A small sphere of mass m and charge +q with zero velocity falls under the influence of gravity (acceleration = g from height h onto an infinite uniformly charged plane with positive charge density sigma as shown below. What will the velocity of the sphere be when it hits the plane? Note that the electric field above an infinite plane of charge is constant everywhere, pointing away from the plane (for a positive charged plane) and equal to sigma/2*episilon . make the downward direction positive (acceleration due to gravity positive) and the repuslive electrostatic force up the negative direction

Question

A small sphere of mass m and charge +q with zero velocity falls under the influence of gravity (acceleration = g from height h onto an infinite uniformly charged plane with positive charge density sigma as shown below. What will the velocity of the sphere be when it hits the plane? Note that the electric field above an infinite plane of charge is constant everywhere, pointing away from the plane (for a positive charged plane) and equal to sigma/2*episilon . make the downward direction positive (acceleration due to gravity positive) and the repuslive electrostatic force up the negative direction

Answer 1

To find the velocity of the sphere when it hits the plane, we need to consider the forces acting on it: gravity and the electric force.

1. Gravity (Fg): Since the downward direction is taken as positive, the force due to gravity is given by Fg = mg, where m is the mass of the sphere and g is the acceleration due to gravity.

2. Electric force (Fe): The electric force experienced by the charged sphere in an electric field is given by Fe = qE, where q is the charge on the sphere and E is the electric field.

The electric field above an infinite plane of charge is constant and equal to E = σ/(2ε₀), where σ is the charge density of the plane and ε₀ is the permittivity of free space.

Since the electric force is acting in the opposite direction to gravity, we take it as negative. So Fe = -qE.

Now, using Newton's second law of motion, the net force acting on the sphere is given by the sum of the gravitational and electric forces:

Fnet = Fg + Fe

Since the sphere is falling, we can assume there is no air resistance, and the net force equals the mass times the acceleration:

Fnet = mg - qE = ma

Simplifying the equation, we get:

mg - q(σ/(2ε₀)) = ma

Solving for a, we have:

a = (mg - q(σ/(2ε₀)))/m

Now, let's analyze the situation when the sphere hits the plane. At that moment, its acceleration will be zero as it reaches its maximum velocity.

Thus, we have:

0 = (mg - q(σ/(2ε₀)))/m

Simplifying the equation for the velocity:

0 = mg - q(σ/(2ε₀))

qσ/(2ε₀) = mg

The initial velocity of the sphere is assumed to be zero.

Therefore, when the sphere hits the plane, its velocity will be zero.