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

In summary, the scenario involves a charged object on a frictionless incline, where Coulomb's Law describes the electrostatic force between the charge and another charge or mass. The repulsion caused by the electric charge affects the motion of the mass along the incline, illustrating the interplay between gravitational and electrostatic forces. The analysis highlights how the angle of the incline and the magnitude of the charges influence the resultant motion, emphasizing Coulomb's Law's role in predicting the behavior of charged objects in various configurations.
  • #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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FAQ: Coulomb's Law Application: A charge repelling a mass on a frictionless incline

What is Coulomb's Law and how does it apply to a charge repelling a mass on a frictionless incline?

Coulomb's Law describes the electrostatic force between two charged objects. It states that the magnitude of the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. In the context of a charge repelling a mass on a frictionless incline, Coulomb's Law helps determine the electrostatic force acting on the mass due to the charge, which affects the mass's motion up the incline.

How do you calculate the electrostatic force between two charges on a frictionless incline?

The electrostatic force \( F \) between two charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by Coulomb's Law: \( F = k \frac{|q_1 q_2|}{r^2} \), where \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)). On a frictionless incline, this force will act along the line joining the two charges.

How does the incline angle affect the motion of the mass under the influence of Coulomb's force?

The incline angle \( \theta \) affects the components of forces acting on the mass. The electrostatic force has to be resolved into components parallel and perpendicular to the incline. The component of the electrostatic force parallel to the incline will affect the acceleration of the mass up or down the incline, while the perpendicular component is balanced by the normal force from the incline surface.

What role does the mass of the object play in the scenario of a charge repelling a mass on a frictionless incline?

The mass of the object determines its inertia, which affects how it responds to the electrostatic force. According to Newton's second law, the acceleration \( a \) of the mass \( m \) is given by \( a = F_{\text{parallel}} / m \), where \( F_{\text{parallel}} \) is the component of the electrostatic force parallel to the incline. A larger mass will result in a smaller acceleration for the same force.

How do you determine the equilibrium position of the mass on the incline?

The equilibrium position is found when the net force acting on the mass along the incline is zero. This occurs when the component of the gravitational force down the incline \( mg \sin \theta \) is balanced by the component of the electrostatic repulsive force up the incline. Mathematically, \( mg \sin \theta = k \frac{|q_

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