Getting the electric potential function using the electric field components

AI Thread Summary
To obtain the electric potential function V(x,y,z) from the electric field components Ex, Ey, and Ez, one cannot simply integrate and sum the components due to the nature of partial derivatives and the constants of integration involved. The integration of Ex and Ey must account for a "constant" that can depend on the other variables, which is crucial for maintaining consistency when taking partial derivatives. When integrating, the resulting potential function may differ from the original due to these constants, leading to discrepancies in the electric field components upon differentiation. A trial-and-error approach is often necessary to refine the solution, as direct integration may not yield correct results in multi-dimensional cases. Understanding the relationship between the electric field and potential is essential for accurate calculations in electrostatics.
ahmeeeeeeeeee
Messages
21
Reaction score
0
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 seperately 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 ?
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top