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Homework Help: Electric Potential insulating rod

  1. Nov 24, 2005 #1
    1.) An insulating rod having linear charge density 40 uC/m and linear mass desnity u = 0.1 kg/m is released from rest in a uniform electric field E = 100V/m directed perpendicular to the rod.

    What is speed of the rod after travelling 2.00m?

    In previous questions, i used the formulas V = -Ed, where E is the magnitude of the electric field and d is the distance from the point charge. V would give me the potential (or electric potential) of the field. However, here i have strange values like "linear charge density" and "unfiform electric field". How do i deal with these ??

    2.) Two insulating spheres have radii 0.30 cm and 0.50 cm with masses 0.1kg and 0.7kg. The charges are evenly distributed with values -2.00 uC and 3.00 uC. They are released from rest when their centers are seperated by 1.0m. a.) How fast are they moving when they collide?

    I immediately thought that somehow i will have to use conservation of momentum. But setting up my equation:

    [tex]m_1v_1 + m_2v_2 = m_t * v_t [/tex] (where mt and vt are combinded mass and velocity).

    Where do i go from here?

    Thank you.
  2. jcsd
  3. Nov 24, 2005 #2


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    In #1, to get how fast a mass is traveling, one needs to know the acceleration, or force and mass.

    Now think of F = qE and F = ma or a = F/m = qE/m or (q/m)E.

    Electric field intensity has units V/m = (J/C)/m = (N-m/C)/m = N/c.

    One has the charge/unit length and mass/unit length, so use the charge/unit mass.

    In #2, the centers of the spheres are separated by 1 m, and the radii are 0.3cm and 0.5cm, so they will contact in 0.921 m.

    The electric force between the two is initially F=kq1q2/d2 where d is the separation.

    So the accelerate toward each other after starting simultaneously, and they must meet at the same time.

    Each is subject to the same force which is increasing as they appoach, but the acceleration is different - a1=F/m1 and a2=F/m2, and they each start with zero velocity.
  4. Nov 24, 2005 #3
    Thank you. I solved the first question and am planning to solve the other one later.
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