To tackle this problem, we need to analyze the forces acting on the charged particle as it falls towards the charged disc. The disc has a uniform positive surface charge density, which creates an electric field that influences the motion of the particle. Let's break this down step by step.
Understanding the Electric Field of the Disc
The electric field \( E \) generated by a uniformly charged disc at a point along its axis can be expressed as:
E = \frac{\sigma}{2\epsilon_0} \left(1 - \frac{z}{\sqrt{z^2 + a^2}}\right)
where:
- \( \sigma \) is the surface charge density of the disc.
- \( \epsilon_0 \) is the permittivity of free space.
- \( z \) is the distance from the center of the disc along its axis.
- \( a \) is the radius of the disc.
Finding the Height H
When the particle of mass \( m \) and charge \( q \) is dropped from height \( H \), it experiences gravitational force and the electric force due to the electric field of the disc. The gravitational force \( F_g \) acting on the particle is given by:
F_g = mg
where \( g \) is the acceleration due to gravity. The electric force \( F_e \) acting on the particle is:
F_e = qE
Substituting the expression for \( E \), we get:
F_e = q \cdot \frac{\sigma}{2\epsilon_0} \left(1 - \frac{z}{\sqrt{z^2 + a^2}}\right)
As the particle falls, it will reach a point where the electric force balances the gravitational force. Setting \( F_g = F_e \):
mg = q \cdot \frac{\sigma}{2\epsilon_0} \left(1 - \frac{z}{\sqrt{z^2 + a^2}}\right)
We can rearrange this equation to find the height \( z \) at which the forces balance. However, we need to find the height \( H \) from which the particle is dropped such that it just reaches the disc. To find \( H \), we can use the relationship between the charge-to-mass ratio \( \frac{q}{m} \) given as \( 4 \, \text{g/s} \). This can be expressed in SI units as \( 4 \times 10^{-3} \, \text{C/kg} \).
Substituting \( \frac{q}{m} \) into our force balance equation, we can derive the height \( H \) by integrating the motion equations under the influence of both forces. The exact calculation will depend on the values of \( \sigma \) and \( a \), but the general approach involves finding the point where the net force is zero.
Potential Energy and Equilibrium Position
The potential energy \( U \) of the particle in the electric field can be expressed as:
U(z) = mgz + qV(z)
where \( V(z) \) is the electric potential due to the disc at height \( z \). The potential \( V(z) \) can be found by integrating the electric field:
V(z) = -\int E \, dz
To sketch the potential energy as a function of height, we would plot \( U(z) \) against \( z \). The equilibrium position is found where the derivative of the potential energy with respect to height is zero:
\frac{dU}{dz} = 0
This point indicates where the forces acting on the particle are balanced, leading to stable or unstable equilibrium depending on the second derivative test.
Summary
In summary, to find the height \( H \) from which the particle is dropped, we need to set up the equations of motion considering both gravitational and electric forces. The equilibrium position can be determined by analyzing the potential energy function. This problem beautifully illustrates the interplay between electric fields and gravitational forces in charged particle dynamics.
