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Chapter 23: Problem 14

A particle with charge \(+4.20 \mathrm{nC}\) is in a uniform electric field \(\vec{E}\) directed to the left. The charge is released from rest and moves to the left; after it has moved \(6.00 \mathrm{~cm},\) its kinetic energy is \(+2.20 \times 10^{-6} \mathrm{~J}\). What are (a) the work done by the electric force, (b) the potential of the starting point with respect to the end point, and (c) the magnitude of \(\overrightarrow{\boldsymbol{E}}\) ?

Short Answer

Expert verified
The work done by the electric force is \(+2.20 \times 10^{-6} J\). The potential of the starting point with respect to the end point is \(524.76 V\). The magnitude of the electric field is \(8746.0 N/C\).

Step by step solution

01

Calculate work done by electric force

The work done by the electric field is equal to the change in kinetic energy of the particle, which is \(2.20 \times 10^{-6} J\). The work done by the electric field is \(W = +2.20 \times 10^{-6} J\) as kinetic energy cannot be negative.
02

Calculate the potential difference

The potential difference \(V\) between two points in an electric field is given by \(V = W/Q\), where \(Q\) is the charge. By substituting the given charge \(Q = +4.20 nC = 4.20 \times 10^{-9} C\) and the work done, we get \(V = (2.20 \times 10^{-6} J) / (4.20 \times 10^{-9} C) = 524.76 V\). This is the potential at the starting point with respect to the end point.
03

Calculate the magnitude of electric field

The magnitude of electric field \(E\) is obtained from the relationship \(E = V/d\), where \(d\) is the distance the particle has moved. The given displacement is \(d = 6.00 cm = 0.06 m\). Substituting these values, we get \(E = (524.76 V) / (0.06 m) = 8746.0 N/C\).

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Key Concepts

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

Work-Energy Principle
The work-energy principle is a powerful concept that helps us understand how energy is transferred in a system. In simple terms, it states that the work done on an object is equal to the change in its kinetic energy.
For the given problem, the particle was initially at rest, which means its initial kinetic energy was zero. After moving in the electric field, its kinetic energy became \(2.20 \times 10^{-6} \, \text{J}\). This change in kinetic energy matches the work done by the electric field force on the particle.
Using this principle, we can determine that the work done by the electric force is \(+2.20 \times 10^{-6} \, \text{J}\). This principle simplifies the study of motion by focusing on the energy changes rather than forces and accelerations over specific paths.
Electric Potential
Electric potential is a way to express how much potential energy a charge would have at a specific point in an electric field. It helps us understand how much work would be needed to move a charge within the field.
In the exercise, we calculate the potential difference between two points using the formula \(V = W/Q\). Here, \(W\) is the work done, and \(Q\) is the charge of the particle.
- Given that the work done is \(+2.20 \times 10^{-6} \, \text{J}\) and the charge is \(+4.20 \, \text{nC}\), or \(4.20 \times 10^{-9} \, \text{C}\), we find that the potential difference \(V\) is \(524.76 \, \text{V}\).
This potential difference represents how much energy would be released or required to move the charge from its starting point to the endpoint in the electric field.
Kinetic Energy
Kinetic energy is the energy of motion. When an object moves, it possesses kinetic energy depending on its mass and velocity, calculated by the formula \(KE = \frac{1}{2}mv^2\). For a charged particle moving in an electric field, the kinetic energy helps us understand how fast it's moving after being acted on by the field.
In this exercise, the particle was released from rest, indicating initial kinetic energy was zero. The electric force did work on the particle, increasing its kinetic energy to \(+2.20 \times 10^{-6} \, \text{J}\).
This increase demonstrates how work done by an electric field is transformed into kinetic energy, causing the particle to accelerate. This transformation is crucial in many applications, including capacitors and electric motors.
Potential Difference
Potential difference, or voltage, is a measure of the work needed to move a charge between two points in an electric field. It tells us how much potential energy a charge gains or loses as it moves.
We calculated the potential difference in this exercise using the formula \(V = W/Q\). Here, it quantifies the energy change per charge unit as the particle moves.
- Substituting the known values, the potential difference was found to be \(524.76 \, \text{V}\).
This positive potential difference indicates energy was released by the field, highlighting how electric fields can do work on charges, altering their potential energy.

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Most popular questions from this chapter

A very small sphere with positive charge \(q=+8.00 \mu \mathrm{C}\) is released from rest at a point \(1.50 \mathrm{~cm}\) from a very long line of uniform linear charge density \(\lambda=+3.00 \mu \mathrm{C} / \mathrm{m} .\) What is the kinetic energy of the sphere when it is \(4.50 \mathrm{~cm}\) from the line of charge if the only force on it is the force exerted by the line of charge?

A small sphere with mass \(5.00 \times 10^{-7} \mathrm{~kg}\) and charge \(+7.00 \mu \mathrm{C}\) is released from rest a distance of \(0.400 \mathrm{~m}\) above a large horizontal insulating sheet of charge that has uniform surface charge density \(\sigma=+8.00 \mathrm{pC} / \mathrm{m}^{2} .\) Using energy methods, calculate the speed of the sphere when it is \(0.100 \mathrm{~m}\) above the sheet.

A thin spherical shell with radius \(R_{1}=3.00 \mathrm{~cm}\) is concentric with a larger thin spherical shell with radius \(R_{2}=5.00 \mathrm{~cm}\). Both shells are made of insulating material. The smaller shell has charge \(q_{1}=+6.00 \mathrm{nC}\) distributed uniformly over its surface, and the larger shell has charge \(q_{2}=-9.00 \mathrm{nC}\) distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. (a) What is the electric potential due to the two shells at the fol- (ii) \(r=4.00 \mathrm{~cm}\) lowing distance from their common center: (i) \(r=0\) (iii) \(r=6.00 \mathrm{~cm} ?\) (b) What is the magnitude of the potential difference between the surfaces of the two shells? Which shell is at higher potential: the inner shell or the outer shell?

A solid conducting sphere of radius \(5.00 \mathrm{~cm}\) carries a net charge. To find the value of the charge, you measure the potential difference \(V_{A B}=V_{A}-V_{B}\) between point \(A,\) which is \(8.00 \mathrm{~cm}\) from the center of the sphere, and point \(B\), which is a distance \(r\) from the center of the sphere. You repeat these measurements for several values of \(r>8.00 \mathrm{~cm} .\) When you plot your data as \(V_{A B}\) versus \(1 / r,\) the values lie close to a straight line with slope \(-18.0 \mathrm{~V} \cdot \mathrm{m}\). What does your data give for the net charge on the sphere? Is the net charge positive or negative?

Charge \(Q=5.00 \mu \mathrm{C}\) is distributed uniformly over the volume of an insulating sphere that has radius \(R=12.0 \mathrm{~cm} .\) A small sphere with charge \(q=+3.00 \mu \mathrm{C}\) and mass \(6.00 \times 10^{-5} \mathrm{~kg}\) is projected toward the center of the large sphere from an initial large distance. The large sphere is held at a fixed position and the small sphere can be treated as a point charge. What minimum speed must the small sphere have in order to come within \(8.00 \mathrm{~cm}\) of the surface of the large sphere?
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