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!

Homework Help: Electric Field Intensity and Potential Difference

  1. May 31, 2009 #1
    1. The problem statement, all variables and given/known data
    A man walks into an electric field with a small charged object that has a charged value of 4.2 E-6C. At a distance of 10m from the object that is responsible for the electric field, he notices a force of 8.0 N on the object he is carrying.

    What is the electric field intensity?
    What is the potential difference?

    2. Relevant equations
    E = kc(q/r^2)
    F = kc(q1q2/r^2)
    ∆V = -Ed

    3. The attempt at a solution
    F = kc(q1q2/r^2)
    8.0 = 9 E9 • (4.2 E-6 • q2) / 10^2
    800 = 9 E9 • 4.2 E-6 • q2
    800/(9 E9 • 4.2 E-6) = q2 = 2.1 E-2

    E = kc(q/r^2)
    E = 9 E9 • (4.2 E-6 / 10^2)
    E = 378
    Round to two significant figures:
    E = 380

    [Interesting, but unnecessary; F/E = q2]

    ∆V = -Ed
    ∆V = -380 • 10
    ∆V = -3800

    I think the first set of equations where I used F were unnecessary.
    The electric field intensity = 380 N/C.
    The potential difference is = -3,800 Volts.

    I really don't think I did this right. The numbers I ended up with look too large, and I don't know if volts are supposed to be negative. I don't understand what "q" is supposed to be. Is it the difference between q1 and q2? What's q0? Does it even have a real value? I went through all the material and it still doesn't make sense!
  2. jcsd
  3. May 31, 2009 #2
    It's late and I'm not sure if I'm looking at this properly, but I would just like to say to take another look at the equations for your electric field calculations.

    There are two different charges in the problem, so be careful distinguishing which one is the "test charge" and which one is the "source charge". The same goes with the charges your using in your electric field equations.

    I hope that helps.
  4. May 31, 2009 #3
    E = Fe/q0
  5. May 31, 2009 #4
    Is this approach correct?

    (F/kc)r^2 = q
    E = kc(q/r^2)
    ∆V = -Ed
  6. May 31, 2009 #5
    q0 = 4.2 E -6?
  7. May 31, 2009 #6
    By q0, do you mean the elementary charge?
  8. Jun 2, 2009 #7
    If your still on this question:

    You could actually use two approaches to calculate the electric field, you've given the equations. You might want to go over your notes a little more to gain some insight and, the potential difference looks right, but as for the negative, I'm not quite sure, sorry.

    Anyhow, like I said before, distinguish between your test charge and your source charge, you seem to be mixing that up.

    This "object that is responsible for the electric field" would be the source charge, the one in the question that's giving the force. So, the other would be the test charge.

    As far as the equations go, E=Fe/q can be used, and so can E=kq1/r^2, but it requires more work. One thing to remember is that E=Fe/q is referring to the test charge, while E=kq1/r^2 is referring to the source charge.

    And I'm not sure what q0 is referring too, but I don't believe you need it here anyways.

    I would suggest you try both -- to make sure I'm right (I'm wrong as often as I'm right) -- and so you get some more practice and see the relationships a little better. And by the way, something you will probably kick yourself about, you did calculate the source charge to do the calculation the long way, but then you mixed up the charges when you went to solve for the electric field, so fe=kq1q2/r^2 isn't worthless here.

    Hope I helped, good luck!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook