Electric Field/Electric Potential (Gradient Notation)

  1. 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
     
    Last edited: Oct 27, 2007
  2. jcsd
  3. Chegg
    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]
     
  4. cristo

    cristo 8,411
    Staff Emeritus
    Science Advisor

    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]
     
  5. 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
     
  6. 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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