Problem 53 - University Physics with Modern Physics

Question

A particle with charge \(+7.60 \mathrm{nC}\) 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 \(\mathrm{cm}\) , the additional force has done \(6.50 \times 10^{-5} \mathrm{J}\) of work and the particle has \(4.35 \times 10^{-5} \mathrm{J}\) 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) -2.15×10^{-5} J, (b) -2830.26 V, (c) 35378.25 V/m.

Step 1: Calculate the work done by the electric force

The total work done on the particle is the sum of the work done by the electric force and the work done by the additional force. Since the particle has gained kinetic energy, this energy is equal to the total work done on it. Thus, we have this equation:\[ W_{total} = W_{electric} + W_{additional} \]where \( W_{additional} = 6.50 \times 10^{-5} \, \mathrm{J} \) and \( W_{total} = \text{Kinetic Energy} = 4.35 \times 10^{-5} \, \mathrm{J} \). Substitute these values into the equation:\[ 4.35 \times 10^{-5} = W_{electric} + 6.50 \times 10^{-5} \]Solving for \( W_{electric} \) gives:\[ W_{electric} = 4.35 \times 10^{-5} - 6.50 \times 10^{-5} = -2.15 \times 10^{-5} \, \mathrm{J} \].

Step 2: Determine the electric potential difference

The work done by the electric force is related to the potential difference \( \Delta V \) between the starting and ending points by the formula:\[ W_{electric} = q \Delta V \]Rearranging gives:\[ \Delta V = \frac{W_{electric}}{q} \]Substitute \( W_{electric} = -2.15 \times 10^{-5} \, \mathrm{J} \) and \( q = 7.60 \times 10^{-9} \, \mathrm{C} \) into the equation:\[ \Delta V = \frac{-2.15 \times 10^{-5}}{7.60 \times 10^{-9}} = -2830.26 \, \mathrm{V} \].Thus, the starting point is at a potential 2830.26 V lower than the end point.

Step 3: Calculate the magnitude of the electric field

The electric potential difference \( \Delta V \) is related to the electric field \( E \) and the distance \( d \) the particle traveled by the formula:\[ \Delta V = -E \times d \]Solving for the magnitude of the electric field gives:\[ E = -\frac{\Delta V}{d} \]Substitute \( \Delta V = -2830.26 \, \mathrm{V} \) and \( d = 0.08 \, \mathrm{m} \) (8.00 cm converted to meters) into the equation:\[ E = -\frac{-2830.26}{0.08} = 35378.25 \, \mathrm{V/m} \]. Thus, the magnitude of the electric field is 35378.25 V/m.

Key Concepts

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. When a force acts on a particle causing it to move, its kinetic energy changes. In our problem, when the particle moves to the right, it gains kinetic energy due to the work done on it. The total work done results in a conversion of energy, often manifesting as kinetic energy.

Here’s how it works in simpler terms:

• Work done on the particle increases its kinetic energy.
• This energy transformation obeys the principle of conservation of energy.
• The formula for kinetic energy involves both the mass and the velocity of the object, expressed as \( KE = \frac{1}{2}mv^2 \), but in our case, we focus on the work-energy principle instead.

The given exercise indicates that the kinetic energy gained is \(4.35 \times 10^{-5} \mathrm{J}\). This energy reflects how much work was effectively converted into the particle's motion.

Electric Potential Difference

Electric potential difference (also known as voltage) is a measure of the work needed to move a charge from one location to another within an electric field. It indicates how much potential energy will change as the charge moves between these points.

The formula used is:

• \(\Delta V = \frac{W_{electric}}{q}\)

This formula shows the link between the work done by the electric force and the charge itself. In the exercise, the electric force does \(-2.15 \times 10^{-5} \mathrm{J}\) of work as the particle moves, leading to a potential difference of \( -2830.26 \mathrm{V} \). This suggests that energy has been expended to counteract the electric field’s influence, moving the particle to a region of lower potential power.

Uniform Electric Field

A uniform electric field is one where the electric force is constant in magnitude and direction at every point. This means that a charge within this field will experience a consistent force no matter its position.

In a uniform electric field, the potential difference \(\Delta V\) between two points is directly related to the electric field strength \(E\) and the distance \(d\) moved by the particle:

• \(\Delta V = -E \times d\)

This equation helps us understand how the potential difference varies across distances. From the problem, we calculated the electric field's magnitude to be \(35378.25 \mathrm{V/m}\), using the particle movement distance of 0.08 meters. The particle experiences a uniform force while traversing this electric field, making calculations straightforward whenever the setup adheres to the uniform field condition.