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Electrostatics problem - accelerations of charged beads
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[QUOTE="OckhamsHammer, post: 4513188, member: 488882"] [b]Electrostatics problem -- accelerations of charged beads[/b] 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 I am 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 [/QUOTE]
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Electrostatics problem - accelerations of charged beads
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