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Calculate electric field at origin, with 3 charges

  1. May 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Three charges, +2.5[tex]\mu[/tex]C, -4.8[tex]\mu[/tex]C & -6.3[tex]\mu[/tex]C
    are located at (-0.20m, 0.15m), (0.50m, -0.35m) and (-0.42m, -0.32m) respectively. What is the electric field at the origin?
    q1 = +2.5[tex]\mu[/tex]C
    q2 = -4.8[tex]\mu[/tex]C
    q3 = -6.3[tex]\mu[/tex]C

    2. Relevant equations
    a[tex]^{}2[/tex] + b[tex]^{}2[/tex] = c[tex]^{}2[/tex]
    v[tex]_{}x[/tex] = magnitude [tex]\times[/tex]cos([tex]\theta[/tex])
    v[tex]_{}y[/tex] = magnitude [tex]\times[/tex]sin([tex]\theta[/tex])
    E = [tex]\frac{kq}{r^{}2}[/tex]
    law of cosines
    k= 9x10^9
    3. The attempt at a solution
    first i found the hypotenuse for the three charges.
    q1 = .25m
    q2 = .6103m
    q3 = .5280m

    then i used the formula for magnitude of an electric field
    where k is the constant, q was my three charges, and the radius were my three hypotenuses
    my results were:
    q1 = 360000
    q2 = 115983
    q3 = 203383

    i used the law of cosines to get [tex]\theta[/tex]
    my three angles were:
    1 = 36.8
    2 = 35
    3 = 37.3

    to find Ex, i multiplied the product of my magnitudes by the cosine of its respective angle:
    my results:
    1 = 288263
    2 = 95007
    3 = 161785
    i added these up and got 545055, the book says 2.2x10^5!
    i didnt bother doing y, since im completely lost!
    Please help me!
     
  2. jcsd
  3. May 17, 2009 #2

    LowlyPion

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    Homework Helper

    Doesn't the direction of q3 carry a negative sign ... i.e. pointing toward the right from the origin?

    Hence |E1| + |E2| - |E3| along x?

    288 + 95 - 161 = 222
     
  4. May 17, 2009 #3
    yeah q3 has a negative sign.
    So is the way i did the problem correct?
     
  5. May 17, 2009 #4
    I got another problem im trying to solve for y, but when i add everything up i get + 443958, not -4.1x10^5 like the book says.

    P.S. how do i know which ones to add, and which ones to subtract?
     
  6. May 17, 2009 #5

    LowlyPion

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    Homework Helper

    Remember the E-Field is a vector field.

    So you not only need to account for the sign of the charge, but you also must take into account where the point that you are taking the E-Field at is relative to the charge.

    A positive charge has radial outward field. A negative charge is radial inward. So depending on which side you are and whether it is a + or - is what determines the sign of the |E|, not simply which quadrant it may lay in.
     
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