Electrostatics problem - accelerations of charged beads

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Homework Help Overview

The discussion revolves around an electrostatics problem involving two charged beads: a plastic bead and a glass bead, with specific charges and masses. The participants are attempting to calculate the magnitudes of the accelerations of each bead based on their interactions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply equations related to electric fields and forces to determine the accelerations of the beads. Some participants question the interpretation of the equations used, particularly regarding the application of Coulomb's Law and the relationship between force and acceleration.

Discussion Status

Participants are actively engaging with the problem, with some offering clarifications on the equations involved and the principles of electrostatics. There is an exploration of the forces acting on each bead and how they relate to their respective accelerations, but no consensus has been reached on the correct approach or interpretation.

Contextual Notes

There is a noted confusion regarding the application of Coulomb's Law and the distinction between electric field and force. Participants are also considering the implications of Newton's Third Law in the context of the forces acting on the beads.

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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