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Homework Help: Question about Charge

  1. Jul 7, 2007 #1
    Does anyone know how to calculate these questions?

    01.jpg

    02.jpg
     
  2. jcsd
  3. Jul 7, 2007 #2

    Doc Al

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    Do your own work, show what you did and where you got stuck, and then you'll get plenty of help.
     
  4. Jul 7, 2007 #3
    E = V / d (E is electric field strength; V is potential difference; d is separation distance);

    W = Q x V (W is workdone; Q is charge; V is potential difference);

    Thus, E = W / Q / d = W / (Q x d) and d = 4.0 m;

    But I don't know the value of W. How can I find the value of electric field strength at P?
     
  5. Jul 7, 2007 #4

    Doc Al

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    You are dealing with point charges here: Look up expressions for the electric field and electric potential at a given distance from a point charge. Note that electric field is a vector, but potential is a scalar.
     
  6. Jul 7, 2007 #5
    E = Q / (4 pi eo r^2)

    For +Q, E1 = Q / (4 pi eo 4^2) = Q / (64 pi eo)

    For -Q, E2 = -Q / (4 pi eo 4^2) = -Q / (64 pi eo)

    E1 + E2 = ?

    V = Q / (4 pi eo r)

    For +Q, V1 = Q / (4 pi eo 4) = Q / (16 pi eo)

    For -Q, V2 = -Q / (4 pi eo 4) = -Q / (16 pi eo)

    V1 + V2 = 0 and so this is the correct answer?
     
  7. Jul 7, 2007 #6

    Doc Al

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    Electric field is a vector, so direction counts. Add them like vectors.

    Good!
     
  8. Jul 7, 2007 #7
    |E| = sqrt (E1^2 + E2^2)

    E1 = 2 / (64 x 3.14 x 8.854 x 10^-12)

    E2 = -E1

    Thus, |E|

    = sqrt (E1^2 + (-E1)^2)

    = sqrt (2 x E1^2)

    = sqrt (2 x (2 / (64 x 3.14 x 8.854 x 10^-12))^2)

    = 15.90 x 10^8 V/m

    (But the answer isn't correct)
     
  9. Jul 7, 2007 #8

    Doc Al

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    That would be true if E1 and E2 were perpendicular, but they are not. Consider the horizontal & vertical components of each.
     
  10. Jul 7, 2007 #9
    Let theta be the angle between the horizontal line and the hypotenuse,

    sin theta = 1 / 4 = 0.25

    Take right/upward directions as positive and left/downward directions as negative,

    Horizontally,

    For +Q, component of E = E cos theta

    For -Q, component of E = -E cos theta

    Their sums = E cos theta - E cos theta = 0

    Vertically,

    For +Q, component of E = -E sin theta

    For -Q, component of E = -E sin theta

    Their sums = -E sin theta -E sin theta = -2E sin theta

    |E| = |Horizontal Component of E| + |Vertical Component of E|

    |E|

    = sqrt ((Horizontal Component of E)^2 + (Vertical Component of E)^2)

    = sqrt (0^2 + (-2E sin theta)^2)

    = 2E sin theta

    = 2E (0.25)

    = E / 2

    = 5.6 x 10^8 V/m

    Am I correct?
     
    Last edited: Jul 7, 2007
  11. Jul 7, 2007 #10

    Doc Al

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    You got it!
    Typo here; this should be:
    |E|^2 = |Horizontal Component of E|^2 + |Vertical Component of E|^2
     
  12. Jul 7, 2007 #11
    Thank you very much!
     
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