Electric Potential Difference and Position Vectors: Finding Vba

AI Thread Summary
The discussion focuses on calculating the electric potential difference, Vba, using the relationship between electric field E and potential V. The user has derived E from the potential function but is confused about integrating E to find Vba. They explore the idea of directly calculating Vb and Va from their respective position vectors instead of integrating E, noting that the constant term in V does not affect the potential difference. Clarifications are provided on how to set up the integral correctly, emphasizing that only potential differences matter in this context. Ultimately, the conversation highlights the importance of understanding the relationship between electric potential and field in solving related problems.
Arman777
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Homework Statement


I have a potential Value like ##V=V(x,y,z)+C##
I found ##\vec E## using partial derivative, like ##\vec E=((-∂V/∂x)i+(-∂V/∂y)j+(-∂V/∂z) k)##
Theres two position vectors,
##\vec r_{a}=2i##
##\vec r_{b}=j+k##
We need to find ##V_{ba}=?##

Homework Equations


##V_b-V_a=-\int_{r_{a}}^{r_{b}} \vec E⋅d\vec r##
##V_r=V(x_i,y_i,z_i)## where ##r=(x_i,y_i,z_i)##

The Attempt at a Solution


Ok I found E but since we are taking partial derivative the constant term disappeared.

I can find from ##V_b=V(0,1,1)## and ##V_a=V(2,0,0)## and the difference will be ##V_b-V_a=V_{ba}##

But If ı try to do this from ##V_b-V_a=-\int_{r_{a}}^{r_{b}} \vec E⋅d\vec r## using this.How can I approach the question.##\vec E## is a function of ##x,y,z## but we need a function of ##\vec r##

I mean the confusing part is,
##V_b-V_a=-\int_{r_{a}}^{r_{b}} ((-∂V/∂x)i+(-∂V/∂y)j+(-∂V/∂z) k)⋅d\vec r##

How can I take integral in this case ?

I ll do ##\vec E⋅\vec r_a-\vec E⋅\vec r_b## ??

And is my approach or answer is true..? , Is a constant term here makes a diffference ?
 
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Hi,

If you have an expression for V, why go the long way via ##\vec E## and integration if you can simply take ##V_b-V_a## ?
 
BvU said:
Hi,

If you have an expression for V, why go the long way via ##\vec E## and integration if you can simply take ##V_b-V_a## ?

Just curiosity :)
 
Arman777 said:
How can I take integral in this case ?
Write ##d\vec r## as ##dx \, {\bf\hat\imath} + dy \, {\bf\hat\jmath}+ dz\, {\hat k}## and write out the dot product to give you three terms of the integrand in three integrals...
 
BvU said:
Write ##d\vec r## as ##dx \, {\bf\hat\imath} + dy \, {\bf\hat\jmath}+ dz\, {\hat k}## and write out the dot product to give you three terms of the integrand in three integrals...

so ##V_b-V_a=-\int_{x=0}^{2}\int_{y=0}^{1}\int_{z=0}^{1} ((-∂V/∂x)+(-∂V/∂y)+(-∂V/∂z))dxdydz##
 
or maybe
##V_b-V_a=-\int_{x=0}^{2}((-∂V/∂x)dx+\int_{y=0}^{1}(-∂V/∂y)dy+\int_{z=0}^{1}(-∂V/∂z))dz##
 
Are these two integral same ?
 
Try an example, e.g. ##V(x, y, z) = 2x+3y+4z + 1199## :smile:
 
Arman777 said:
Are these two integral same ?
No. dxdydz is a volume integral. That's not the idea...
 
  • #10
BvU said:
No. dxdydz is a volume integral. That's not the idea...
I see you are right
But it must be the
Arman777 said:
or maybe
##V_b-V_a=-\int_{x=0}^{2}((-∂V/∂x)dx+\int_{y=0}^{1}(-∂V/∂y)dy+\int_{z=0}^{1}(-∂V/∂z))dz##

I did your example and it gave me the same thing.I understand the idea.But the constant dissapered which that's bothers.Or maybe it didnt.

And thanks :smile::smile:
 
  • #11
I think in the calculating V difference, the constant has no affect.Of course it doesn't that's the logical thing...
 
  • #12
Ok thanks a lot again
 
  • #13
Correct: a potential is basically unnoticeable. Only potential differences bring about something that can be sensed.
 
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