Work From Electric Force: Videos & Practice Problems

Work done by the electric force relates to changes in electric potential energy as charges move. The work can be expressed as $w=-q∆v$ for point charges, while in a constant electric field, it is $w=qEdcos\theta$. The work is path-independent, depending only on the initial and final positions, and approaches zero as charges move infinitely far apart.

Work due to Electric Force

Understanding the work done by electric forces is crucial in the study of electrostatics. When a charge moves within an electric field, its position relative to other charges changes, leading to variations in potential energy. This change in energy is a result of the electric force acting on the charge, which is responsible for its acceleration and movement.

The work done on a charge can be expressed using the formula: \( W = F \cdot d \cdot \cos(\theta) \), where \( W \) is work, \( F \) is the force, \( d \) is the distance moved, and \( \theta \) is the angle between the force and the direction of movement. In the context of electric forces, it is important to note that the electric force is a conservative force. This means that the work done is related to the change in potential energy, where a loss in potential energy corresponds to a gain in kinetic energy, and vice versa.

From the work-energy principle, we can derive that the work done on an object is equal to the change in its kinetic energy. This relationship can also be expressed as the negative change in potential energy: \( W = -\Delta U \). The change in potential energy for a charge moving through a potential difference can be calculated using the equation: \( \Delta U = q \Delta V \), leading to the work done being expressed as: \( W = -q \Delta V \).

For point charges, the work done when moving a charge from one point to another is given by the formula: \( W = k \cdot q \cdot Q \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \), where \( k \) is Coulomb's constant, \( q \) is the moving charge, \( Q \) is the source charge, and \( r_1 \) and \( r_2 \) are the initial and final distances from the source charge, respectively. This formula accounts for the fact that the electric force varies with distance, making it unsuitable to use the standard work formula involving constant force.

In contrast, when dealing with a constant electric field, the work done can be calculated using the same formula for work: \( W = qE \cdot d \cdot \cos(\theta) \), where \( E \) is the magnitude of the electric field and \( d \) is the displacement of the charge. The angle \( \theta \) is the angle between the direction of the electric field and the direction of the displacement.

One key concept in electrostatics is path independence, which states that the work done by an electric force depends only on the initial and final positions of the charge, not the path taken to get there. This means that regardless of the route taken, the work done remains the same as long as the starting and ending points are unchanged.

As charges move far away from each other, the electric potential energy and potential approach zero, which can be observed from the equations governing electric forces. This concept is essential for understanding the behavior of charges in various configurations.

In practical applications, such as calculating the work done when moving charges in an electric field or between point charges, it is important to apply the correct formulas based on the nature of the electric field (constant or variable) and to consider the implications of the results, such as the significance of positive or negative work.

An electron moves from point A to point B. The potential difference between these two points is 100 V. What is 
a. the point of higher potential?
b. the work done on the electron?
c. the final speed of the electron if its initial speed is zero?

• A. (a) B (b) 1.6 × 10^-17 J (c) 5.93 × 10⁶ m/s
• B. (a) B (b) -1.6 × 10^-17 J (c) 5.93 × 10⁶ m/s
• C. (a) A (b) 1.6 × 10^-17 J (c) 5.93 × 10⁶ m/s
• D. (a) A (b) -1.6 × 10^-17 J (c) 5.93 × 10⁶ m/s

Answer: (A)

Work to Bring Two Charges From Infinity

In this problem, we are tasked with calculating the work done by the electric force when bringing a charge from an infinite distance to a specific point in a coordinate system. We start with a 5 coulomb charge located infinitely far away and aim to move it to the origin of the coordinate system, where a -2 coulomb charge is positioned.

To determine the work done, we utilize the formula for work done by electric forces on point charges, which is given by:

W = k \cdot Q \cdot q \left( \frac{1}{r_{\text{initial}}} - \frac{1}{r_{\text{final}}} \right)

Here, k is Coulomb's constant, Q is the charge at the origin (5 coulombs), q is the charge being moved (-2 coulombs), r_initial is the initial distance (infinity), and r_final is the final distance (5 meters).

Since the initial distance is infinite, the term 1/r_initial approaches zero. Thus, the equation simplifies to:

W = k \cdot Q \cdot q \left( 0 - \frac{1}{5} \right)

Substituting the values, we have:

W = (8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) \cdot (5 \, \text{C}) \cdot (-2 \, \text{C}) \left( -\frac{1}{5} \right)

Calculating this gives:

W = 1.8 \times 10^{10} \, \text{J}

This positive value indicates that the work done by the electric force is positive, which aligns with the natural attraction between the positive and negative charges. The electric force acts in the direction of the displacement, resulting in positive work. If both charges were of the same sign, the work would be negative, as energy would need to be supplied to move the charges closer together.

