Electrostatics problem - accelerations of charged beads

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SUMMARY

The discussion centers on calculating the accelerations of two charged beads: a 1.5g plastic bead with a charge of -3.9nC and a 3.8g glass bead with a charge of 17.6nC, positioned 2.2 cm apart. The participant initially misapplied Coulomb's law and the equations for electric field and force. Correctly applying the equations reveals that the forces acting on both beads are equal in magnitude but opposite in direction, leading to different accelerations due to their differing masses. The final calculated accelerations are 3.3 x 10^-9 m/s² for the plastic bead and 5.9 x 10^-9 m/s² for the glass bead.

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  • Understanding of Coulomb's law and electric forces
  • Knowledge of Newton's second law of motion
  • Familiarity with basic electrostatics concepts
  • Ability to perform unit conversions (grams to kilograms, nanocoulombs to coulombs)
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  • Study Coulomb's law and its application in electrostatics
  • Learn about electric fields generated by point charges
  • Explore Newton's laws of motion in the context of charged particles
  • Practice problems involving forces and accelerations of charged objects
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OckhamsHammer
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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 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
 
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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.
 
That sounds like Newtons 3rd law...but I am 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?
 
OckhamsHammer said:
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 <---- Coulomb's law: yields the force, not the field.
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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