Problem 53 - University Physics with Modern Physics

Question

A particle with charge \(+\)7.60 nC is in a uniform electric field directed to the left. Another force, in addition to the electric force, acts on the particle so that when it is released from rest, it moves to the right. After it has moved 8.00 cm, the additional force has done 6.50 \(\times 10^{-5}\) J of work and the particle has 4.35 \(\times 10^{-5}\) J of kinetic energy. (a) What work was done by the electric force? (b) What is the potential of the starting point with respect to the end point? (c) What is the magnitude of the electric field?

Step-by-Step Solution

Answer

(a) -2.15 x 10^{-5} J (b) 2829 V (c) 3544 N/C

Step 1: Understand the Work-Energy Principle

The total work done on a particle is equal to the change in its kinetic energy. In this case, the initial kinetic energy is 0 because the particle starts from rest.

Step 2: Calculate Total Work Done on the Particle

The particle has final kinetic energy of 4.35 \( \times 10^{-5} \) J. Since it starts from rest, the total work done on the particle is also 4.35 \( \times 10^{-5} \) J.

Step 3: Apply Conservation of Energy

According to the conservation of energy principle, the total work done is the sum of the work done by the electric force and the additional force. Therefore, \[ W_{\text{total}} = W_{\text{electric}} + W_{\text{additional}} \].

Step 4: Find the Work Done by the Electric Force

We know the additional force does 6.50 \( \times 10^{-5} \) J of work. Using the equation from Step 3: \[ 4.35 \times 10^{-5} = W_{\text{electric}} + 6.50 \times 10^{-5} \] Solving for \( W_{\text{electric}} \), we find \[ W_{\text{electric}} = 4.35 \times 10^{-5} - 6.50 \times 10^{-5} = -2.15 \times 10^{-5} \text{ J} \].

Step 5: Calculate Potential Difference

Work done by the electric force is related to the potential difference: \[ W_{\text{electric}} = q \cdot \Delta V \] \[ \Delta V = \frac{W_{\text{electric}}}{q} = \frac{-2.15 \times 10^{-5}}{7.60 \times 10^{-9}} = -2829 \text{ V} \].Thus, the potential at the starting point is higher by 2829 V with respect to the end point.

Step 6: Determine the Magnitude of the Electric Field

The work done by the electric field can also be calculated using \[ W_{\text{electric}} = -E \cdot q \cdot d \]Substituting the known values: \[ -2.15 \times 10^{-5} = -E \cdot 7.60 \times 10^{-9} \cdot 0.08 \] Solving for \( E \), we get \[ E = \frac{2.15 \times 10^{-5}}{7.60 \times 10^{-9} \times 0.08} = 3544 \text{ N/C} \].

Key Concepts

Electric Field

An electric field is a region around a charged particle where other charged particles experience a force. Imagine it as invisible lines stretching into space, indicating the direction and strength of the force.

• Direction of Force: A positive charge in an electric field will always feel a force in the direction of the field lines.
• Magnitude of Electric Field: It is measured in Newtons per Coulomb (N/C). The stronger the electric field, the greater the force on any charged particle within it.

In this problem, the uniform electric field pushes the charged particle to the left. However, the particle moves to the right due to the presence of an additional force acting in that direction.
Determining the magnitude of this electric field involves the relationship between the work done by the electric force and the distance the particle travels. By reformulating the equation as \[ W_{\text{electric}} = -E \cdot q \cdot d \] and solving for the electric field, we find a magnitude of 3544 N/C.

Potential Difference

Potential difference, or voltage, is the difference in electric potential energy per unit charge between two points in space. It's like a measure of how much "push" a charged particle would need to move from one point to the other.

• Relation to Work: The work done by the electric force is related to potential difference. If a charge moves in the direction of an electric field, the potential energy decreases, which aligns with doing negative work.
• Calculating Potential Difference: We calculate it using the formula \[ \Delta V = \frac{W_{\text{electric}}}{q} \].

In our example, the negative work done by the electric force (\(-2.15 \times 10^{-5} \) J) indicates that the starting point has a higher potential energy. The potential difference of 2829 V suggests a drop in electric potential from start to end, which means the starting point is at 2829 volts higher potential than the endpoint.

Conservation of Energy

The conservation of energy principle states that energy cannot be created or destroyed, only transformed from one form to another. When considering mechanical systems, like this particle in an electric field, it's crucial to account for all types of work done.

• Total Work Done: The change in kinetic energy reflects the total work done on the particle, which is the sum of all individual works acting on it.
• Balancing Forces: In situations involving electric forces and additional forces, conservation of energy helps us see how these forces contribute to the particle's movement.

Applying conservation of energy here means that the work done by the electric field and the additional force collectively results in the particle's increase in kinetic energy. Initially at rest, the particle achieves a final kinetic energy of \(4.35 \times 10^{-5} \) J, signifying that the sum of all work, including negative work from the electric force and positive work from the additional force, equals this kinetic energy change.