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

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Homework Help Overview

The discussion revolves around applying Coulomb's Law to determine the distance at which a charged object creates equilibrium for a mass on a frictionless incline. The problem involves gravitational forces and electrostatic forces acting on the mass, with specific parameters given, including the angle of the incline.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculations involving gravitational and electrostatic forces, questioning the use of cosine versus sine in resolving forces along the incline. There are attempts to clarify the correct approach to finding equilibrium and the implications of unit conversions.

Discussion Status

Several participants have identified potential mistakes in the initial calculations and assumptions. There is ongoing exploration of the correct trigonometric functions to use in the context of the incline, with some guidance provided on resolving forces into components. The discussion reflects a mix of confusion and clarification regarding the application of physics principles.

Contextual Notes

Participants note discrepancies in the initial calculations and the importance of using the correct units. The discussion also highlights the need to understand the direction of forces in relation to the incline.

DanielGuh
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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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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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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.
 
DanielGuh said:
However, I don't know why do we use sin instead of cos.
Why did you use ##\cos## in the first place?
 
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.
 
DeBangis21 said:
Sin is for vertical axis, cos for horizontal.
That doesn't sound right.
 
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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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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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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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