Problem 4 - Physics

Question

A particle has a charge of \(+1.5 \mu \mathrm{C}\) and moves from point \(A\) to point \(B\), a distance of \(0.20 \mathrm{~m}\). The particle experiences a constant electric force, and its motion is along the line of action of the force. The difference between the particle's electric potential energy at \(A\) and \(B\) is \(\mathrm{EPE}_{A}-\mathrm{EPE}_{B}=+9.0 \times 10^{-4} \mathrm{~J}\). (a) Find the magnitude and direction of the electric force that acts on the particle. (b) Find the magnitude and direction of the electric field that the particle experiences.

Step-by-Step Solution

Answer

(a) Electric force: \(4.5 \times 10^{-3} \mathrm{~N}\) along the motion. (b) Electric field: \(3.0 \times 10^{3} \mathrm{~N/C}\) in the force direction.

Step 1: Understand the given data

The particle has a charge of \(+1.5 \mu \mathrm{C}\) which is \(1.5 \times 10^{-6} \mathrm{C}\). The distance moved is \(0.20 \mathrm{~m}\). The change in electric potential energy \(\Delta \mathrm{EPE}\) is given as \(+9.0 \times 10^{-4} \mathrm{~J}\). From this, we see that the potential energy decreases as the particle moves, which indicates that the force and motion are in the same direction.

Step 2: Calculate the electric force

The electric potential energy difference can be expressed in terms of force as \(\Delta \mathrm{EPE} = - F \cdot d\), where \(d\) is the distance. Solve for \(F\): \[ F = - \frac{\Delta \mathrm{EPE}}{d} \]Substitute \(\Delta \mathrm{EPE} = +9.0 \times 10^{-4} \mathrm{~J}\) and \(d = 0.20 \mathrm{~m}\):\[ F = - \frac{+9.0 \times 10^{-4} \mathrm{~J}}{0.20 \mathrm{~m}} = -4.5 \times 10^{-3} \mathrm{~N} \]The negative sign indicates that the force acts in the direction of motion; thus the magnitude of the force is \(4.5 \times 10^{-3} \mathrm{~N}\) in the direction of the particle’s motion.

Step 3: Calculate the electric field

The relationship between electric force and electric field \(E\) is given by \(F = qE\), where \(q\) is the charge. Rearrange this to find \(E\):\[ E = \frac{F}{q} \]Substitute \(F = 4.5 \times 10^{-3} \mathrm{~N}\) and \(q = 1.5 \times 10^{-6} \mathrm{C}\): \[ E = \frac{4.5 \times 10^{-3} \mathrm{~N}}{1.5 \times 10^{-6} \mathrm{C}} = 3.0 \times 10^{3} \mathrm{~N/C} \]The direction of the electric field is the same as the direction of the force.

Key Concepts

Electric Field

An electric field is a region around a charged object where other charges experience a force. It is a vector field and has both magnitude and direction.
The strength of the electric field, represented by the symbol \( E \), is measured in newtons per coulomb (N/C).
The electric field can be calculated using the formula:

• \( E = \frac{F}{q} \)

Here, \( F \) is the force experienced by a charge \( q \). Since electric fields exert a force on any charged object placed within them, they are crucial for understanding how charged particles move in a region. In our problem, a charged particle moves along the line of the force, meaning the electric field acts in the same direction as the particle's motion. This is why we were able to calculate the electric field using the known force and charge.

Electric Potential Energy

Electric potential energy (EPE) is the energy a charged particle possesses due to its position in an electric field. This concept is similar to gravitational potential energy, where objects have energy because of their position in a gravitational field. When a charge moves in an electric field, its electric potential energy changes. This change, \( \Delta \mathrm{EPE} \), can be expressed as:

• \( \Delta \mathrm{EPE} = - F \cdot d \)

Here, \( F \) is the magnitude of the electric force, and \( d \) is the distance moved by the charge.
In our problem, the positive value of \( \Delta \mathrm{EPE} \) indicates that the electric force is aiding the motion, decreasing the particle's potential energy. This means as the particle moves from point A to point B, it releases energy, akin to a ball rolling down a hill.

Particle Motion

When a particle moves in an electric field, its motion is influenced by the field's force. In our scenario, the particle follows the direction of force, maintaining linear motion. The force acting on it causes acceleration, according to Newton's second law. This direct motion along the field lines is termed as linear motion.
Understanding particle motion in electric forces involves:

• Identifying the direction of force.
• Calculating the acceleration based on the particle's mass, if needed.
• Recognizing that energy transformation occurs during motion.

In the given exercise, the particle's linear motion is a direct result of the constant electric force and the continuous decrease in electric potential energy.

Electric Charge

Electric charge is a basic property of matter and is the source of electric fields. It is denoted by \( q \) and measured in coulombs (C). Charges can be either positive or negative, and they interact with electric fields to produce forces.Basic properties of electric charge include:

• Like charges repel, while opposite charges attract.
• Electric charge is conserved; it cannot be created or destroyed.
• Charge is quantized, existing in discrete amounts (i.e., multiples of the elementary charge).

In our problem, the particle has a charge of \(+1.5 \mu \mathrm{C}\), or \(1.5 \times 10^{-6} \mathrm{C}\).
This charge interacts with the electric field to generate a force, directing the particle's movement from point A to B.