Problem 16 - University Physics with Modern Physics

Question

A particle with a charge of \(+4.20 \mathrm{nC}\) is in a uniform electric field \(\overrightarrow{\boldsymbol{E}}\) directed to the left. It is released from rest and moves to the left; after it has moved \(6.00 \mathrm{cm},\) its kinetic energy is found to be \(+1.50 \times 10^{-6} \mathrm{J}\) . \(\mathrm{J}\) (a) What work was done by the electric force? (b) What is the potential of the starting point with respect to the end point? (c) What is the magnitude of \(\overrightarrow{\boldsymbol{E}} ?\)

Step-by-Step Solution

Answer

(a) Work done is \(1.50 \times 10^{-6} \text{ J}\). (b) Potential difference is 357.14 V. (c) Electric field magnitude is 5952.33 V/m.

Step 1: Understand the Problem

We need to compute different properties related to a charged particle moving in an electric field. Specifically, we will find the work done by the electric field, the potential difference, and the magnitude of the electric field given certain initial and final conditions.

Step 2: Define the Work-Energy Principle

The work done by an electric force on a charged particle is equal to the change in kinetic energy of the particle. Mathematically, this is stated as: \[ W = \Delta KE = KE_{\text{final}} - KE_{\text{initial}}. \] Since the particle is released from rest, its initial kinetic energy is zero.

Step 3: Calculate the Work Done by the Electric Force

Use the work-energy principle to compute the work done:\[ W = KE_{\text{final}} - KE_{\text{initial}} = 1.50 \times 10^{-6} \text{ J} - 0 = 1.50 \times 10^{-6} \text{ J}. \] Thus, the work done by the electric force is \(1.50 \times 10^{-6} \text{ J}.\)

Step 4: Relate Work to Potential Difference

The work done by the electric force is also related to the potential difference between two points as:\[ W = q \Delta V, \]where \(q\) is the charge. Rearrange to find the potential difference:\[ \Delta V = \frac{W}{q}. \]

Step 5: Calculate the Potential Difference

Substitute the known values into the equation:\[ \Delta V = \frac{1.50 \times 10^{-6} \text{ J}}{4.20 \times 10^{-9} \text{ C}} = 357.14 \text{ V}. \] The potential of the starting point with respect to the end point is 357.14 V.

Step 6: Relate Work, Potential Difference, and Electric Field

The potential difference \(\Delta V\) can also be expressed in terms of the electric field and distance as:\[ \Delta V = E \cdot d, \]where \(E\) is the magnitude of the electric field and \(d\) is the distance moved (\(6.00 \text{ cm} = 0.06 \text{ m}\)). Rearrange to find \(E\):\[ E = \frac{\Delta V}{d}. \]

Step 7: Calculate the Magnitude of the Electric Field

Substitute the calculated potential difference and the distance into the equation:\[ E = \frac{357.14 \text{ V}}{0.06 \text{ m}} = 5952.33 \text{ V/m}. \] Thus, the magnitude of the electric field is approximately 5952.33 V/m.

Key Concepts

Electric Potential Difference

The electric potential difference, often known as voltage, indicates the difference in electric potential energy per charge between two points in an electric field. It's like comparing two spots where one has more potential or stored energy. A charged particle, such as the one in our exercise, moves due to this potential difference. Think of it as rolling a ball down a hill - the potential difference is the height of the hill. For a charge moving through an electric field, this concept helps us understand how much energy is required or released. In mathematical terms, the potential difference \( \, \Delta V \, \) is expressed as the work \( \, W \, \) done per unit charge \( \, q \, \). This gives us the relationship, \( \, \Delta V = \frac{W}{q} \). In our example, using a work value and a charge, we found the potential difference to be \( 357.14 \text{ V} \). This tells us how much potential energy changes per charge and gives insight into the strength of the electric force involved.

Understanding electric potential difference is key to grasping how electric fields influence charged particles. It's a central concept in circuits and ground to electrostatic principles.

Work-Energy Principle

The work-energy principle provides a great insight into how energy changes in a system. It states that the work done by all forces acting on a particle equals the change in the particle's kinetic energy. For a charged particle in an electric field, this tells us how much energy the field transfers to the particle.

In our case, the particle was initially at rest, making its initial kinetic energy zero. As it moved in response to the uniform electric field, it gained kinetic energy noted to be \( 1.50 \times 10^{-6} \text{ J} \). This entire increase in kinetic energy comes from the work done by the electric field. Mathematically, we express this as \[ W = \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \].

This principle aids in solving for unknowns, such as work done or kinetic energy changes, by providing a clear energy balance criterion. For students, understanding this link bridges concepts of force and energy, illuminating how fields impart motion through energy transfer.

Kinetic Energy

Kinetic energy is the energy an object possesses because of its motion. Whenever a charged particle moves within an electric field, it possesses kinetic energy. This energy is crucial as it tells us how swiftly the particle is moving and how much work the field has done in getting it to move.

The formula for kinetic energy is \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the particle. However, in problems dealing with electric fields and charges, we often bypass finding mass or velocity directly and instead focus on energy changes. In our problem, the charged particle's movement resulted in a kinetic energy gain of \( 1.50 \times 10^{-6} \text{ J} \), without needing to detail its mass or speed.

Kinetic energy change directly relates to how electric potential energy is converted as the particle moves. Understanding kinetic energy in this context highlights how electric fields do work on charged particles, transforming potential energy into motion-causing kinetic energy.

Uniform Electric Field

A uniform electric field is characterized by consistent strength and direction at every point within the field. This regularity simplifies calculations, as the effect on charges is predictably uniform across distances.

In our exercise, the uniform electric field exerts a constant force on the charged particle as it moves a specific distance. This consistency allows us to directly relate potential difference and work to the field's strength, using \( E = \frac{\Delta V}{d} \) as seen in the solution. With the potential difference of \( 357.14 \text{ V} \) and distance moved of \( 0.06 \text{ m} \), the field's magnitude calculates to approximately \( 5952.33 \text{ V/m} \), demonstrating a straightforward way to quantify the field's effect.

Understanding uniform electric fields supports students in developing a clear sense of how electric forces impact charged particles, giving consistent results over known distances and offering a controlled setting for exploring basic electrostatic principles.