Problem 17 - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

(a) 0 J, (b) 0.00075 J, (c) 0.00026 J.

Step 1: Understanding Work Done by Electric Force

Work done by an electric force on a charge in an electric field is calculated using the formula \[ W = qEd \cos \theta \]where \( W \) is the work done,\( q \) is the charge,\( E \) is the electric field magnitude,\( d \) is the distance moved by the charge,and \( \theta \) is the angle between the direction of the electric field and the direction of movement.

Step 2: Convert Charge Units

The charge given is 28.0 nC (nanocoulombs). Convert it to coulombs by multiplying by \( 10^{-9} \). Thus,\[ q = 28.0 \times 10^{-9} \text{ C} \]

Step 3: Calculate Work Done in Rightward Movement (a)

For movement to the right, \( \theta = 90^{\circ} \) because the electric field is vertical. Thus, \[ \cos 90^{\circ} = 0 \].Substitute in the formula\[ W = (28.0 \times 10^{-9} \text{ C})(4.00 \times 10^4 \text{ V/m})(0.450 \text{ m}) \cdot 0 = 0 \text{ J} \].No work is done since the charge moves perpendicularly to the field.

Step 4: Calculate Work Done in Upward Movement (b)

When the charge moves upward, it is moving in the same direction as the electric field. Here, \( \theta = 0^{\circ} \) and \( \cos 0^{\circ} = 1 \).Use the formula:\[ W = (28.0 \times 10^{-9} \text{ C})(4.00 \times 10^4 \text{ V/m})(0.670 \text{ m}) \cdot 1 \]\[ W = 0.00075 \text{ J} \].

Key Concepts

Electric Force

Electric force is the force exerted by an electric field on a charged particle. It's similar to how gravitational force acts on a mass. This force can be calculated using Coulomb's law, which states that the force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.
Electric force is always directed along the line connecting the two charges. When we place a charge in an electric field, like the 28.0 nC charge mentioned, the field exerts a force on it. The strength and direction of this force are determined by both the magnitude of the charge and the electric field.

• The direction of the electric force depends on the nature of the charge: positive charges experience force in the direction of the field, while negative charges experience force in the opposite direction.
• The magnitude of this force can be thought of as the product of the electric field and the charge's magnitude (F = qE).

Work Done

Work is done when a force moves an object over a distance. In the context of electric fields, work done refers to the energy transferred by the electric force when a charge is moved within the field.
This can be calculated with the formula:\[ W = qEd \cos \theta \]where \( W \) is the work done, \( q \) is the charge, \( E \) is the electric field magnitude, \( d \) is the distance moved by the charge, and \( \theta \) is the angle between the direction of the electric field and the movement of the charge.It's essential to understand that:

• If the charge moves in the same direction as the electric field, maximum work is performed because \( \theta = 0^{\circ} \) and \( \cos 0^{\circ} = 1 \). This means the entire force is used to move the charge.
• If the charge moves perpendicular to the electric field, no work is done because \( \theta = 90^{\circ} \) and \( \cos 90^{\circ} = 0 \). Here, the force doesn't contribute to the movement in the direction of the field.

Charge Movement

The movement of a charge through an electric field depends on the direction and magnitude of the forces acting upon it. Using the given question, let's break down how different movements affect the work done:

• **Rightward movement:** Here, the charge moves perpendicular to the field (\( \theta = 90^{\circ} \)). Thus, no work is done as the movement does not align with the electric field direction.
• **Upward movement:** This is in the same direction as the electric field, making \( \theta = 0^{\circ} \). The force fully contributes to moving the charge, resulting in maximum work done.
• **Movement downward at an angle:** This scenario involves an angle of \( 45^{\circ} \) down from the horizontal. Calculating work involves determining the effective component of the force in the electric field's direction using \( \cos 45^{\circ} = \frac{1}{\sqrt{2}} \).

In each scenario, the angle \( \theta \) plays a crucial role in determining how much of the energy the electric force transfers into moving the charge along the path.

Uniform Electric Field

A uniform electric field has the same strength and direction at every point within it. This is an ideal model often used for simplifying calculations in physics and is akin to having a constant and evenly distributed force across an area.
One way to create a uniform electric field is between two parallel plates with a constant voltage applied.In this uniform field:

• All charges experience the same magnitude of force regardless of their position within the field.
• Movement of charges directly along or directly opposite the field lines results in clear calculations using the formula \( W = qEd \cos \theta \).
• Understanding the uniformity of the field helps predict the behavior of charges and allows for easy computation of work done, as seen in the original exercise.

This predictable environment is usually depicted with parallel equally spaced lines, indicating that the electric field strength does not change with position.