Coulomb's Law Application: A charge repelling a mass on a frictionless incline

  • #1
DanielGuh
3
2
Homework Statement
At what distance from the 4 x 10-6 C charge is the 2.30 gram mass shown below (file attached) in equilibrium?

The angle of the frictionless incline is 28 degrees.

a) 1.31cm
b) 4.51m
c) 1.60m
d) 0.451m
Relevant Equations
Fg=mgcos(theta)
Fc=kq1q2/r^2
electrostatic constant (k) = 8.99*10^9
Diagram.png
Fg = Fc
Fg = 2.3g*9.8m/s^2*cos(28)=19.90N
19.90 = (8.99*10^9)*(4*10^-6)*(6*10^-6)/(r^2)
1/(r^2) = 92.23
r = 0.104m

However, it's not one of the option...
 
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  • #2
DanielGuh said:
Homework Statement: At what distance from the 4 x 10-6 C charge is the 2.30 gram mass shown below (file attached) in equilibrium?

The angle of the frictionless incline is 28 degrees.

a) 1.31cm
b) 4.51m
c) 1.60m
d) 0.451m
Relevant Equations: Fg=mgcos(theta)
Fc=kq1q2/r^2
electrostatic constant (k) = 8.99*10^9

View attachment 331275
Fg = Fc
Fg = 2.3g*9.8m/s^2*cos(28)=19.90N
19.90 = (8.99*10^9)*(4*10^-6)*(6*10^-6)/(r^2)
1/(r^2) = 92.23
r = 0.104m

However, it's not one of the option...
A couple of mistakes.
The downslope component of the gravitational force is not mgcos(theta).
Grams are not kilograms.
There may be more.
 
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  • #3
haruspex said:
A couple of mistakes.
The downslope component of the gravitational force is not mgcos(theta).
Grams are not kilograms.
There may be more.
Ohh I got it, the unit should be change, and when I use sin I find the answer is 4.51m. However, I don't know why do we use sin instead of cos.
 
  • #4
DanielGuh said:
However, I don't know why do we use sin instead of cos.
Why did you use ##\cos## in the first place?
 
  • #5
DanielGuh said:
Ohh I got it, the unit should be change, and when I use sin I find the answer is 4.51m. However, I don't know why do we use sin instead of cos.
Sin is for vertical axis, cos for horizontal.
 
  • #6
DeBangis21 said:
Sin is for vertical axis, cos for horizontal.
That doesn't sound right.
 
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  • #7
DeBangis21 said:
Sin is for vertical axis, cos for horizontal.
It doesn't always work like that. In this problem, the direction is the slope's - not vertical or horizontal.

Being able to resolve a force (or any vector) into components is necessary for many sorts of problems. You can click here for a video explaining it from the basics. The last part of the video covers how to deal with an object on a slope.
 
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  • #8
DanielGuh said:
Ohh I got it, the unit should be change, and when I use sin I find the answer is 4.51m. However, I don't know why do we use sin instead of cos.
A useful check is to think about the ##\theta=0## case. No downslope force, so you want the function that gives 0 at 0.
 
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  • #9
Steve4Physics said:
It doesn't always work like that. In this problem, the direction is the slope's - not vertical or horizontal.

Being able to resolve a force (or any vector) into components is necessary for many sorts of problems. You can click here for a video explaining it from the basics. The last part of the video covers how to deal with an object on a slope.
This video helps a lot, thanks!
 
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  • #10
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1. What is Coulomb's Law?

Coulomb's Law is a fundamental law of electrostatics that describes the force between two charged particles. It states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

2. How does Coulomb's Law apply to a charge repelling a mass on a frictionless incline?

In this scenario, the charge and the mass are both considered to be point particles, and the force between them can be calculated using Coulomb's Law. The charge will exert an electrostatic force on the mass, causing it to repel and move along the incline.

3. What is the formula for calculating the force between a charge and a mass using Coulomb's Law?

The formula is F = (k * q * m) / d^2, where F is the force, k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2), q is the charge of the particle, m is the mass of the particle, and d is the distance between them.

4. How does the distance between the charge and the mass affect the force?

According to Coulomb's Law, the force between two particles is inversely proportional to the square of the distance between them. This means that as the distance increases, the force decreases, and vice versa.

5. What is the significance of the frictionless incline in this application of Coulomb's Law?

The frictionless incline ensures that there are no other external forces acting on the mass besides the electrostatic force from the charge. This allows for a more accurate calculation of the force between the charge and the mass using Coulomb's Law.

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