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Homework Help: Electric Field Q

  1. Aug 7, 2007 #1
    The problem statement, all variables and given/known data

    Two charges -1nC and +1nC are positioned at (-2,0) and (2,0) respectively, where distances are in cm.

    a) Sketch 5-6 field lines E in the region bounded by x=-2, x=2, y=0 and y=8

    b) Caclculate the horizontal and verticle components of the E field at (-2,6), Draw a diagram showing the charges, the fields at P due to both charges and the resultant field. Find magnitude E and the angle phi (measured from the +x-axis)

    The attempt at a solution

    a)http://img241.imageshack.us/img241/1310/untitledxg6.jpg [Broken]

    b) I'm not sure how to calculate the hotizontal and ver components of the E at (-2,6)!
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Aug 7, 2007 #2

    Doc Al

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    What's the electric field at that point due to each point charge? Add the two fields--they are vectors--to get the net field at that point. (Hint: Find and add the components.)
  4. Aug 7, 2007 #3
    Do I use coulombs law, |F| = (k)(|q1|)(|q2|)/(r^2)

    and then E = F/Qo ?

    I found the second formula in my notes, but not sure how Qo comes into it?
  5. Aug 7, 2007 #4

    Doc Al

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    Let q1 be your point charge (for which you are finding the field) and q2 be a test charge at point P. q2 = Qo.

    When all is said and done, the magnitude of the field from a point charge Q is:
    |F| = k|Q|/(r^2)
  6. Aug 8, 2007 #5
    P doesn't appear to have a charge, so would I use
    |F| = (k)(|q1|)(|q2|)/(r²)
    |F| = k|Q|/(r²)?
  7. Aug 10, 2007 #6
    I think I pretty much managed to do it..
    Basically I used formula E = kq/r2 to find the E on P by point A(2,0) and the force on P by point B(-2,0).

    Then I found the x and y components of the E on P by A.
    Then calculated resultant in x direction = 1025.95V/m (this is equal to the E on P by A as E on P by B does not contribute to x direction.)


    resultant in y direction = 2497.22(E on P by B in y direction) - 1521.03 (E on P by A in y direction)

    Then I found magnitude (1416.17V/m) by pythag and angle using trig (44deg)
    Heres a diagram, I know its messy

    http://img402.imageshack.us/img402/83/img0351afn1.jpg [Broken]
    Last edited by a moderator: May 3, 2017
  8. Aug 10, 2007 #7


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    okay, except that phi is supposed to be from the +x -axis.

    Watch your language, though ...
    the E-field is caused by Charges, not points,
    and the E doesn't "act on" a point (only acts on a charge).
  9. Aug 10, 2007 #8
    so phi = 180+44 = 224deg?
  10. Aug 10, 2007 #9


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    I'm getting different numbers. For the x-component due to the +1 n C... I get

    [tex]\frac{kq}{r^2} = \frac{(9*10^9)(1*10^{-9})}{52}[/tex]

    So that comes out to 9/52 = 0.173 V/m

    then take the x - component... [tex]\frac{4}{\sqrt{52}}*0.173[/tex] which comes out to 0.096 V/m.

    And for the y - component [tex]\frac{6}{\sqrt{52}}*0.173[/tex] which comes out to 0.144 V/m

    I'm just giving the magnitudes, I'm ignoring the signs here...
  11. Aug 10, 2007 #10
    I was partially wrong, but I think you are too. (?)

    Units are in cm, so root(52)=0.07cm
    0.07^2 = 0.052

    But im confused. Do you convert to m after squaring the square root of (52) or before?
  12. Aug 10, 2007 #11


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    Oops. Sorry about that.

    I made a mistake earlier. You should convert before squaring. You meant to write 0.07m right?
    Last edited: Aug 10, 2007
  13. Aug 10, 2007 #12
    yes lol.
    so radius should actually be..

    root (52) = 7.211cm
    7.211^2 = 52cm
    52cm = 0.52m

    But the other calculation is right for Epb
    ((8.99*10^9)*(1*10^-9))/(0.06^2) = 2497.22V/m.

    One is so big, one is so small, could this be right?
    Last edited: Aug 10, 2007
  14. Aug 10, 2007 #13


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    distances are in centi-meters ... not meters, so
    9e9 [Vm/C]*1e-9[C]/.0052[m^2] = 1731 V/m ,
    . . . yielding an x-component of 960 V/m ... hmm. roundoff error?
  15. Aug 10, 2007 #14
    How did you manage to get 0.0052?
  16. Aug 10, 2007 #15


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    Sorry... I made a mistake in my post earlier that I edited afterwards. You should definitely convert to m before squaring.
  17. Aug 10, 2007 #16
    Sweet, thanks! Finally got that all sorted..

    The x component then becomes 966.76 V/m
    and y component then becomes -1433.28V/m

    Resultant thus becomes
    x direction = 966.76 V/m
    y direction = 2497.22-1433.28 = 1063.94

    and magnitude becomes 1437.57V/m.

    My phi now changes to 180+48=228deg measured from positive x axis
  18. Aug 10, 2007 #17


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    Cool! remember to convert your units before you do any squaring or anything... sorry about my earlier blunder.
  19. Aug 10, 2007 #18

    Not to worry! Thanks alot for helping in this thread and my other one! *bows to physics god*
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