Getting the electric potential function using the electric field components

In summary, integration of partial derivatives is not as straightforward as integration of simple functions. There is a "constant" of integration that is not really a constant, but rather a function of the other variables. This can lead to errors when adding the separate integrated partial derivatives, which is why the answer was wrong in this scenario. Additionally, solving multiple-dimensional differential equations requires clever trial solutions and is not as simple as direct integration.
  • #1
ahmeeeeeeeeee
21
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 ?
 
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  • #2
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 .
 
  • #3
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 ??
 
  • #4
I mean , every repeated (xy)or (xy^2) or (x^2y) or etc.. I just take one of them
 
  • #5
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 .
 
  • #6
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.
 

FAQ: Getting the electric potential function using the electric field components

1. How do you calculate the electric potential function using the electric field components?

The electric potential function can be calculated using the formula V = -∫E·dl, where E is the electric field vector and dl is the infinitesimal displacement vector along the path of integration. This integral can be evaluated using calculus.

2. What are the units of the electric potential function?

The units of the electric potential function are volts (V) or joules per coulomb (J/C).

3. Can the electric potential function be negative?

Yes, the electric potential function can be negative. This indicates that the electric potential energy of the system is negative, meaning that work must be done to bring a positive charge from infinity to that point in the electric field.

4. What is the relationship between the electric potential function and the electric field?

The electric potential function and the electric field are related by the formula E = -∇V, where ∇V is the gradient of the electric potential function. This means that the direction of the electric field is in the direction of the steepest decrease in the electric potential function.

5. How can the electric potential function be used in practical applications?

The electric potential function can be used to calculate the potential difference between two points in an electric field, which is important in understanding the flow of electricity. It is also used in the design and analysis of electronic devices, such as capacitors and resistors.

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