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Electric Field/Electric Potential (Gradient Notation)

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PFStudent
#1
Oct27-07, 07:17 PM
P: 171
1. The problem statement, all variables and given/known data

Hey,

I have a question about Electric Field/Electric Potential gradient notation.

Since,

[tex]
{\vec{E}} = {-}{\nabla}{V(r)}
[/tex]

Which reduces to,

[tex]
\vec{E} = {-}{\nabla}{V(x, y, z)}
[/tex]

When expanded is,

[tex]
\vec{E} = {-}{\left[{\frac{\partial[V]}{\partial{x}}}{\hat{i}} + {\frac{\partial[V]}{\partial{y}}}{\hat{j}} + {\frac{\partial[V]}{\partial{z}}}{\hat{k}}\right]}
[/tex]

So using partial derivative notation can I write,

[tex]
{\vec{E}} = {-}{\vec{V}'_{xyz}}
[/tex]

So, is the above correct notation?

The reason I am hesitant is, because formally the gradient is defined as a vector operator that takes a scalar field (such as the electric potential) and changes it to a vector field (such as the electric field) through: partial differentiation with the addition of unit vectors ([tex]\hat{i}, \hat{j}, \hat{k}[/tex]).

However, writing it as below sort of implies the potential is a vector (which it isn't), but gives the impression that it is because of how the gradient is defined.

[tex]
{\vec{E}} = {-}{\vec{V}'_{xyz}}
[/tex]

So, is the above notation correct?

-PFStudent
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Poop-Loops
#2
Oct27-07, 07:24 PM
P: 863
Quote Quote by PFStudent View Post
However, writing it as below sort of implies the potential is a vector (which it isn't), but gives the impression that it is because of how the gradient is defined.

[tex]
{\vec{E}} = {-}{\vec{V}'_{xyz}}
[/tex]

So, is the above notation correct?

-PFStudent
I wouldn't use it. I would just leave it as:

[tex]
{\vec{E}} = {-}{\nabla}{V(r)}
[/tex]

Or

[tex]
{\vec{E}} = {-}{\nabla}{V}
[/tex]
cristo
#3
Oct28-07, 10:31 AM
Mentor
cristo's Avatar
P: 8,310
Quote Quote by PFStudent View Post
[tex]
{\vec{E}} = {-}{\vec{V}'_{xyz}}
[/tex]

So, is the above notation correct?
No, it is not correct. That is, there is no notation I know of that looks like that that is defined as the gradient of a scalar field.

As the above poster says, there is nothing wrong with [itex]\vec{E}=-\nabla V[/itex]

PFStudent
#4
Oct28-07, 11:37 AM
P: 171
Electric Field/Electric Potential (Gradient Notation)

Hey,

Yea, thanks for the input, I can see why that notation,

[tex]
{\vec{E}} = {-}{\vec{V}'{xyz}}
[/tex]

is wrong. Since, we are adding the components of a vector that is not the same as taking the partial derivative of a function with respect to each of the variables.

Since, all the gradient is doing is the following,

[tex]
\vec{E} = {-}{\nabla}{V(x, y, z)} = {-}{\left[{\frac{\partial}{\partial{x}}{\left[V\right]}} + {\frac{\partial}{\partial{y}}}{\left[V\right]}} + {\frac{\partial}{\partial{z}}}{\left[V\right]}}\right]}{\hat{r}}
[/tex]

Thanks,

-PFStudent
PFStudent
#5
Oct29-07, 12:58 PM
P: 171
Hey,

I've been thinking about this and I have a follow up question.

Since,

[tex]
\vec{E} = {-}{\nabla}{V(r)} = {-}{\left[{\frac{\partial}{\partial{x}}{\left[V\right]}} + {\frac{\partial}{\partial{y}}}{\left[V\right]}} + {\frac{\partial}{\partial{z}}}{\left[V\right]}}\right]}{\hat{r}}
[/tex]

and also,

[tex]
E = {-}{\frac{\partial}{\partial{r}}}{\left[{V(r)}\right]}
[/tex]

So then,

[tex]
\vec{E} = {-}{\nabla}{V(r)} = {-}{\frac{\partial}{\partial{r}}}{\left[{V(r)}\right]}{\hat{r}}
[/tex]

Now can I rewrite the above as below?

[tex]
\vec{E} = {-}{\frac{\partial}{\partial{(x, y, z)}}}{\left[{V(x, y, z)}\right]}{\hat{r}}
[/tex]

Which for [tex]{E}[/tex] can also be written as,

[tex]
{E} = {-}{\frac{\partial}{\partial{(x, y, z)}}}{\left[{V(x, y, z)}\right]}
[/tex]

So, is the notation for the above two equations correct?

Thanks,

-PFStudent


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