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!

Electric Field help

  1. Mar 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Alright i have two problems i need help with

    1st.

    Find the electric field at a point midway between two charges of +43.1 10^-9 C and +61.7 10^-9 C separated by a distance of 24.7 cm.

    ________ N/C

    2nd.

    A Van de Graaff generator is charged so that the magnitude of the electric field at its surface is 2.8 104 N/C.

    (a) What is the magnitude of the electric force on a proton released at the surface of the generator?
    ________ N
    (b) Find the proton's acceleration at this instant.
    ________ m/s^2


    2. Relevant equations

    E= K*q/r^2


    3. The attempt at a solution
    For the first one, I dont really know how to get the point in the middle. I got 9.1 x 10^3 N/C for the electric frield of the 61.7 c x 10^-9 and 6.4x10^3 N/C for the 43.1 c x10-9. So basicly what i need help with is the electric field at the mid-point.

    For the 2nd one. I got part (a) right, i got 4.48 x10 ^-15 N. I need help getting the acceleration of the proton at that instant.
     
  2. jcsd
  3. Mar 15, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    (I'm not doing your math so if that's wrong you're still in trouble, but ...) Recall that the electric field is a vector field. At the point of interest you have calculated the individual scalar magnitudes of the vectors presumably at the midpoint. All you need to do now is determine the orientation of the two field contributions and add. By inspection they are both + charges so the field will necessarily be a difference of the 2, but directed away from the greater of the 2. (+ charges have outward directed fields.)

    For 2)b) Doesn't F = m*a still?
     
  4. Mar 15, 2009 #3
    For your first part add the two fields vectorially,they point in opposite directions.
    For your second part use Newtons second law,you will need the mass of the proton.
     
  5. Mar 15, 2009 #4
    oh, ok i forgot about the mass of a proton, thanks, but im still stuck in the first question. i dont know if i did the math wrong but im not getting it right
     
  6. Mar 15, 2009 #5

    LowlyPion

    User Avatar
    Homework Helper

    Maybe show your calculation?

    Distance is cm. Did you make it (.247/2) for your distance? That's a common oversight.
     
  7. Mar 15, 2009 #6
    Heres my calculations, and i did make it .247 before.

    9x10^9 * 43.1 x10-9/.247^2 + 9x10^9 * 61.7 x10^-9/.247^2

    is that correct?

    comes out to be 6.4 x10^3 + 9.1 x 10^3 = 15500?

    Then what? Do i divide by two?
     
  8. Mar 15, 2009 #7

    LowlyPion

    User Avatar
    Homework Helper

    No. Unfortunately you need to divide by 2 before you square your distances. So you could multiply your answer by 4 and that should undo not halving the denominator before you squared it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Electric Field help
  1. Electric Field HW Help (Replies: 1)

  2. Electric field help (Replies: 3)

  3. Electric Field Help (Replies: 3)

Loading...