Problem 14 - University Physics with Modern Physics

Question

A particle with charge \(+\)4.20 nC is in a uniform electric field \(\overrightarrow{E}\) directed to the left. The charge is released from rest and moves to the left; after it has moved 6.00 cm, its kinetic energy is \(+2.20 \times 10^{-6}\) 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 \(\overrightarrow{E}\) ?

Step-by-Step Solution

Answer

(a) +2.20 x 10^-6 J, (b) 523.8 V, (c) 8729.2 V/m

Step 1: Understanding the Work-Energy Principle

According to the work-energy principle, the work done by a force is equal to the change in kinetic energy of a particle. Since the particle is released from rest, its initial kinetic energy is zero. Thus, the work done by the electric force on the particle is equal to its final kinetic energy. Here, the final kinetic energy is given as \(+2.20 \times 10^{-6}\) J.

Step 2: Calculating Work Done by Electric Force

The work done by the electric force \( W \) is equal to the change in kinetic energy of the particle. Thus, we have:\[W = \Delta KE = +2.20 \times 10^{-6}\; \text{J}\]So the work done by the electric force is \(+2.20 \times 10^{-6}\) J.

Step 3: Calculating Electric Potential Difference (Voltage)

The work done by an electric field is related to the electric potential difference \( V \) by the equation:\[W = qV\]where \( q = 4.20 \times 10^{-9}\; \text{C} \) is the charge of the particle. Rearranging for \( V \), we find:\[V = \frac{W}{q} = \frac{2.20 \times 10^{-6}}{4.20 \times 10^{-9}} = 523.8\; \text{V}\]Thus, the electric potential of the starting point with respect to the endpoint is 523.8 V.

Step 4: Determining the Magnitude of Electric Field \( \overrightarrow{E} \)

The relationship between the electric potential difference \( V \), the electric field \( E \) and the distance \( d \) is given by:\[V = Ed\]Rearranging for \( E \), we have:\[E = \frac{V}{d} = \frac{523.8}{0.06} = 8729.2\; \text{V/m}\]So, the magnitude of the electric field is 8729.2 V/m.

Key Concepts

Work-Energy Principle

The Work-Energy Principle is a fundamental concept in physics that relates the work done by a force on an object to its change in kinetic energy. In simple terms, when you exert a force on an object and it moves, you are doing work on it. The result of this work is a change in the object's energy state, specifically its kinetic energy.
For example, if an object starts from rest, like the charged particle in our problem, its initial kinetic energy is zero. When work is performed on it, its kinetic energy changes, because it starts moving. This transformation is perfectly illustrated when the kinetic energy of the particle becomes \(+2.20 \times 10^{-6}\) J after it moved through the electric field.
Always remember, the work-energy principle is incredibly useful for solving problems where you need to find the speed, displacement, or energy transformation of an object.

Electric Potential Difference

Electric potential difference, also known as voltage, is the measure of the work done to move a charge from one point to another in an electric field. It is crucial in understanding how electric charges interact within a field.
In our scenario, the electric potential difference was calculated after determining the work done by the electric field. The formula used is:

• \( W = qV \)
• Here, \(W\) is the work done, \(q\) is the charge, and \(V\) is the potential difference.

Substituting known values into the formula, we determined that the potential difference between the start and endpoint for the particle is 523.8 V. This quantity tells us how much energy per charge unit is used in moving from one point to another, providing insight into the strength and efficacy of the electric field.

Electric Force

Electric force is the interaction between charged particles due to their electric charge. It's a fundamental force that can either attract or repel particles.
In this problem, the electric force acts on a particle with a positive charge, causing it to move in the direction of the electric field.
Simply put, the force is what causes the particle to accelerate and gain kinetic energy. The vantage of such problems is that once you know the force applied and the distance moved, you can determine the work done by the force, which directly aligns with the Work-Energy Principle. These fundamental interactions of forces and fields are key to understanding broader concepts in electric circuits and other physics applications.

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. It depends on two key factors:

• the mass of the object, and
• its velocity.

If an object starts from rest, its initial kinetic energy is zero. When an electric force acts on it, the object begins to move, gaining speed - and consequently, gaining kinetic energy. For the particle in our exercise, this change was calculated to be \(+2.20 \times 10^{-6}\) J once it had moved a certain distance within the electric field. Understanding the link between force, displacement, and kinetic energy helps illustrate how energy is transferred and transformed in a system. Kinetic energy is pivotal in many areas of physics and engineering, from simple motion studies to complex energy management systems.