Problem 46 - University Physics with Modern Physics

Question

A point charge \(q_1 = +\)5.00 \(\mu\)C is held fixed in space. From a horizontal distance of 6.00 cm, a small sphere with mass 4.00 \(\times 10^{-3}\) kg and charge \(q2 = +\)2.00 \(\mu\)C is fired toward the fixed charge with an initial speed of 40.0 m\(/\)s. Gravity can be neglected. What is the acceleration of the sphere at the instant when its speed is 25.0 m\(/\)s?

Step-by-Step Solution

Answer

Compute the new distance using energy conservation, then use this in Coulomb's Law to find force and Newton's 2nd Law for acceleration.

Step 1: Determine Electric Force Between Charges

To find the electric force acting on the small sphere due to the fixed charge, use Coulomb's Law:\[ F = k \frac{|q_1 q_2|}{r^2} \]where:- \(q_1 = +5.00 \, \mu\text{C} = 5.00 \times 10^{-6} \, \text{C}\)- \(q_2 = +2.00 \, \mu\text{C} = 2.00 \times 10^{-6} \, \text{C}\)- \(k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2\)- \(r\) is the distance between the charges.To find \(r\), use the conservation of mechanical energy to determine how far the sphere has traveled by the time it slows down to 25.0 m/s.

Step 2: Apply Conservation of Energy

Initially, the sphere has kinetic energy and potential energy due to the fixed charge. Set initial energy equal to energy at the instant speed is 25.0 m/s:Initial kinetic energy: \[ KE_i = \frac{1}{2} m v_i^2 \]where \(v_i = 40.0 \, \text{m/s}\) and \(m = 4.00 \times 10^{-3} \, \text{kg}\).Initial potential energy between charges:\[ U_i = k \frac{q_1 q_2}{r_i} \]where \(r_i = 0.06 \, \text{m}\).Final kinetic energy:\[ KE_f = \frac{1}{2} m v_f^2 \]where \(v_f = 25.0 \, \text{m/s}\).Final potential energy:\[ U_f = k \frac{q_1 q_2}{r_f} \]where \(r_f\) is the new distance we need to find.Setting up the energy conservation equation:\[ KE_i + U_i = KE_f + U_f \]Solve for \(r_f\).

Step 3: Calculate Final Distance

Substitute the values into the conservation of energy equation:\[ \frac{1}{2} \times 4.00 \times 10^{-3} \times (40.0)^2 + 8.99 \times 10^9 \times \frac{5.00 \times 10^{-6} \times 2.00 \times 10^{-6}}{0.06} = \frac{1}{2} \times 4.00 \times 10^{-3} \times (25.0)^2 + 8.99 \times 10^9 \times \frac{5.00 \times 10^{-6} \times 2.00 \times 10^{-6}}{r_f} \]Solve this for \(r_f\).

Step 4: Calculate the Electric Force at New Distance

Now that \(r_f\) has been determined, substitute it back into the Coulomb's Law equation to find \(F\):\[ F = 8.99 \times 10^9 \times \frac{5.00 \times 10^{-6} \times 2.00 \times 10^{-6}}{r_f^2} \]

Step 5: Determine the Acceleration of the Sphere

With the force found, use Newton's second law to calculate the acceleration:\[ F = m a \]\[ a = \frac{F}{m} \]Substitute \(F\) from Step 4 and mass \(m = 4.00 \times 10^{-3} \, \text{kg}\) into this equation to get acceleration \(a\).

Key Concepts

Coulomb's Law

Coulomb's Law is a fundamental principle in electrostatics, describing the force between two charged objects. It explains how like charges repel each other, while opposite charges attract. The law is expressed mathematically as:

• \[ F = k \frac{|q_1 q_2|}{r^2} \]

Here, \( F \) represents the electric force between the charges, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, \( r \) is the distance between the charges, and \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \). This equation highlights that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Applying Coulomb's Law in practice involves calculating these forces in various configurations. In our problem, we used it to determine the force exerted by a fixed charge on a moving charge at different points in its path. This understanding allows us to predict the behavior of charged particles in electric fields.

Mechanical Energy Conservation

Mechanical energy conservation is a pivotal concept that guides the understanding of how energy is transferred and transformed within a system. It states that the total mechanical energy of a system remains constant if no external forces are doing work on it. In our problem, we consider both kinetic and potential energy.

• Kinetic energy is expressed as \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity.
• Potential energy between charges is given by \( U = k \frac{q_1 q_2}{r} \), which changes as the distance \( r \) varies.

By setting the initial total energy equal to the total energy at a later point in time, we can solve for unknown parameters, such as the distance between charges when the speed changes. This allows us to track how energy shifts between kinetic and potential forms without any loss in the closed system, except when work is done by or against the field itself.

Electric Force

Electric force is an essential concept in electrostatics, representing the interaction between charged particles due to electric fields. The force experienced by a charge in an electric field is given by Coulomb's Law. This force can cause acceleration, affecting a particle's motion, especially when other forces like gravity can be neglected.

• As charges move nearer or further apart, the electric force changes according to \( F = k \frac{|q_1 q_2|}{r^2} \).
• The direction of the force depends on the signs of the charges involved: like charges push away, while opposite charges pull towards each other.

Understanding electric force is crucial for calculating how charged particles accelerate. In scenarios like our problem, calculating this force at a specific instant, given the distances and charges, provided us with the necessary values to apply to further calculations.

Newton's Second Law

Newton's Second Law of Motion is a cornerstone of classical mechanics, relating the net force acting on an object to its acceleration. The law is succinctly expressed with the formula:

• \[ F = ma \]

where \( F \) represents the force acting on an object, \( m \) is the object's mass, and \( a \) is the acceleration. In the realm of electrostatics, this law is invaluable when analyzing how charged particles move under the influence of electric forces.
In our exercise, after determining the electric force exerted on the small sphere, Newton's Second Law was applied to find the sphere's acceleration. By rearranging the formula to solve for acceleration \( a \), we can set \( a = \frac{F}{m} \), providing a direct method to find how the sphere's velocity changes due to the electric force. This relationship simplifies predicting motion in systems where electric forces are predominant.