Two indentical beads w/ mass m and charge q in hemispherical

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SUMMARY

This discussion focuses on calculating the charge q on two identical beads, each with mass m and charge q, placed in a hemispherical bowl of radius R. At equilibrium, the beads are separated by a distance d. The key equations utilized include the electric force equation Fe = (Ke*q^2)/d^2 and the balance of forces involving gravitational and normal forces. The final expression derived for the charge q is q = sqrt((d^2 * (mg/sin(x)) * cos(x))/ Ke), where x is the angle between the radius R and the distance d.

PREREQUISITES
  • Understanding of electrostatics, specifically Coulomb's law.
  • Knowledge of free body diagrams and equilibrium conditions.
  • Familiarity with trigonometric functions and their applications in geometry.
  • Basic principles of forces acting on objects in a gravitational field.
NEXT STEPS
  • Learn how to express trigonometric functions in terms of geometric parameters, specifically using geometry to relate sin(x) and cos(x) to d and R.
  • Study the concept of electric force in more complex systems, including multiple charges and their interactions.
  • Explore the principles of equilibrium in physics, focusing on static and dynamic scenarios.
  • Investigate the applications of Coulomb's law in real-world scenarios, such as in particle physics or electrostatic applications.
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Students studying physics, particularly those focusing on electrostatics and mechanics, as well as educators seeking to clarify concepts related to forces and equilibrium in charged systems.

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Homework Statement


Two identical beads each have a mass m and charge q. When placed in a hemispherical bowl of radius R with frictionless, nonconducting walls, the beads move, and at equilibrium, they are a distance d apart (see figure below). Determine the charge q on each bead.

Homework Equations


Fe = (Ke*q^2)/d^2)

The Attempt at a Solution


I drew a free body diagram first. Since the beads are at equilibrium, the net force must be zero.
So, considering the electric force, the gravitational force, and the normal force of the hemisphere (where x is the angle between R and d) I got..

Fx = Ncos(x) - Fe = 0
Fy = Nsin(x) - mg = 0

Ncos(x) = Fe
Nsin(x) = mg

N = mg/sin(x)
(mg/sin(x))*cos(x) = (Ke*q^2)/d^2
d^2*(mg/sin(x))*cos(x) = Ke * q^2
q = sqrt((d^2 * (mg/sin(x)) * cos(x))/ Ke)

Now, I've seen problems similar to this where d is equal to R, giving an equilateral triangle, however that is not the case in this problem. I don't know how to eradicate the sin and cos of x or how to put x into terms of R and D.
Help!
 
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ortiz123 said:

Homework Statement


Two identical beads each have a mass m and charge q. When placed in a hemispherical bowl of radius R with frictionless, nonconducting walls, the beads move, and at equilibrium, they are a distance d apart (see figure below). Determine the charge q on each bead.

Homework Equations


Fe = (Ke*q^2)/d^2)

The Attempt at a Solution


I drew a free body diagram first. Since the beads are at equilibrium, the net force must be zero.
So, considering the electric force, the gravitational force, and the normal force of the hemisphere (where x is the angle between R and d) I got..

Fx = Ncos(x) - Fe = 0
Fy = Nsin(x) - mg = 0

Ncos(x) = Fe
Nsin(x) = mg

N = mg/sin(x)
(mg/sin(x))*cos(x) = (Ke*q^2)/d^2
d^2*(mg/sin(x))*cos(x) = Ke * q^2
q = sqrt((d^2 * (mg/sin(x)) * cos(x))/ Ke)

Now, I've seen problems similar to this where d is equal to R, giving an equilateral triangle, however that is not the case in this problem. I don't know how to eradicate the sin and cos of x or how to put x into terms of R and D.
Help!
Use a bit of geometry to express sin(x), cos(x) in terms of d and R.
 

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