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Electric field between two uniformly, linear charged rods

  1. Jun 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Problem 27.9 Two 10-cm-long thin glass rods uniformly charged to +10nC are placed side by side, 4.0 cm apart.
    What are the electric field strengths E1, E2, E3 and at distances 1.0 cm, 2.0 cm, and 3.0 cm to the right of the rod on the left, along the line connecting the midpoints of the two rods?

    2. Relevant equations

    Eline= 1/4πε0 * 2|λ|/r


    3. The attempt at a solution

    I understand that the Electric field at the midpoint would be zero. Point one and two will have the same strength but different directions.

    I believe point one could be calculated by E1-(E3/3
    Since the linear, uniform charge is inversely, linearly porportional.

    For E1 I get 1.8E5 N/C; so, 1.8E5 N/C -(1.8E5N/C / 3) = 1.2E5 N/C

    Mastering Physics reports the answer as 1.25E5 N/C
     
  2. jcsd
  3. Jun 29, 2012 #2

    collinsmark

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    Hello tetris4life,

    Welcome to Physics Forums!

    The equation which you used,
    Eline= 1/(4πε0) * 2|λ|/r
    is only valid for an infinitely long line (or infinitely long cylinder).

    It also works as a very good approximation for the electric field near the midpoint of a finite line (or finite cylinder) if r << L, where r is the distance to the line and L is the length of the line. But in this problem, the distance to one of the rods is 3 cm, and the length of the rod is only 10 cm. It's true that 3 cm < 10 cm. But I would not go so far as to say 3 cm << 10 cm.

    Starting with Coulomb's law (in vector form), if you go through the vector calculus to find the electric field, you will find that Mastering Physics is correct*.

    *(correct in this particular instance, anyway.)
     
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