1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Charge density and electric field

  1. Jan 14, 2016 #1

    I am wanting to see why charge density divided by e nought is equal to F/q and V/d. Unit cancellation makes it easy to equate F/q = V/d, but why is charge density alone enough to be equal to the electric field? I feel like something is missing here but I can't reconcile it nicely in my head or on paper. To be more clear, I don't understand how we get units of N/C from charge density which is C / m2. Hopefully this makes sense and someone can point me in the right direction.
  2. jcsd
  3. Jan 14, 2016 #2
    Charge density is not enough to have an electric field. You also need the distance from your charge.
    For point charges, electric field is [tex] E = k \frac{Q} {d^2} [/tex]
    where k is Coulomb's constant, Q is charge, and d is distance.

    If you have another charge q at distance d, the force between charges will be F = q*E, that's where E = F/q comes from.
    Unit cancellation isn't enough to relate quantities though. V/d only works for point charges. To actually relate V and E, in general, you need to use calculus.

    N/C is not a unit for charge density, that is a unit for an electric field. Either way, the units don't make sense to you because you are forgetting Coulomb's constant (the k in the formula above), which has units of Nm^2/(C^2).
    Depending on what unit system you use, you can express electric field with or without this constant, but it changes you units for charge.
    Last edited: Jan 14, 2016
  4. Jan 14, 2016 #3
    Ok, I follow the units here. What am I missing to show that sigma / e nought is equivalent to F / q...?
  5. Jan 14, 2016 #4
    ## \epsilon_0 ##has his own units. Did you consider this?
    It's not charge density alone.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook