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
To find the velocity of the sphere when it hits the plane, we need to consider the forces acting on the sphere.
1. Gravitational Force: The sphere is subject to the force of gravity since it's falling. The gravitational force acting on the sphere can be calculated using the equation F_gravity = m * g, where m is the mass of the sphere and g is the acceleration due to gravity.
2. Electric Force: The sphere also experiences an electric force due to the presence of the uniformly charged plane. The electric force can be calculated using the equation F_electric = q * E, where q is the charge on the sphere, and E is the electric field above the infinitely charged plane.
Since the electric field above an infinitely charged plane is constant, we can substitute E = sigma / (2 * epsilon), as mentioned in the question, where sigma is the charge density (positive in this case), and epsilon is the permittivity of free space.
Now, let's analyze the forces acting on the sphere. The gravitational force acts downward (-) while the electric force acts upward (negative direction). The net force acting on the sphere is given by the sum of these forces:
Net force = F_gravity + F_electric
Since the forces act in opposite directions, we can write the equation as:
Net force = m * g - q * E
Now, using Newton's second law, we know that the net force acting on an object is equal to its mass multiplied by its acceleration:
Net force = m * a
Equating the two expressions for net force, we can solve for the acceleration (a):
m * a = m * g - q * E
Now, the acceleration a for this case is negative because the downward direction is taken as positive. So we have:
-a = -g + (q * E) / m
Now, we can rearrange the equation to solve for the acceleration:
a = g - (q * E) / m
Finally, we can use kinematic equations to find the velocity of the sphere when it hits the plane. The initial velocity is zero (since the sphere starts from rest). The final velocity can be found using the equation:
v^2 = u^2 + 2 * a * s,
where u is the initial velocity (zero), a is the acceleration we just calculated, and s is the distance the sphere falls (the height h mentioned in the question).
Plugging in the values, the equation becomes:
v^2 = 0 + 2 * (g - (q * E) / m) * h
Simplifying further:
v^2 = 2 * g * h - (2 * q * E * h) / m
Finally, taking the square root of both sides gives us the velocity of the sphere:
v = sqrt(2 * g * h - (2 * q * E * h) / m)
So, the velocity of the sphere when it hits the plane is given by this equation, taking into account the gravitational and electric forces acting on the sphere.
To find the final velocity of the sphere when it hits the plane, we need to consider the forces acting on it and apply Newton's laws.
1. Force due to gravity (Fg): The force due to gravity can be calculated as Fg = mg, where m is the mass of the sphere and g is the acceleration due to gravity.
2. Force due to the electric field (Fe): The force due to the electric field can be calculated as Fe = qE, where q is the charge of the sphere and E is the electric field strength.
3. Net force (Fnet): The net force acting on the sphere is the sum of the gravitational force and the force due to the electric field, given by Fnet = Fg + Fe. Since the gravitational force is acting in the positive downward direction, and the force from the electric field is acting in the negative upward direction, we can write Fnet = Fg - Fe.
4. Newton's second law: According to Newton's second law, the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the acceleration of the sphere can be calculated as a = Fnet / m.
5. Solving for the acceleration: Substitute the expressions for Fg and Fe into the equation for Fnet: a = (mg - qE) / m. Mass cancels out, leaving a = g - (qE / m).
6. Final velocity (Vf): We can use the kinematic equation v^2 = u^2 + 2as to solve for the final velocity, where u is the initial velocity (which is zero in this case) and s is the distance traveled (h, the initial height).
7. Solving for the final velocity: Rearrange the kinematic equation to solve for Vf: Vf = sqrt(2as). Substitute the expression for a obtained in step 5, Vf = sqrt(2gh - 2(qEh / m)), simplifying to Vf = sqrt(2gh - (2qEh / m)).
Therefore, the final velocity of the sphere when it hits the plane can be calculated using the equation Vf = sqrt(2gh - (2qEh / m)).