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Chapter 19: Problem 4

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.

Short Answer

Expert verified
(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 by step solution

01

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.
02

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鈥檚 motion.
03

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A charge of \(-3.00 \mu \mathrm{C}\) is fixed in place. From a horizontal distance of \(0.0450 \mathrm{~m}, \mathrm{a}\) particle of mass \(7.20 \times 10^{-3} \mathrm{~kg}\) and charge \(-8.00 \mu \mathrm{C}\) is fired with an initial speed of \(65.0 \mathrm{~m} / \mathrm{s}\) directly toward the fixed charge. How far does the particle travel before its speed is zero?

Charges \(q_{1}\) and \(q_{2}\) are fixed in place, \(q_{2}\) being located at a distance \(d\) to the right of \(\mathrm{q}_{1} . \mathrm{A}\) third charge \(q_{3}\) is then fixed to the line joining \(q_{1}\) and \(q_{2}\) at a distance \(d\) to the right of \(q_{2}\). The third charge is chosen so the potential energy of the group is zero; that is, the potential energy has the same value as that of the three charges when they are widely separated. Determine the value for \(q_{3}\), assuming that (a) \(q_{1}=q_{2}=q\) and (b) \(q_{1}=q\) and \(q_{2}=-q .\) Express your answers in terms of \(q\)

During a lightning flash, a potential difference \(V\) cloud \(-V\) ground exists between a cloud and the ground. As a result, negative electric charge is transferred from the ground to the cloud. (a) As the charge moves from the ground to the cloud, work is done on the charge by the electric force. How is this work related to the potential difference and the charge \(q ?(\mathrm{~b})\) If this work could be used to accelerate an automobile of mass \(m\) from rest, what would be the automobile's final speed? Express your answer in terms of the potential difference, the charge, and the mass of the automobile. (c) If this work could all be converted into heat, what mass \(m\) of water could be heated so that its temperature increases by \(\Delta T ?\) Write your answer in terms of the potential difference, the charge, the mass, the specific heat capacity \(c\) of water, and \(\Delta T\) Suppose a potential difference of \(V_{\text {cloud }}-V_{\text {ground }}=1.2 \times 10^{9} \mathrm{~V}\) exists between the cloud and the ground, and \(q=-25 \mathrm{C}\) of charge is transferred from the ground to the cloud. (a) How much work \(W\) ground-cloud is done on the charge by the electric force? (b) If the work done by the electric force were used to accelerate a 1100 \(\mathrm{kg}\) automobile from rest, what would be its final speed? (c) If the work done by the electric force were converted into heat, how many kilograms of water at \(0{ }^{\circ} \mathrm{C}\) could be heated to \(100^{\circ} \mathrm{C}\) ?

Concept Questions Two capacitors have the same plate separation. However, one has square plates, while the other has circular plates. The square plates are a length \(L\) on each side, and the diameter of the circular plates is \(L\). (a) If the same dielectric material were between the plates in each capacitor, which one would have the greater capacitance? (b) By putting different dielectric materials between the capacitor plates, we can make the two capacitors have the same capacitance. Which capacitor should contain the dielectric material with the greater dielectric constant? Give your reasoning in each case. The capacitors have the same capacitance because they contain different dielectric materials. The dielectric constant of the material between the square plates has a value of \(\kappa_{\text {square }}=3.00 .\) What is the dielectric constant \(\kappa_{\text {circle }}\) of the material between the circular plates? Be sure that your answer is consistent with your answers to the Concept Questions.

Refer to Interactive Solution 19.39 at for one approach to this problem. The electronic flash attachment for a camera contains a capacitor for storing the energy used to produce the flash. In one such unit, the potential difference between the plates of an \(850-\mu \mathrm{F}\) capacitor is \(280 \mathrm{~V}\). (a) Determine the energy that is used to produce the flash in this unit. (b) Assuming that the flash lasts for \(3.9 \times 10^{-3} \mathrm{~s},\) find the effective power or "wattage" of the flash.
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