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Homework Help: Electric Fields (vectors)

  1. Apr 12, 2009 #1
    1. The problem statement, all variables and given/known data

    The x-y plane contains point charge with magnitudes and positions as follows:

    A charge of Q lies at co-ordinates (0, a)
    A charge of -Q lies at co-ordinates (0,-a)

    Show that the magnitude of the electric field at (b, 0) is given by

    E = Qa / [2πε0(a²+b²)^(3/2)]

    ε0 = Epsilon nought = 8.85 x10^-12

    2. Relevant equations

    E = Q/ 4πε0r²

    3. The attempt at a solution

    E due to Q is

    E1 = Q/ 4πε0r²
    E1 = Q/ 4πε0(a²+b²)

    Therefore the vector E'

    E' = E1*Cos(x) i^ + E1*Sin(x) j^
    E' = QCos(x) / 4πε0(a²+b²) i^ + QSin(x)/ 4πε0(a²+b²) j^

    E due to -Q is

    E2 = Q/ 4πε0r²
    E2 = Q/ 4πε0(a²+b²)

    Therefore the vector E''

    E'' = E2*Cos(x) i^ + E2*Sin(x) j^
    E'' = QCos(x) / 4πε0(a²+b²) i^ + QSin(x)/ 4πε0(a²+b²) j^

    Vector E = E' + E''
    E = QCos(x) / 4πε0(a²+b²) i^ + QSin(x)/ 4πε0(a²+b²) j^
    + QCos(x) / 4πε0(a²+b²) i^ + QSin(x)/ 4πε0(a²+b²) j^
    E = QCos(x) / 2πε0(a²+b²) i^ + QSin(x)/ 2πε0(a²+b²) j^

    Magnitude of E
    E = sqrt{Q²Cos²(x) / [2πε0(a²+b²)]² + Q²Sin²(x)/ [2πε0(a²+b²)]²}

    E = Q / 2πε0(a²+b²)

    As you can see its close lol, need Q*a not Q by itself and (a²+b²) should be(a²+b²)^(3/2)?
  2. jcsd
  3. Apr 13, 2009 #2


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    Science Advisor
    Gold Member

    remember that one charge is +Q and another is -Q. Also remember that electric fields are vectors and therefore direction matters, so your use of "cos(x)" and "sin(x)" in both equations begs the question "what is x referring to here?" So, my advice is to draw a diagram. It will make the question much simpler.
  4. Apr 13, 2009 #3
    Sorry i forgot to explain what x is. After drawing a diagram with resolved forces at (b,0), x would be the angle between the resultant E fields and the x axis. I am not sure which signs need to be negative any clues/advice/lecture will be appreciated very much.
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