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Electric field using dielectric constant

  1. May 24, 2015 #1
    1. The problem statement, all variables and given/known data
    What is the magnitude of the electric field due to a 6.0 x 10-9 C charge at a point located 0.025m from the charge? The charge and the point in question are located underwater (κ(water) = 80).

    2. Relevant equations
    I know that E = magnitude of electric field = kq/r2, where k = Coulomb's constant of 9 x 109, and I also know that κ = kappa = dielectric constant aka relative permittivity of a dielectric material, which we use in the equation C = capacitance = κεoA/d.

    However, I can't workout how to use κ to find electric field strength, because I don't see how a capacitor is involved in this situation? And with the three variables I have been given (the size and electric polarity of charge, the distance, and the dielectric constant of water), I also don't know of any formulas I can use that don't require any other variables.

    The practise test I'm working on is multichoice, and the answers state that the final answer is equal to 1.1 x 103 N/C, but I'm not sure how to work backwards from that either. Any help would be greatly appreciated! We have a test later this week, and I'm sure Murphy's Law will make sure that unless I get my head around it, a question like this will no doubt be in the exam, so any and all advice will be welcomed!
     
  2. jcsd
  3. May 24, 2015 #2
    use this equation E=Kq/kr^2,where

    K= 9 * 10^9
    k=dielectric constant
    q=charge
    r=distance in question

    Edit: Answer is 1.08 * 10^3 N/C

    Equation reference: H C Verma Concepts Of Physics
     
    Last edited: May 24, 2015
  4. May 24, 2015 #3
    Do you need any proof of this equation?
     
  5. May 24, 2015 #4
    Thank you very much! So, is it fair to say that dielectric constant (ie Kappa) is actually always used in the E=kq/r2 equation when the surrounding environment is not specified? Which would mean it's assumed that it's just air, which has a dielectric constant of 1, so doesn't actually affect the equation?
    ...I guess what I'm asking, in a roundabout way, is should I use that equation for any question where the surrounding environment of the charge is not air?
    Thank you again for your help, I had no idea how to approach it, you've saved me a lot of confusion!
     
  6. May 24, 2015 #5
    Yes.. You are right! Thank you! you are welcome!
     
  7. May 24, 2015 #6
    There is one thing that you should know... Once you use kappa in the equation.. you should never use the bound charge appearing due to polarization of the dielectric.

    only free charge should be used..
     
  8. May 25, 2015 #7

    rude man

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    Use a gaussian surface around your charge, and Gauss's law.
    Now you can handle any combination of charge and observation point being surrounded by various dielectric material.
    .
     
  9. May 25, 2015 #8
    I'm going to bookmark this thread, because while I currently have no idea about Gauss' Law, no doubt I'll find out about it next semester, or soon enough. Thank you for your advice! Sometimes it seems like every time I learn a new physics concept, I become aware of half a dozen others that I also need to learn - but it makes my degree more entertaining if I think of it as fighting a Hydra! Thank you again
     
  10. May 25, 2015 #9
    The constant K we use in finding the electric field is= $$\frac {1}{4\pi \epsilon_0}$$
    So , when you are talking about an dielectric just replace the ##\epsilon_0## part with ##k\epsilon_0## where k is the dielectric constant , just the way you do for the capacitors. Thats why you get that k in the denomimator of the formula.
     
  11. May 25, 2015 #10

    rude man

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    Har! Are you going for a physics major or something like "physical sciences" where calculus-level physics is not covered? My school had that option & most others do too I'm sure. So you may or may not run into Dr. Gauss.
     
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