Getting the electric potential function using the electric field components

[ahmeeeeeeeeee]

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 ?

 

[SammyS]

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 .

 

[ahmeeeeeeeeee]

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 ??

 

[ahmeeeeeeeeee]

I mean , every repeated (xy)or (xy^2) or (x^2y) or etc.. I just take one of them

 

[SammyS]

I mean , every repeated (xy)or (xy^2) or (x^2y) or etc.. I just take one of them

There is no special V_x, or V_y, or V_z. Only V.

E_x = -∂V/∂x , E_y = -∂V/∂y , E_z = -∂V/∂z

When you integrate -E_x 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 E_y & E_z .

 

[chrisbaird]

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.