a large insulating thick sheet of thickness 2 d carries a uniform charge per unit volume ρ. a particle of mass m, carrying a charge q having a sign opposite to that of the sheet, is released from the surface of the sheet. the sheet does not offer any mechanical resistance to the motion of the particle. find the oscillation frequency v of the particle inside the sheet.
a large insulating thick sheet of thickness 2 d carries a uniform charge per unit volume ρ. a particle of mass m, carrying a charge q having a sign opposite to that of the sheet, is released from the surface of the sheet. the sheet does not offer any mechanical resistance to the motion of the particle. find the oscillation frequency v of the particle inside the sheet.
Answer: To find the oscillation frequency of the particle inside the sheet, we can consider the equilibrium position and use the principles of simple harmonic motion.
- Equilibrium Position:
When the particle is at its equilibrium position inside the sheet, the electrical force due to the charged sheet must balance the gravitational force acting on the particle.
Electrical force (Fe) = Gravitational force (Fg)
The electrical force is given by Coulomb’s Law:
Fe = (1/4πε₀) * (q * ρ * A) / d²
Here, ε₀ is the vacuum permittivity, q is the charge of the particle, ρ is the charge density of the sheet, A is the area of the sheet, and d is the thickness of the sheet.
The gravitational force is given by:
Fg = mg
where m is the mass of the particle and g is the acceleration due to gravity.
Setting Fe equal to Fg, we have:
(1/4πε₀) * (q * ρ * A) / d² = mg
- Oscillation Frequency:
To find the oscillation frequency, we need to consider the effective spring constant of the system. Since the sheet does not offer any mechanical resistance, the electrical force acts as the restoring force, and the system behaves like an electric spring.
The electrical force can be written as:
Fe = k * x
where k is the effective spring constant and x is the displacement of the particle from the equilibrium position.
By comparing this with Fe = (1/4πε₀) * (q * ρ * A) / d², we can equate the two expressions:
k * x = (1/4πε₀) * (q * ρ * A) / d²
From Hooke’s law, we know that k = mω², where ω is the angular frequency. Therefore, we can rewrite the equation as:
mω²x = (1/4πε₀) * (q * ρ * A) / d²
The angular frequency ω can be related to the oscillation frequency v by ω = 2πv.
Substituting this into the equation, we have:
m(2πv)²x = (1/4πε₀) * (q * ρ * A) / d²
Simplifying the equation, we can solve for v:
v² = (1/4πε₀m) * (q * ρ * A) / (d²x)
v = √[(1/4πε₀m) * (q * ρ * A) / (d²x)]
This equation gives the oscillation frequency v of the particle inside the sheet, given the relevant parameters and the displacement x from the equilibrium position.