Understanding the relationship between charge interactions and work done is crucial in electrostatics, as it illustrates how forces influence the movement of charges in an electric field.

What work is needed to assemble an equilateral triangle of side length 5 cm, with a 5µC charge at each vertex?

• A. - 0.045 J
• B. - 0.135 J
• C. - 4.5 J
• D. - 13.5 J

Answer: (D)

Speed of Electron in Electric Field

In this scenario, we analyze the motion of an electron that starts from rest and moves within a uniform electric field. To begin, we assume the electric field points to the right, which is crucial for determining the direction of the force acting on the electron. The charge of the electron, denoted as \( q \), is equal to \(-1.6 \times 10^{-19}\) coulombs, and the strength of the electric field \( E \) is \( 500 \, \text{N/C} \). The displacement \( d \) of the electron is given as \( 10 \, \text{cm} \) or \( 0.1 \, \text{m} \).

To find the final speed \( v \) of the electron, we utilize the work-energy principle, which states that the work done on the electron is equal to the change in its kinetic energy. The work \( W \) done by the electric field can be expressed as:

\[ W = qEd \cos(\theta) \]

Here, \( \theta \) is the angle between the direction of the electric field and the displacement. Since the electron is negatively charged, it moves opposite to the direction of the electric field. Thus, if the electric field points to the right and the displacement is to the left, the angle \( \theta \) is \( 180^\circ \). The cosine of \( 180^\circ \) is \(-1\), leading to:

\[ W = -qEd \]

Substituting the known values, we have:

\[ W = -(-1.6 \times 10^{-19} \, \text{C})(500 \, \text{N/C})(0.1 \, \text{m}) = 8 \times 10^{-18} \, \text{J} \]

Next, we relate the work done to the kinetic energy of the electron. The change in kinetic energy is given by:

\[ W = \Delta KE = \frac{1}{2} m v^2 - 0 \]

Since the electron starts from rest, we can simplify this to:

\[ W = \frac{1}{2} m v^2 \]

Rearranging this equation to solve for the final velocity \( v \), we get:

\[ v^2 = \frac{2W}{m} \]

Substituting the work we calculated and the mass of the electron \( m = 9.11 \times 10^{-31} \, \text{kg} \), we find:

\[ v = \sqrt{\frac{2 \times 8 \times 10^{-18} \, \text{J}}{9.11 \times 10^{-31} \, \text{kg}}} \]

Calculating this yields a final speed of:

\[ v \approx 4.19 \times 10^6 \, \text{m/s} \]

This result indicates the speed of the electron after being accelerated through the electric field. Understanding the relationship between electric fields, forces, and motion is essential in physics, particularly in the study of charged particles.

Here’s what students ask on this topic:

What is the work done by an electric force on a charge moving in a uniform electric field?

The work done by an electric force on a charge moving in a uniform electric field can be calculated using the formula:

$W=qEdcos\theta$

Here, $q$ is the charge, $E$ is the magnitude of the electric field, $d$ is the displacement of the charge, and $\theta$ is the angle between the direction of the electric field and the displacement. This formula applies to situations where the electric field is constant, and the work done depends on the initial and final positions of the charge, not the path taken.

How is the work done by an electric force related to electric potential energy?

The work done by an electric force is related to the change in electric potential energy. For a charge $q$ moving through a potential difference $\mathrm{∆V}$, the work done is given by:

$W=-q\mathrm{∆V}$

This equation indicates that the work done by the electric force is equal to the negative change in electric potential energy. When a charge moves in an electric field, its potential energy changes, and this change is directly related to the work done by the electric force.

Why is the work done by an electric force path-independent?

The work done by an electric force is path-independent because the electric force is a conservative force. This means that the work done depends only on the initial and final positions of the charge, not on the path taken between these positions. Mathematically, this is expressed as:

$W=-q\mathrm{∆V}$

Since the potential difference $\mathrm{∆V}$ depends only on the initial and final positions, the work done by the electric force is the same regardless of the path taken by the charge.

How do you calculate the work done by an electric force on a point charge?

To calculate the work done by an electric force on a point charge, you can use the formula:

$W=kqQ(\frac{1}{r}-\frac{1}{r})$

Here, $k$ is Coulomb's constant, $q$ and $Q$ are the charges, and $r\_\mathrm{initial}$ and $r\_\mathrm{final}$ are the initial and final distances between the charges. This formula accounts for the fact that the electric force between point charges varies with distance.

What happens to the electric potential energy as charges move infinitely far apart?

As charges move infinitely far apart, the electric potential energy approaches zero. This is because the electric potential energy between two charges is given by:

$U=\frac{k}{q}/r$

where $k$ is Coulomb's constant, $q$ and $Q$ are the charges, and $r$ is the distance between them. As $r$ approaches infinity, the value of $U$ approaches zero, indicating that the potential energy becomes negligible at large distances.