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Calculating work done by an Electric field on a positive charge

  1. Jun 12, 2017 #1

    psy

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    • Poster reminded to use the Homework Help Template in the future for schoolwork posts
    Hey

    How can i calculate the work done
    by a homogeneous electric field on a positive charge q
    = 7 × 10^-8 C when the charge moves from point 1 to point 2
    . I have two cases:
    a) the charge is parallel to the
    Electric field;
    b) the displacement of the charge takes place at the angle α = 60 °
    to the
    Electric field.
    The electric field strength is E = 6 × 105 V / m, the displacement
    of the charge is s = 10 cm

    https://www.flickr.com/photos/155324944@N02/35099011562/in/dateposted/




    So the mechanical work equation is equivalent to the electrical

    case a)
    The force done on the particle is the electric field strength multiplied with its charge :

    F = q * E = 7* 10^-8 C * 6 * 10^5 V/m = 42 * 10^-3 CV/m

    The work is given by W = F*s = 42 * 10^-3 CV/m * 0.1 m = 4.2 *10^-3 CV ( m*kg*A^2 / s^2)

    case b)
    the particle is still moving in the same direction as the field force ,but a with an angle,
    so i would use

    F = q * E * cos(60°)

    Is this a proper way to calculate it?


     
  2. jcsd
  3. Jun 12, 2017 #2

    Doc Al

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    Staff: Mentor

    The first case (a) looks good to me. (The standard unit for work is the Joule.)
     
    Last edited: Jun 13, 2017
  4. Jun 13, 2017 #3

    ehild

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    Homework Helper
    Gold Member

    No, the equation is wrong. The electric force is proportional to the electric field, ##\vec F = q \vec E##
    The problem text says that "b) the displacement of the charge takes place at the angle α = 60 °to the Electric field."
    The charge is scalar, it does not have direction.
    What is the work if the force and the displacement make an angle α ?
     
  5. Jun 13, 2017 #4

    Doc Al

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    Staff: Mentor

    As ehild points out, that equation is incorrect. But it's a step in the right direction if you meant to give the component of the force in the direction of the displacement.
     
  6. Jun 13, 2017 #5

    gneill

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    Staff: Mentor

    The link above requires one to have a flickr account, and even if one does it does not resolve to a valid destination. Please upload your image to the PF server so that members here don't have to follow offsite links or have accounts on other services in order to view them.
     
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