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Need confirmation on a problem

  1. Feb 12, 2005 #1
    point charges, Q1 (4,0,-3) and Q2 (2,0,1). Q2 = 4 nC, find Q1 such that the a. electric field at (5,0,6) has no z component.
    b. force on a test charge at (5,0,6) has no x component.

    heres what i did for a:

    i found the distance from Q1 to (5,0,6) to be the square root of 82
    distance from Q2 to (5,0,6) to be square root of 34.

    i than found the unit vectors.
    multipled by the charges and set the z components equal.

    so i had 9Q1/sqr rt 82 = - 4nc*5/sqr rt 34
    ended up with Q1 = -3.41 nc. answer in the back of the book says -3.463

    on the second part i found the two forces using (Q*Qt)/4*pi*Eo*R^2.
    i than multiplied these by the unit vectors in the x direction and set one of them equal to negative the other and solved for Q1. i ended up with an answer of -48nC, the book says -18nC.
     
  2. jcsd
  3. Feb 12, 2005 #2

    dextercioby

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    Why are there square roots in the denominators in the first equality.Shouldn't they have disappeared when squared...?

    Daniel.
     
  4. Feb 12, 2005 #3
    thats how the unit vector is found. divide the change in x or y or z and than divide by the distance between the two points. distance between the two points is the square root of (x2-x1) + (y2-y1) + (z2-y1)
     
  5. Feb 12, 2005 #4

    dextercioby

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    Yes,but you should have equated "z" components of the fields ALTOGETHER...Does that change anything...?

    Daniel.
     
  6. Feb 12, 2005 #5
    so i find electric field due to Q1 and Q2 by using this formula right:
    E = F/Q
    where F = (Q*Qt)/(4*pi*Eo*R^2). i dont see how thats going to work because the Qs are going to cancel.
     
  7. Feb 12, 2005 #6

    dextercioby

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    No,they're not.What's the general expression of an electric field generated by a point charge "q" situated in the origin of a coordinate system,field created at point of position vector [itex] \vec{r} [/itex] ??

    Daniel.
     
  8. Feb 12, 2005 #7
    E = (q*Q/4*pi*Eo*R^2)/q Ar
    right?
     
  9. Feb 12, 2005 #8

    dextercioby

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    No.It's more like a vector,of this form
    [tex] \vec{E}=\frac{q}{4\pi\epsilon_{0}|\vec{r}|^{2}} \frac{\vec{r}}{|\vec{r}|} [/tex]

    For your problem,chose a system of coordinates and compute the total field at the point you're interested.Just then,u can set the component of the resulting field equal to zero.

    Daniel.
     
  10. Feb 12, 2005 #9
    k, what about the second part? im pretty sure i did that right
     
  11. Feb 12, 2005 #10
    and i tred it the way you told me and i got an answer of -8.3 nC for Q1. not the same as the back of the book.
     
  12. Feb 12, 2005 #11

    dextercioby

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    For the second part,you need to have done first part right...Because it uses the charge calculated at point "a"...

    Daniel.
     
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