1. Not finding help here? Sign up for a free 30min 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 and electron acceleration

  1. Jul 4, 2010 #1
    What are the magnitude and direction of the acceleration of an electron at a point where the electric field has magnitude 6900 N/C and is directed due north?
     
  2. jcsd
  3. Jul 4, 2010 #2
    label the north direction as [tex]\hat{y}[/tex] so [tex]\vec{E}=E \hat{y} [/tex]. The "electric force" equals the charge times the electric field. that is,

    [tex]m \vec{a} = -eE \hat{y} [/tex]

    [tex] \vec{a} = \frac{-eE}{m} \hat{y} [/tex]

    where e is the charge of the electron and m is its mass.
     
  4. Jul 4, 2010 #3
    that would then make the direction south???? i found my answer by using F=qE and the F=ma. and i have another question.....
    Two point charges, q1 = +20.0 nC and q2 = +11.0 nC, are located on the x-axis at x = 0 and x = 1.00 m, respectively. Where on the x-axis is the electric field equal to zero? i wanted to use the equation E=k[q]/r^2 but i'm confused as to what to do with the x=0 and x=1
     
  5. Jul 4, 2010 #4
    You can use the x values to determine the distance (r) between the two charges.
     
  6. Jul 4, 2010 #5
    i am not sure how to encorporate the the one formula with the various values
     
  7. Jul 4, 2010 #6
    Note that electric fields are vectors. For the field to cancel at a point, the field due to A should cancel out the field due to B. Where do you think that could happen (between A and B / left of A / right of B)? Once you figure that out, take a point 'x', and find out x for Eax = Ebx.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook