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Homework Help: Coulomb's law vs Electric Field

  1. Oct 4, 2016 #1
    1. The problem statement, all variables and given/known data
    I just wanted to make sure I understood the difference between the two.
    Coulombs law is the force between two charges. Two charges that are physical(maybe not the right word) located on some axes. We can use the equation
    F=k(q1 * q2 * r(hat))/r^2

    r^2=|r| = |x-x'|
    where r hat equal= r/|r|= (x-x')/|x-x')|

    if we only have one charge and we want to know what the force would be at some random point q. We know that F=qE(x). So, F/q=E(x). which would give us
    E(x)=k(q1* r(hat))/r^2

    They both calculate the force, coulombs is the force between two charges and the electric field is the force a some point q. (q would be a field point.)
    As the distance grows on both the force gets weaker and weaker.

    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Oct 4, 2016 #2
    I know that you can sum the forces with Coulombs Law but can you sum up the electric fields?
     
  4. Oct 4, 2016 #3
    A coulomb is a unit of measure (quantity of electricity).
    Force is a physical quantity.
    An electric field models the network of electrostatic forces charged bodies would exert on each other.
     
  5. Feb 4, 2018 #4
    whats the difference between an electric field an electric force? They seem synonymous
     
  6. Feb 4, 2018 #5

    gneill

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    Staff: Mentor

    The electric field is what is in the surrounding environment that produces a force on a charge. One way to express the units associated with an electric field is N/C (Newtons per Coulomb). The electric field and force have different units.

    Compare with the gravitational field: The acceleration due to gravity (meters per second2) can also be expressed in units of N/kg (Newtons per kilogram).
     
  7. Feb 4, 2018 #6
    If there is only one charge, call it q1, there is no force at any random point q. A force acts on a material object. There is no force "at" an empty point in space. It is only when you place another charge q at some point in space, that the charge q has a force acting on it, because of your original charge q1.
    If there is only one charge, it generates an electric field, not force, at every point in space, given by your formula:
    E(x)=k(q1* r(hat))/r^2
    This is different from a force. If now you place a charge q2 at the point x, then there is a force on q2 given by
    F = q2 E(x)
     
  8. Feb 4, 2018 #7

    Ray Vickson

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    The electric (and magnetic) fields have existence that is (essentially) separated from the charges that created them. For example, take
    the force expression
    $$\mathbf{F} = \frac{k q_1 q_2 \: \hat{r}}{r^2}.$$
    If the particles are quite far apart, then when you suddenly wiggle ##q_2##, the expression above for ##\mathbf{F}## changes (because ##\hat{r}## and ##r## change), but for a while the actual force on particle #1 does not change. Rather, the influence of particle #2 is propagated at the speed of light, so there would be a bit of a delay before the wiggling effects are felt by particle #1. The effects arise through the influence of the electromagnetic field. The field is very real, the electrostatic force is very real, and certainly the two are closely linked; but they are not really the same.
     
  9. Feb 6, 2018 #8
    Got it. Thanks. Another question super basic, but entertain me.

    If E=M. Which is agreed upon. How is it possible that E=Mc^2?
     
  10. Feb 7, 2018 #9

    Ray Vickson

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    Nobody agrees that E=M. Why do you think such a thing could be true?
     
  11. Feb 7, 2018 #10
    What is E and what is M?
     
  12. Feb 7, 2018 #11
    Thats the basis of quantum mechanics.
     
  13. Feb 7, 2018 #12
    I would really appreciate being told what is E and what is M. There is no reference to M in any of the earlier posts, and E is the electric field in all the earlier posts in this thread. So I do not understand the relation E = M at all.
     
  14. Feb 7, 2018 #13

    Ray Vickson

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    No, it is not. If you claim it is, please cite references---for example, a book or an article.
     
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