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

In summary: Then substitute into the equation for q and solve for q. In summary, the problem involves two identical beads with mass m and charge q placed in a hemispherical bowl of radius R. At equilibrium, the beads are a distance d apart. Using a free body diagram and considering the electric force, gravitational force, and normal force, an equation is derived to solve for the charge q on each bead. By expressing sin(x) and cos(x) in terms of d and R and substituting into the equation, the value of q can be determined.
  • #1
ortiz123
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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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  • #2
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.
 

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

1. What is the electric potential at the center of the hemispherical cavity?

The electric potential at the center of the hemispherical cavity is zero. This is because the electric field created by the two identical beads with opposite charges cancels out at the center, resulting in a net electric potential of zero.

2. How do the electric fields created by the two beads differ inside and outside of the hemispherical cavity?

Inside the hemispherical cavity, the electric field created by the two beads is zero. This is because the electric fields from the two beads cancel out inside the cavity. Outside of the cavity, the electric field is the same as that of a single point charge, with the magnitude given by Coulomb's law.

3. What is the force between the two beads?

The force between the two beads is given by Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Therefore, the force between the two beads will be repulsive if they have the same charge and attractive if they have opposite charges.

4. How does the mass of the beads affect the electric potential and electric field inside the hemispherical cavity?

The mass of the beads does not affect the electric potential or electric field inside the hemispherical cavity. These quantities depend only on the charge of the beads and their distance from the center of the cavity.

5. Can the beads be moved inside the cavity without affecting the electric potential at the center?

Yes, the beads can be moved inside the cavity without affecting the electric potential at the center. As long as the beads remain at the same distance from the center and maintain their charges, the electric potential at the center will remain zero.

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