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Calculating E due infinite line and point charge?

  1. Apr 13, 2015 #1
    1. The problem statement, all variables and given/known data
    Here is the problem I am stuck on. I have checked my process multiple times, but have come up with the same wrong result. I would like to find out where by error of thinking lies.

    "An infinitely long line charge of uniform linear charge density λ = -1.30 µC/m lies parallel to the y axis at x = -1.00 m. A point charge of 1.20 µC is located at x = 2.00 m, y = 3.00 m. Find the electric field at x = 3.00 m, y = 2.50 m."
    The answer form is Evecbolditalic.gif = [ ] kN/C

    Side note: since it's in vector form, does that mean they want the answer to be written in i hats and j hats?

    2. Relevant equations
    For a infinite line of charge a distance R away, the electric field produced is E = 2kλ / R.
    For just a point charge some distance r away, the electric field is E = kq / r2.

    3. The attempt at a solution
    I drew a picture in the x and y coordinate plane with the line of charge at x = -1, the point charge at (2, 3), and the analyze point at (3, 2.5).

    First, I calculated the electric field due to the infinite line of charge. Since the y-component cancels, the net electric field at (3, 2.5) due to the line of charge is just the horizontal x-component. So R = 3 - (-1) = 4.
    So E = 2kλ / R = 2(9*109)(-1.3*10-6) / 4 = -5850 N/C.

    Second I calculated the electric field due to the point charge. I used the distance formula to find the distance r between the point charge and (3, 2.5).
    r = sqrt[(3-2)2+(2.5-3)2] = sqrt[1.25]
    r2 = 1.25 meters
    So E = kq / r2 = (9*109)(1.2*10-6) / 1.25 = 8640 N/C.

    But this is the total. To get the x-direction of this electric field, I drew a vector triangle formed by the line between (2, 3) and (3, 2.5). The hypotenuse is sqrt(1.25) meters, the short side is |2.5-3| = 0.5 meters, and the long side is |3-2| = 1 meter. Using inverse trig, I got that the angle between the angle between the two points was 26.565 degrees.

    Using the angle, I then calculated the x and y components of the E due to the point charge.
    For the x-component I did Ecos(26.565) = 8640cos(26.565) = 7727.854 N/C.
    For the y-component I did Esin(26.565) = 8640sin(26.565) = 3863.92 N/C. It's actually -3863.92, since the force would be in the negative y direction.

    So the total E at the point (3, 2.5) is the total x and y-components.
    The total x-component is -5850 + 7727.854 = 1877.854 N/C
    The total y-component is -3863.92 N/C.

    This is where I am unsure about how to answer. Do I do the square root? Like E = sqrt[(1877.854)2 + (-3863.92)^2]? This result gives 4296.07 N/C. But wouldn't this be just the magnitude of the electric field at (3, 2.5), thus excluding direction? Or is the question wanting me to answer in some other form? Is some part of my process wrong?
     
  2. jcsd
  3. Apr 13, 2015 #2

    Simon Bridge

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    The electric field is a vector so you always have to specify a direction as well as a magnitude. How you do that is up to you - since all the information is given in Cartesian coordinates, you should probably use a Cartesian representation (it'll be easier).

    The square-root of the sum of the squares of the components would give you the magnitude, that is correct.
    Your process seems fine - work out the fields individually ##\vec E_{tot} = \vec E_{line} + \vec E_{point}## ... double check the arithmetic and make sure you got the signs right in the vector sum ...

    It can help to do the math in vector from from the get-go: $$\vec E_{tot} = \frac{2k\lambda}{R}\hat R + \frac{kQ}{r^3}\vec r$$ ... may hep eliminate rounding errors since you know ##\vec r = 2\hat\imath + 0.5\hat\jmath##(check) and ##\hat R## is, in this case, ##\hat\imath##.

    Lastly: how do you know the answer you got was wrong?
     
  4. Apr 13, 2016 #3
    Give the answer in vector components, and divide your answers by 1000 since the requested units are kN/C. Good luck on the exam tomorrow
     
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