Electric Potential Difference and Position Vectors: Finding Vba

In summary, the conversation discusses finding the potential difference between two position vectors using the equation Vb-Va = -∫E⋅dr, where V is a function of x, y, and z and E is found using partial derivatives. The approach of using integration is compared to simply taking Vb-Va, and it is determined that the constant term in the potential has no effect on the calculation of the potential difference.
  • #1
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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  • #2
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## ?
 
  • #3
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 :)
 
  • #4
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...
 
  • #5
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##
 
  • #6
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##
 
  • #7
Are these two integral same ?
 
  • #8
Try an example, e.g. ##V(x, y, z) = 2x+3y+4z + 1199## :smile:
 
  • #9
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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Related to Electric Potential Difference and Position Vectors: Finding Vba

1. What is an electric field?

An electric field is a physical quantity that describes the strength and direction of the force exerted on a charged particle by other charged particles. It is represented by a vector that points in the direction of the force and has a magnitude determined by the charge of the particles and their distance from each other.

2. What is electric potential?

Electric potential is a measure of the potential energy per unit charge at a given point in an electric field. It is often described as the amount of work needed to move a unit charge from one point to another against the electric field.

3. How are electric field and electric potential related?

Electric field and electric potential are closely related. Electric fields are created by differences in electric potential, and the strength of an electric field is directly proportional to the rate of change of electric potential. In other words, the steeper the change in electric potential, the stronger the electric field.

4. How can electric potential be calculated from an electric field?

Electric potential can be calculated from an electric field using the equation V = -∫E • dr, where V is the electric potential, E is the electric field, and the integral is taken over the path between two points. This equation takes into account the magnitude and direction of the electric field along the path.

5. What are the units of electric potential?

The SI unit for electric potential is volts (V), which is equivalent to joules per coulomb (J/C). In atomic units, the unit for electric potential is the hartree (Eh).

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