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Electrostatics problem - accelerations of charged beads

  1. Sep 23, 2013 #1
    Electrostatics problem -- accelerations of charged beads

    Hi everyone

    Im reworking my MP homework and I can't seem to get the right answer for this problem(26.32). The set up says:

    A 1.5g plastic bead charged to -3.9nC and a 3.8g glass bead charged to 17.6nC are 2.2 cm apart(center to center).

    The question im having trouble answering is:

    What are the magnitudes of the accelerations of the plastic bead and glass bead? (and they want it in m/s^2)

    converting all the givens so I can use the correct equations it get:

    m(plastic)= 1.5*10^-3 kg
    q(plastic) = -3.9*10^-9 C

    m(glass) = 3.8*10^-3 kg
    q(glass) = 17.6*10^-9 C

    r= 2.2*10^-2 m


    Im using these 3 equations:

    (1)Electric field = (K q1*q2)/r^2

    (2)F= (Electric field)*(charge)

    (3)a= F/m

    where the mass and charge are different for each object

    Im getting for plastic:

    Electric field = 1.27*10^-3 N/C (in - i direction)

    F(plastic)= 4.97*10^-12 N (+ i)

    a(plastic)= 3.3*10^-9 m/s^2 (+i)

    For glass:

    E-field = same

    F(glass)= 22.4*10^-12 (-i)

    a(glass)= 5.9*10^-9 (-i)

    This is a really easy problem and I didn't have any trouble the 1st time through...what am I missing???

    -OH
     
  2. jcsd
  3. Sep 23, 2013 #2

    CAF123

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    Gold Member

    The force on the plastic bead and the glass bead should be the same. Why is this the case? Your eqns (1) and (2) combine to give Coulombs Law. After you have found the force on each of the beads, you can then determine each of their acceleration via N2, which will not be the same for both beads because they have different masses.
     
  4. Sep 23, 2013 #3
    That sounds like newtons 3rd law...but Im still a little confused.. equation(2) says I multiply the electro static force (given in N/C..which is the same for both charges) by the individual charge I'm studying to get the Force in newtons. Am I interpreting the equation wrong?
     
  5. Sep 23, 2013 #4

    gneill

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    Staff: Mentor

    Bit of a misunderstanding of the formula there. Your formula (1) gives the force acting on the charges (it's equal magnitude and opposite in direction on each charge). You could use the formula for the electric field, which involves a single charge, but then you multiply by the charge its acting on to find the force and you're back to Coulomb's law!

    So. Work out the magnitude of the forces from (1). Determine the directions of the forces by considering the charge signs and their relative locations.
     
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