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Scalar's form of Coulomb's Law

  1. Sep 22, 2007 #1
    Title: Scalar form of Coulomb's Law?

    1. The problem statement, all variables and given/known data
    2. Relevant equations

    Coulomb's Law

    Vector Form:

    [tex]
    \vec{F}_{12} = {\frac{{{k}_{e}}{{q}_{1}}{{q}_{2}}}{{r}^{2}}{\hat{r}_{21}}}
    [/tex]

    Magnitude Form:

    [tex]
    |\vec{F}_{12}| = {\frac{{{k}_{e}}{\left|{q}_{1}\right|}{\left|{q}_{2}\right|}}{{r}^{2}}}
    [/tex]

    So, what is the scalar form of Coulomb's Law?

    In addition, when is the scalar form of Coulomb's Law negative?

    3. The attempt at a solution

    [tex]
    {F}_{12} = {\pm}{\frac{{{k}_{e}}{\left|{q}_{1}\right|}{\left|{q}_{2}\right|}}{{r}^{2}}}
    [/tex]

    Where: [itex]+[/itex] if [itex]{{q}_{1}}{{q}_{2}} \geq 0[/itex] and [itex]-[/itex] if [itex]{{q}_{1}}{{q}_{2}} < 0[/itex]

    I thought it was the above, but I do not believe that is right.

    Any help is appreciated.

    Thanks,

    -PFStudent
     
  2. jcsd
  3. Sep 22, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The "scalar" form of Coulomb's law is the same as what you've called the "magnitude" form. It's used for calculating the magnitude of the force. (At least that's how I've always used the terms.)
     
  4. Sep 23, 2007 #3
    Hey,

    So, the "Scalar form of Coulomb's Law" is really a misnomer then, as the word “scalar” implys that the result can be positive or negative, however because of how the "scalar" form of Coulomb's Law is written this cannot be. Then in reality there really is no "scalar" form of Coulomb's Law, because that would imply that the result could be negative, but as you have pointed out that cannot be.

    So, then consider the [itex]x[/itex] and [itex]y[/itex] components of Coulomb's Law.

    [tex]
    {\vec{F}_{12}} = {F}_{{12}_{x}}{\hat{i}} + {F}_{{12}_{y}}{\hat{j}}
    [/tex]

    Are they given by,

    [tex]
    {{F}_{{12}_{x}}} = {\left|{\vec{F}_{12}}\right|}{{cos}{\theta}_{12}}
    [/tex]

    [tex]
    {{F}_{{12}_{y}}} = {\left|{\vec{F}_{12}}\right|}{{sin}{\theta}_{12}}
    [/tex]

    Or are they given by,

    [tex]
    {{F}_{{12}_{x}}} = {{F}_{12}}{{cos}{\theta}_{12}}
    [/tex]

    [tex]
    {{F}_{{12}_{y}}} = {{F}_{12}}{{sin}{\theta}_{12}}
    [/tex]

    Note that for the two above equations, the sign of [itex]{F}_{12}[/itex] depends on the product of [itex]{{q}_{1}}{{q}_{2}}[/itex].

    Thanks for the reply Doc Al.

    Thanks,

    -PFStudent
     
  5. Sep 23, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    ambiguous notation

    If you use [itex]{F}_{12}[/itex] to stand for the magnitude of [itex]\vec{F}_{12}[/itex], then I see no difference in your equations and [itex]{F}_{12}[/itex] would not depend on the signs of the charges.
     
  6. Sep 23, 2007 #5
    Ahh...ok, makes things much clearer now.

    So, out of curiosity why do (most) physics textbooks refer to a "Scalar Form" of Coulomb's when that is really sort of incorrect?

    Thanks for the clarification, Doc Al.

    Thanks,

    -PFStudent
     
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