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Quesitons on electrostatics

  1. Nov 26, 2009 #1
    1. The problem statement, all variables and given/known data

    A force of [tex]4.30\cdot10^{-2} [/tex]Newtons is needed to move a charge of 56 microCoulombs a distance of 20 cm in the direction of a uniform electric field. What is the potentioal difference that will provide this force

    2. Relevant equations

    Im trying to find one that would work

    3. The attempt at a solution

    i have converted 56 microCoulombs into [tex]5.6\cdot10^{-5} [/tex] Coulombs and 20 centimeters into .2 Meters. but thats the easy stuff. I cant seem to find an equation that would help me solve this.
     
    Last edited: Nov 26, 2009
  2. jcsd
  3. Nov 26, 2009 #2
    Can you write eqs that show relationship between F-E and E-V?
     
  4. Nov 26, 2009 #3
    i only know a few equations:

    [tex]\Delta~V = k_C\frac{q}{r}[/tex]

    [tex]\Delta~V =-E\Delta~d[/tex]

    [tex]\Delta~V =\frac{PE_{electric}}{q}[/tex]

    But i dont see any of these helping, unless im overlooking something huge.
     
  5. Nov 26, 2009 #4
    i have one equation for force, but it requires two stationary charged particles, all i have is one.
     
  6. Nov 26, 2009 #5
    Hi there,

    First, try calculating the E-field, and then use the answer you get to then calculate the p.d. :)
     
  7. Nov 26, 2009 #6
    the only equation i have to calculate E-Field is:

    [tex]E = \frac{k_Cq}{r^2}[/tex]

    and i cant use that because the 56 microcoulomb particle isnt the thing creating the uniform field, it is being acted upon by another field. all i know is that it was moved 20 centimeters with a force of 4.3x10^-2 newtons.
     
  8. Nov 26, 2009 #7
    You should also have the equation F=qE...you can rearrange this to find the electric field, because you know the size of the force that had to act on the charged particle. This field magnitude can then be used to find the potential difference :)
     
  9. Nov 26, 2009 #8
    oh i forgot about {tex]E=\frac{F}{q_0}[/tex]

    solving for E:

    [tex]E=\frac{4.3\cdot10^-2}{5.6\cdot10^-5}[/tex]

    [tex]E= 767.85[/tex]

    Then i plug that into [tex]\Delta~V =E\Delta~d[/tex]

    and get the answer 153.57

    and i checked in the answers section of my lab manual and it is correct.

    Thanks so much for your help!
     
    Last edited: Nov 26, 2009
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