Solve Electric Force II Homework Problem

In summary, the question presents an experiment involving a lightweight spring, a 1.0 g mass, and two small plastic beads to measure charge. The spring constant is found to be 0.981 from the first experiment, and the force between the two beads is calculated to be -0.004905 N. However, the mistake in the calculation of q^2 instead of 2q leads to an incorrect answer. The correct answer is found by taking the square root of the calculated value, resulting in a magnitude of charge of 33.221 nC on each bead.
  • #1
Goldenwind
146
0
[SOLVED] Electric Force II

Homework Statement


You have a lightweight spring whose unstretched length is 4.0 cm. You're curious to see if you can use this spring to measure charge. First, you attach one end of the spring to the ceiling and hang a 1.0 g mass from it. This stretches the spring to a length of 5.0 cm. You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.5 cm.

What is the magnitude of the charge (in nC) on each bead?

Homework Equations


F = -k1 * x
F = k2 * q1q2 / r^2

g = a = -9.81
k1 = Spring constant. Unknown.
k2 = Electrostatic constant = 9*10^9

The Attempt at a Solution


In the first experiment
F = ma = -k1*x
0.001*-9.81 = -k1*(0.05 - 0.04)
Therefore the spring constant k1 = 0.981

Now the second experiment:
F = -k1*x
F = -(0.981)(0.045 - 0.04)
F = -0.004905
This is the force applied on a spring of k1 = 0.981, to stretch it 0.5cm

Similarily, this should also be the force between the two beads:
F = k2*q1q2 / r^2, however q1 = q2 according to the question.
F = k2*(2q) / r^2, solving for q.
Fr^2/(2*k2) = q
(-0.004905)(0.045)^2/(2*9*10^9) = q

q = -0.0000000000000005518125 C
q = -0.0000005518125 nC

This answer is incorrect.
Where is my mistake?
 
Last edited:
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  • #2
Goldenwind said:
Similarily, this should also be the force between the two beads:
F = k2*q1q2 / r^2, however q1 = q2 according to the question.
OK.
F = k2*(2q) / r^2, solving for q.
That should be q^2, not 2q.
 
  • #3
Doc Al said:
OK.

That should be q^2, not 2q.
I almost feel like I'm wasting your time, as my mistakes seem so trivial..

F = k2*q1q2 / r^2, however q1 = q2 according to the question.
F = k2*(q^2) / r^2, solving for q.
Sqrt((Fr^2)/k2) = q
Sqrt((-0.004905 * 0.045^2)/(9*10^9)) = q
Sqrt(-0.000009932625 / 9000000000) = q
q = -0.000000000000001103625 C
q = -0.000001103625 nC
q = -1.103625 * 10^-6 nC

This answer is also incorrect. While what you pointed out was indeed a mistake I made, there appears to be more >.<;
Reading over it again.
 
  • #4
Goldenwind said:
I almost feel like I'm wasting your time, as my mistakes seem so trivial..

F = k2*q1q2 / r^2, however q1 = q2 according to the question.
F = k2*(q^2) / r^2, solving for q.
Sqrt((Fr^2)/k2) = q
Sqrt((-0.004905 * 0.045^2)/(9*10^9)) = q
Sqrt(-0.000009932625 / 9000000000) = q
This is OK. (But get rid of that minus sign!)
q = -0.000 000 000 000 001 103625 C
Looks like you didn't take the square root.

(Please use scientific notation. All those zeroes are opportunities for error. And hard to read! :wink:)
 
  • #5
Doc Al said:
This is OK. (But get rid of that minus sign!)

Looks like you didn't take the square root.

(Please use scientific notation. All those zeroes are opportunities for error. And hard to read! :wink:)


F = k2*q1q2 / r^2, however q1 = q2 according to the question.
F = k2*(q^2) / r^2, solving for q.
Sqrt((Fr^2)/k2) = q
Sqrt((0.004905 * 0.045^2)/(9*10^9)) = q
Sqrt(9.932625*10^-6 / 9*10^9) = q
q = Sqrt(1.103625*10^15)
q = 3.3220851885525151294968428473179*10^-8 C
q = 33.221 nC

Checking... correct!

Doc Al is saving my education.
<3
 

1. How do I calculate electric force?

To calculate the electric force between two charged particles, you can use the formula F = (k * q1 * q2) / r^2, where k is the Coulomb's constant, q1 and q2 are the charges of the particles, and r is the distance between them. Make sure to use consistent units for charge and distance.

2. Can I use this formula for any type of charge distribution?

No, this formula is only applicable for point charges or spherically symmetric charge distributions. For other types of charge distributions, you may need to use integration to calculate the electric force.

3. What is the unit of electric force?

The unit of electric force is Newtons (N), which is a derived unit in the SI system. It can also be expressed in terms of other units such as kg*m/s^2 or C*m/s^2.

4. How does distance affect the electric force?

According to the inverse square law, as the distance between two charged particles increases, the electric force between them decreases. This means that the force is inversely proportional to the square of the distance between the particles.

5. Can electric force be repulsive or attractive?

Yes, electric force can be both repulsive and attractive. If two particles have the same charge, they will repel each other. If they have opposite charges, they will attract each other. This is described by Coulomb's law, which states that like charges repel and opposite charges attract.

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