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Homework Help: Charge within an electric field

  1. Feb 3, 2010 #1

    JJBladester

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    Gold Member

    1. The problem statement, all variables and given/known data

    A -10.0 nC charge is located at position (x,y) = (2.0cm, 1.0cm). At what (x,y) position(s) is the electric field:

    161,000i - 80,500j N/C

    x = ??? cm
    y = ??? cm

    2. Relevant equations

    E = Kq/r2

    3. The attempt at a solution

    I know that the magnitude of E is [tex]\sqrt{161,000^{2}+80,500^{2}}[/tex] = 1.8e5.

    So, to find the distance between the electric charge and where the field is:

    1.8e5 = Kq/r2

    Therefore, r = [tex]\sqrt{(9e^{9})(10e^{-9})/1.8e^{5}}[/tex] = .02m = 2cm

    So, I know the distance between the charge and the electric field. The angle that the electric field makes with the x-axis is tan-1(-80,500/161,000) = -26.57

    I was thinking that I have to do something with polar coordinates to finish the problem but I can't seem to make the mental leap.

    I know in polar coordinates, if you're given a distance (2cm) and an angle (-26.57), you should be able to find the rectangular coordinates by:

    x = 2cos(-26.57)
    y = 2sin(-26.57)

    But isn't this the position of the electric field in relation to the origin (not the charge)?
     
  2. jcsd
  3. Feb 3, 2010 #2

    Delphi51

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    Homework Helper

    I played with this a bit and may have found an insight. First, I noticed that for the (x,y) answer, x > 2 (because the Ex is positive) and y > 1 because Ey is negative. I sketched the vector and saw that at (x,y) the E field is
    kq/r²[(x-2)/r*i - (y-1)/r*j]
    If you write that the x component is 161000 and the y component is -80500 you get a pair of equations that can be partly solved very quickly. Solve one for kq/r² and sub that into the other. You get the equation of a straight line on the x,y plane. Combine that with your value for the radius and you'll soon have the point on that line with the right radius and x > 2, y > 1.
     
  4. Feb 3, 2010 #3

    JJBladester

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    Gold Member

    Hey Delphi51,

    I actually just figured it out...

    Since the charge is negative, the electric field would point toward the charge. In layman's terms, the point we're looking for is somewhere up and to the left of the charge.

    After realizing the above stuff, the angle (theta) is 26.57 degrees (notice it's not negative, which was a mistake I made the first time around).

    The distance I came up with above was correct, but not to the right amount of sig. figs. To the correct sig figs, the point is 2.236cm from the charge. Then, d_x = rcos(theta) = 2cm and d_y = rsin(theta) = 1cm. So, the point ends up being at (0cm, cm). Bada bing!
     
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