Problem 15 - University Physics with Modern Physics

Question

A charge of 28.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.00 \(\times 10^4\) V\(/\)m. What work is done by the electric force when the charge moves (a) 0.450 m to the right; (b) 0.670 m upward; (c) 2.60 m at an angle of 45.0\(^\circ\) downward from the horizontal?

Step-by-Step Solution

Answer

(a) 0 J; (b) 750.4 μJ; (c) -2057.8 μJ

Step 1: Understanding the Problem

We are given a charge of 28.0 nC placed in a uniform electric field pointing vertically upward with a magnitude of 4.00 \( \times 10^4 \text{ V/m} \). We need to find the work done by this electric force on the charge when it moves in different directions.

Step 2: Equation for Work Done by Electric Field

The work done by an electric field when a charge moves is given by the formula: \[ W = qEd \cos \theta \] where \( W \) is the work done, \( q \) is the charge, \( E \) is the electric field strength, \( d \) is the distance moved, and \( \theta \) is the angle between the direction of the field and the direction of motion.

Step 3: Calculating Work for 0.450 m to the Right

Here, the distance \( d = 0.450 \text{ m} \), the angle \( \theta = 90^\circ \) since the charge is moving perpendicular to the field. Substitute into the formula:\[ W = (28.0 \times 10^{-9} \text{ C})(4.00 \times 10^4 \text{ V/m})(0.450 \text{ m}) \cos 90^\circ \]\[ W = 0 \text{ J} \]The work done is zero because the motion is perpendicular to the field.

Step 4: Calculating Work for 0.670 m Upward

Now, \( d = 0.670 \text{ m} \) and \( \theta = 0^\circ \) because the motion is along the field direction (upward). Substitute into the formula:\[ W = (28.0 \times 10^{-9} \text{ C})(4.00 \times 10^4 \text{ V/m})(0.670 \text{ m}) \cos 0^\circ \]\[ W = (28 \times 10^{-9})(4 \times 10^4)(0.670)(1) \]\[ W = 7.504 \times 10^{-4} \text{ J} \]Convert to \( \mu \text{J} \):\[ W = 750.4 \mu \text{J} \]Thus, the work done is 750.4 \( \mu \text{J} \).

Step 5: Calculating Work for 2.60 m at 45 Degrees Downward

In this case, \( d = 2.60 \text{ m} \), and \( \theta = 135^\circ \) since it's 45 degrees downward from the horizontal (90 + 45 degrees from the vertical field).Substitute into the formula:\[ W = (28.0 \times 10^{-9} \text{ C})(4.00 \times 10^4 \text{ V/m})(2.60 \text{ m}) \cos 135^\circ \]\[ W = (28 \times 10^{-9})(4 \times 10^4)(2.60)(-0.7071) \]\[ W = -20.578 \times 10^{-4} \text{ J} \]Convert to \( \mu \text{J} \):\[ W = -2057.8 \mu \text{J} \]Thus, the work done is -2057.8 \( \mu \text{J} \).

Key Concepts

Work Done by Electric Force

When a charge moves through an electric field, work is done by the electric force. This is similar to how other forces do work in physics. The work done essentially represents the energy used to move the charge over a distance. To calculate the work done in an electric field, we use the formula:

• \( W = qEd \cos \theta \)
• Here, \( W \) is the work done, \( q \) is the charge, \( E \) is the electric field strength, \( d \) is the distance moved, and \( \theta \) is the angle between the direction of the field and the direction of motion.

The work done by the electric force can be positive, negative, or even zero.
This depends on the displacement's direction relative to the field. If the charge moves parallel to the electric field direction, maximum work is done.
If it moves perpendicular, no work is done.

Uniform Electric Field

A uniform electric field is a space where the electric field strength is the same at every point.
It is often represented by equally spaced parallel lines indicating direction and magnitude. The electric field force on a charge in a uniform electric field is constant, which simplifies calculation of the work done. In this exercise, the field is directed vertically upward, meaning any vertical motion of a charge either aligns or opposes this field.
This uniformity allows us to use the simple work formula:

• \( W = qEd \cos \theta \)

The consistent direction and magnitude make predicting the behavior of a charge straightforward:

• If the charge is positioned upward, it experiences a force along the vertical direction.
• Horizontal movements do not affect the work done by the field.

Angle Between Field and Motion

The angle \( \theta \) between the electric field and the path the charge takes is crucial in determining the work done.
It modifies the impact of the field on the charge's movement, as seen in the cosine term of the formula \( W = qEd \cos \theta \).
An angle of 0° signifies the motion is in the same direction as the field, and the cosine of 0° is 1 thus maximum work is done by the force.
Conversely, an angle of 90° means motion is perpendicular to the field, resulting in no work (since \( \cos 90° = 0 \)).
When the angle is greater than 90°, such as with 135°, the motion opposes the field to some degree, resulting in negative work done by the electric force, as the force moves opposite to the motion direction. Understanding this angle can help predict how different directions and magnitudes of motion affect energy changes on the charge.