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Coulomb's Law - finding the x and y components of an electric field at the origin

  1. Jul 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Two test charges are located in the x–y plane. If q1 = -2.75 nC and is located at x = 0.00 m, y = 0.800 m and the second test charge has magnitude of q2 = 3.20 nC and is located at x = 1.00 m, y = 0.400 m,

    calculate the x and y components, Ex and Ey, of the electric field, , in component form at the origin, (0,0). The Coulomb Force constant is 1/(4*pi*ε) = 8.99 × 10^9 N·m^2/C^2

    given:
    q1 = -2.75 nC; x = 0.00 m, y = 0.800 m
    q2 = 3.20 nC; x = 1.00 m, y = 0.400 m

    2. Relevant equations
    E=kq/r^2

    3. The attempt at a solution
    So what I did was take the q1 and q2 values (given in nC) and convert them to C.
    Next I took found the distance from the origin of the two points, solved using the E=kq/r^2 for each point where k=8.99e+9.
    I found the components of the E1 (has only j component) and E2 (by solving the angle from the origin using tan^-1(y-distance/x-distance) and multiplying that by the E2 value)
    then I added the two vectors together but still do not get the right answer.

    I Calculated:
    q1 = -2.75e-9 C
    q2 = 3.20e-9 C
    r1 = .8
    r2 = sqrt(1^2 + .4^2) = 1.077
    theta = tan^-1(.4/1) = 21.8 deg

    I solved the equation
    so E1 = -38.6 j (N/C)
    and |E2| = 24.8 N/C => separate into components
    E2 = 23.03 i + 9.21 j (N/C)

    thus Ex = 23.03 N/C
    Ey = -15.6 N/C
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 5, 2012 #2

    cepheid

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    Staff Emeritus
    Science Advisor
    Gold Member

    Welcome to PF,

    Basically, the directions of your components are wrong.

    In vector form, the electric field of a charge q is $$\mathbf{E} = \frac{kq}{r^2}\mathbf{\hat{r}}$$Boldface quantities represent vectors, and the unit vector ##\mathbf{\hat{r}}## is a unit vector that points "radially outward" (i.e. away from the charge that is the source of the field). The vector ##-\mathbf{\hat{r}}## points towards the charge that is the source of the field. Unlike the Cartesian unit vectors ##\mathbf{\hat{i}}##, ##\mathbf{\hat{j}}##, and ##\mathbf{\hat{k}}##, whose directions are fixed, the direction of ##\mathbf{\hat{r}}## varies depending on where you are in space. In the case where you're located at the origin, and the source of the field, q1, is located on the y-axis at (0,0.8), the radial unit vector, which points away from the source charge and towards the location where you are evaluating the field, is in the ##-\mathbf{\hat{j}}## direction (i.e. ##\mathbf{\hat{r}}## = ##-\mathbf{\hat{j}}## in this particular case). HOWEVER because q1 is negative, this negative sign cancels out the one on the ##\mathbf{\hat{j}}##, and the E-field ends up pointing in the positive ##\mathbf{\hat{j}}## direction. So E1 is directed upwards along the y-axis, towards the charge q1. This make sense, because q1 is negative, so a positive test charge placed at the origin would be attracted up towards it.

    Using a similar argument, you can reason that the x and y components of E2 should be in the ##-\mathbf{\hat{i}}## and ##-\mathbf{\hat{j}}## directions respectively (again, because ##\mathbf{\hat{r}}## points away from q2 and towards the origin, and q2 is positive this time).
     
    Last edited: Jul 6, 2012
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