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Homework Help: 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.
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