.

Show what you get & we can help .
 
Why can't I sum them ??


the original function of v (given) is

V = 8X+12YX^2-20Y^2

from which I got

Ex and Ey and Ez

when I integrate again and sum I get

V =8X + 24YX^2-20Y^2

so the difference is that it is 24 instead of 12



Well , I didn't study partial derivatives if there were something special about the integrations , But I tried many functions and I noticed that when I get

Vx ----- I just take from it the things without (xy) ( Just X)
vy ----- I just take the "Y"s

and then take just one of the (xy)s from Vx or from Vy

this what I "noticed" when I tried some functions , is it true ?? and why ??
 
I mean , every repeated (xy)or (xy^2) or (x^2y) or etc.. I just take one of them
 
ahmeeeeeeeeee said:
I mean , every repeated (xy)or (xy^2) or (x^2y) or etc.. I just take one of them
There is no special Vx, or Vy, or Vz. Only V.

Ex = -∂V/∂x , Ey = -∂V/∂y , Ez = -∂V/∂z

When you integrate -Ex to find V, there is a "constant" of integration. It's not really a constant, just a constant with respect to x, e.i. when you take the (partial) derivative of this "constant" of integration with respect to x, you must get zero. Therefore, this "constant" of integration, is actually quite possibly a function of y and z. You figure out what this "constant" is, by taking the partial derivatives w.r.t. y & w.r.t. z, and comparing the result with Ey & Ez .
 
You cannot typically solve multiple-dimensional differential equations by direct integration because of the way the dimensions couple. You have to cleverly write a trial solution, plug it into see if it works, then refine your trial solution based on what goes wrong when you plug it in. A typical college course on differential equations is a course in the art of making good guesses on trial solutions.
 

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