Electrostatics problem - accelerations of charged beads

In summary: For plastic: m(plastic)= 1.5*10^-3 kg; q(plastic)= -3.9*10^-9 C; m(in-direction)= 0; m(out-direction)= 1.5*10^-3 kg;For glass: E-field = same; F(glass)= 22.4*10^-12 N (-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
  • #1
OckhamsHammer
2
0
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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  • #2
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.
 
  • #3
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?
 
  • #4
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.
 
  • #5


Hi OH,

It looks like you have set up the problem correctly and your calculations are also correct. Sometimes in physics, there can be small discrepancies due to rounding errors or experimental errors. My suggestion would be to double check your calculations and make sure you are using the correct units for each step. Also, try to use more significant figures in your calculations to reduce rounding errors. If you are still having trouble, you can try to solve the problem using a different method or approach to see if you get the same answer. Keep practicing and don't get discouraged, sometimes small mistakes can happen but it's important to keep trying and learning from them. Good luck!
 

Related to Electrostatics problem - accelerations of charged beads

What is electrostatics?

Electrostatics is the branch of physics that deals with the study of electric charges at rest. It involves the study of the behavior of electric charges and the electric and magnetic fields they produce.

What is an electrostatics problem?

An electrostatics problem is a type of physics problem that involves the analysis of the behavior of charged particles in an electric field. These problems typically involve calculating the forces and accelerations of charged particles in various configurations.

What is the acceleration of a charged bead in an electric field?

The acceleration of a charged bead in an electric field depends on the strength of the electric field, the magnitude and direction of the charge on the bead, and the mass of the bead. It can be calculated using the equation F = qE/m, where F is the force, q is the charge, E is the electric field, and m is the mass.

How does the acceleration of a charged bead change with distance from the source of the electric field?

The acceleration of a charged bead in an electric field decreases as the distance from the source of the electric field increases. This is because the electric field strength decreases with distance, and the force on the charged bead is directly proportional to the electric field strength.

What are some real-world applications of electrostatics?

Electrostatics has many practical applications in our daily lives, such as in electronic devices, power generation and distribution, and medical equipment. It is also used in industrial processes like painting and electroplating, and in environmental control systems such as air purifiers and electrostatic precipitators.

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