Problem 16 - College Physics

Question

A particle with a charge of \(+4.20 \mathrm{nC}\) is in a uniform electric field \(\vec{\boldsymbol{E}}\) 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 \(+1.50 \times 10^{-6} \mathrm{J}\) (a) What work was done by the electric force? (b) What is the potential of the starting point with respect to the endpoint? (c) What is the magnitude of \(\vec{E} ?\)

Step-by-Step Solution

Answer

(a) 1.50 × 10^{-6} J; (b) 357.14 V; (c) 5952.33 V/m.

Step 1: Identify what is given

We have a charged particle with a charge of \(+4.20 \ \text{nC} = 4.20 \times 10^{-9} \ \text{C}\) in a uniform electric field \(\vec{E}\) moving to the left by \(6.00 \ \text{cm} = 0.06 \ \text{m}\). After moving, it has a kinetic energy of \(+1.50 \times 10^{-6} \ \text{J}\).

Step 2: Calculate the work done by the electric force

The work done by the electric force \(W\) is equal to the change in the kinetic energy of the particle, since it was released from rest. The initial kinetic energy is 0, so:\[ W = \Delta KE = KE_{final} - KE_{initial} = 1.50 \times 10^{-6} \ \text{J} - 0 = 1.50 \times 10^{-6} \ \text{J}\]

Step 3: Understand the concept of electric potential difference

The electric potential difference \(\Delta V\) between two points is the work done per unit charge against electrostatic forces when moving the charge between those points. It's given by:\[ \Delta V = \frac{-W}{q} \]where \(W\) is the work done by the electric field and \(q\) is the charge.

Step 4: Calculate the electric potential difference

Using the formula for electric potential difference:\[ \Delta V = \frac{-1.50 \times 10^{-6} \ \text{J}}{4.20 \times 10^{-9} \ \text{C}} \]Calculate the value:\[ \Delta V = -357.14 \ \text{V} \]This means the potential at the starting point is 357.14 V greater than that at the endpoint.

Step 5: Understand the relation between electric field and potential difference

The change in electric potential \(\Delta V\) across a distance \(d\) in a uniform electric field \(\vec{E}\) is related by:\[ \Delta V = -E \cdot d \]

Step 6: Calculate the magnitude of the electric field

Using the relation from Step 5, and knowing that the direction of movement and electric field is in the same direction:\[ E = \frac{\Delta V}{d} \]Substituting the values:\[ E = \frac{357.14}{0.06} = 5952.33 \ \text{V/m} \]

Step 7: Finalize with the results

**(a)** The work done by the electric force is \(+1.50 \times 10^{-6} \ \text{J}\). **(b)** The potential of the starting point with respect to the endpoint is \(357.14 \ \text{V}\). **(c)** The magnitude of the electric field \(\vec{E}\) is \(5952.33 \ \text{V/m}\).

Key Concepts

Electric Potential Difference

Electric potential difference, often referred to as voltage, is a key concept in understanding how electric fields work. Imagine you have a charged particle moving between two points in an electric field. The electric potential difference \( \Delta V \) between these points tells you how much work is done per unit charge to move that particle. In simpler terms, it measures the energy change, due to the electric field, just like how height measures the energy in a gravitational field.
\[ \Delta V = \frac{-W}{q} \]
Where \( W \) is the work done and \( q \) is the charge. The negative sign indicates that the work done by the field reduces the potential energy stored in the charge. In the exercise, the potential difference was calculated as \( -357.14 \, \text{V} \, \), meaning the potential at the starting point differs by this value from the endpoint.

Work Done by Electric Force

Work done in physics often refers to the energy transferred when a force is applied over a distance. For electric forces, the work done \( W \) by an electric field on a charged particle is equivalent to the change in kinetic energy of that particle. If the particle starts from rest, its initial kinetic energy is zero. This makes calculating work straightforward:
\[ W = \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \]
In the given exercise, since the initial kinetic energy was zero, all the work done by the electric force becomes the final kinetic energy, which equates to \( 1.50 \times 10^{-6} \, \text{J} \). This work essentially converts potential energy into kinetic energy, propelling the particle through the field.

Kinetic Energy

Kinetic energy is the energy that a particle possesses due to its motion. When a charged particle is subjected to an electric field, the work done on it changes its kinetic energy. Kinetic energy \( KE \) can be calculated using the formula:
\[ KE = \frac{1}{2}mv^2 \]
This is not needed directly in this exercise, as the particle's mass \( m \) and velocity \( v \) are not given. Instead, we simply know that after moving a certain distance, the particle's kinetic energy \( KE_{\text{final}} \) becomes \( 1.50 \times 10^{-6} \, \text{J} \). The electric field converts potential energy to kinetic energy, allowing the particle to accelerate as it travels through the field.

Charge of Particle

A particle's charge plays a central role in how it interacts with an electric field. The charge, denoted as \( q \, \) determines both the magnitude and direction of the force exerted on it by the field. In this scenario, the particle had a charge of \( +4.20 \, \text{nC} = 4.20 \times 10^{-9} \text{C} \, \). This positive charge meant it experienced a force in the direction of the electric field.

• A positive charge will move in the direction of the field lines.
• A negative charge will move opposite to the direction of the field lines.

This charge influences calculations like electric potential difference, as seen in the formula \( \Delta V = \frac{-W}{q} \, \) where \( q \) directly affects the result. Understanding a particle's charge helps predict both its trajectory and speed in an electric field.