Question

A body of mass \(M\), carrying charge \(Q\), falls from rest from a height \(h\) (above the ground) near the surface of the Earth, where the gravitational acceleration is \(g\) and there is an electric field with a constant component \(E\) in the vertical direction. a) Find an expression for the speed, \(v,\) of the body when it reaches the ground, in terms of \(M, Q, h, g,\) and \(E\). b) The expression from part (a) is not meaningful for certain values of \(M, g, Q,\) and \(E\). Explain what happens in such cases.

Short answer

Question: Using the expression derived for the final velocity v of a falling body of mass M and charge Q from a height h in the presence of a gravitational force due to mass M and an electric field E pointing upwards, discuss the physical significance of the result and explain what happens for certain values of M, g, Q, and E. Answer: The expression for the final velocity v of the falling body is given by v = sqrt(2 * (g - E * (Q / M)) * h). The physical significance of this result indicates the effect of both gravitational and electrostatic forces on the falling body. For certain values of M, g, Q, and E, the electrostatic force (Q * E) might become greater than the gravitational force (M * g), leading to the body moving upwards instead of falling. In such cases, the expression for the final velocity v is not valid or meaningful as it represents a scenario where the body is not undergoing free-fall motion.

Step-by-step solution

Find the net force acting on the body

The gravitational force acting on the body is W = M * g and pointing downward. The electrostatic force acting on the body is F_electrostatic = Q * E and pointing upward. The net force F_net will be the difference between the gravitational force and the electrostatic force. F_net = M * g - Q * E.

Apply Newton's 2nd Law of Motion

Applying Newton's 2nd Law of Motion, F_net = M * a, where a is the net acceleration of the body. Substituting F_net from Step 1, we get: M * a = M * g - Q * E a = g - E * (Q / M)

Use the kinematic equation for motion under constant acceleration

Using the kinematic equation for motion under constant acceleration, v^2 = u^2 + 2*a*s, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled (in this case, it's the height h). Since the body falls from rest, u = 0. We get: v^2 = 2 * a * h Substituting the acceleration 'a' obtained in step 2, we have: v^2 = 2 * (g - E * (Q / M)) * h

Find the expression for final velocity v

Taking the square root of both sides, we obtain the expression for the final velocity v: v = sqrt(2 * (g - E * (Q / M)) * h) This is the expression for the final velocity v of the body when it reaches the ground in terms of M, Q, h, g, and E.

Explain the physical significance of the result obtained in part (a)

The above expression for the final velocity v may not be meaningful for certain values of M, g, Q, and E. This is because, in some cases, the electrostatic force acting on the charged body might become greater than the gravitational force. When the electrostatic force is greater than the gravitational force, the net acceleration 'a' becomes negative, indicating that the charged body will not fall towards the ground but instead move upwards. This happens when: Q * E > M * g In such cases, the body will not undergo free-fall motion, and the expression for the final velocity v obtained in part (a) will not be valid or meaningful.

Key concepts

Electrostatic Force

Electrostatic force is a fundamental interaction between electrically charged particles. It can pull them together or push them apart, depending on the charges involved. In the context of our exercise, the charged body experiences an electrostatic force due to an external electric field, represented by the equation
\( F_{electrostatic} = Q \times E \),
where \( F_{electrostatic} \) is the electrostatic force, \( Q \) is the charge of the body, and \( E \) is the electric field strength. This force acts opposite to the gravitational pull, revealing its potential to counteract gravity under certain conditions.

It's valuable to highlight that the strength of the electrostatic force is directly proportional to the charge and the electric field; thus, greater charge or stronger fields result in more substantial forces. Importantly, in a situation where the electrostatic force exceeds the gravitational force, the body defies the usual downward acceleration one expects due to gravity, possibly moving upwards or staying in equilibrium.

Kinematic Equations

Kinematic equations describe the motion of objects under constant acceleration, without accounting for the forces that cause such acceleration. These equations provide a relationship between distance traveled, initial velocity, final velocity, acceleration, and time taken. In the case of the falling object, one of the core kinematic equations used is
\( v^2 = u^2 + 2as \),
where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the displacement. For the object dropping from rest, the initial velocity \( u \) is zero, simplifying the equation. By applying this equation, we can determine the final velocity of an object just before it hits the ground as a function of its acceleration and the height from which it fell. This equation assumes that the acceleration is constant, which is true only when the net force (and hence acceleration) is not changing with time.

Newton's Second Law of Motion

According to Newton's second law of motion, the net force acting on an object is equal to the mass of the object multiplied by its acceleration, mathematically expressed as
\( F_{net} = m \times a \).
In our study problem, we apply this law to find the net acceleration of the charged body as it falls. By subtracting the upward electrostatic force from the downward gravitational force, we establish the net force acting on the body. The resulting net force, according to this law, determines how swiftly the body accelerates towards the ground—or in the case of a stronger electrostatic force, away from it. Newton's second law highlights a profound truth in mechanics: the acceleration of an object is not just about the forces but also its mass. For identical forces, a more massive body accelerates less compared to a less massive one. This insight is crucial when considering objects of different masses behaving distinctly under identical force conditions.

Gravitational Acceleration

Gravitational acceleration, often denoted as \( g \), is the acceleration due to gravity experienced by an object when it is close to a massive body like Earth. It is approximately \( 9.81 \ m/s^2 \) on Earth's surface. This acceleration is the cause of the constant speed increase of freely falling objects, under the sole influence of gravity. In our exercise scenario, we initially consider the body to be influenced only by gravity, before accounting for the electric field.

It's important to note that gravitational acceleration is considered constant near the surface of the Earth, which allows us to use kinematic equations reliably. In the absence of other forces, any two objects, regardless of their mass, would experience the same gravitational acceleration. However, when additional forces, such as the electrostatic force in this case, are in play, the net acceleration will differ from the standard gravitational acceleration, leading to a more complex motion as vividly described in the problem at hand.