Electric field and electron acceleration

  • Thread starter kong12
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  • #1
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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?
 

Answers and Replies

  • #2
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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.
 
  • #3
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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
 
  • #4
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You can use the x values to determine the distance (r) between the two charges.
 
  • #5
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i am not sure how to encorporate the the one formula with the various values
 
  • #6
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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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