Problem 35 - Physics

Question

Equipotential surface \(A\) has a potential of \(5650 \mathrm{~V},\) while equipotential surface \(B\) has a potential of \(7850 \mathrm{~V}\). A particle has a mass of \(5.00 \times 10^{-2} \mathrm{~kg}\) and a charge of \(+4.00 \times 10^{-5} \mathrm{C} .\) The particle has a speed of \(2.00 \mathrm{~m} / \mathrm{s}\) on surface \(A .\) An outside force is applied to the particle, and it moves to surface \(B\), arriving there with a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) How much work is done by the outside force in moving the particle from \(A\) to \(B ?\)

Step-by-Step Solution

Answer

The work done by the external force is 0.213 J.

Step 1: Identify the change in potential energy

The change in the potential energy (abla PE) of the particle is calculated using the particle's charge and the difference in potentials between surfaces A and B. The formula for potential energy change is: \[ \Delta PE = q \times (V_B - V_A)\]Where:- \(q = +4.00 \times 10^{-5} \mathrm{~C}\) is the charge of the particle- \(V_B = 7850 \mathrm{~V}\) is the potential at surface B- \(V_A = 5650 \mathrm{~V}\) is the potential at surface APlugging in the values:\[ \Delta PE = 4.00 \times 10^{-5} \times (7850 - 5650) = 0.088 \mathrm{~J}\]

Step 2: Determine the change in kinetic energy

The kinetic energy (KE) of a particle is given by the formula:\[ KE = \frac{1}{2} m v^2\]Calculate the change in kinetic energy from surface A to B:- Initial speed on A: \(v_A = 2.00 \mathrm{~m/s}\)- Final speed on B: \(v_B = 3.00 \mathrm{~m/s}\)- Mass of the particle: \(m = 5.00 \times 10^{-2} \mathrm{~kg}\)Initial kinetic energy at A:\[ KE_A = \frac{1}{2} \times 5.00 \times 10^{-2} \times (2.00)^2 = 0.10 \mathrm{~J}\]Final kinetic energy at B:\[ KE_B = \frac{1}{2} \times 5.00 \times 10^{-2} \times (3.00)^2 = 0.225 \mathrm{~J}\]Change in kinetic energy:\[ \Delta KE = KE_B - KE_A = 0.225 - 0.10 = 0.125 \mathrm{~J}\]

Step 3: Calculate the total work done

The work done by the external force is equal to the change in total mechanical energy, which is the sum of changes in potential energy and kinetic energy:\[ W_{external} = \Delta KE + \Delta PE\]Plugging in the values from Step 1 and Step 2:\[ W_{external} = 0.125 + 0.088 = 0.213 \mathrm{~J}\]

Key Concepts

Equipotential Surface

An equipotential surface is a concept in physics where every point on the surface has the same electric potential. This means that no work is required to move a charge along the surface, as the potential energy does not change. In this exercise, surface A has a potential of 5650 V, while surface B is at 7850 V. The difference in potential between these surfaces indicates that an external force is needed to move a charged particle from one to the other.

It's crucial to remember that the electric force acting on the particle is perpendicular to the equipotential surfaces. Therefore, if a particle moves along an equipotential surface, it does not experience a change in potential energy due to electric forces alone. This movement, however, will change as it moves to a surface with a different potential, requiring external force or other energy changes to facilitate this transition.

Change in Kinetic Energy

Kinetic energy describes the energy that an object possesses due to its motion. For any particle with mass, its kinetic energy is given by the formula: \[ KE = \frac{1}{2} m v^2 \]Where \( m \) is the mass and \( v \) is the velocity of the particle. In this exercise, the particle starts with an initial speed of 2.00 m/s at surface A and reaches a final speed of 3.00 m/s at surface B.

The change in kinetic energy is calculated based on these velocity changes:- Initial Kinetic Energy at A: 0.10 J- Final Kinetic Energy at B: 0.225 J

Thus, the change in kinetic energy (\( \Delta KE \)) is 0.125 J. This increase reflects the work done by the external force to speed up the particle as it moves between the two equipotential surfaces.

Change in Potential Energy

Potential energy is related to the position of a particle within a force field. In an electrostatic context, it is determined by both the charge of the particle and the electric potential it is subject to. The change in potential energy when moving from surface A (5650 V) to surface B (7850 V) is calculated using:\[ \Delta PE = q \times (V_B - V_A) \]Where \( q \) is the charge of the particle.

Given:- Charge \( q = +4.00 \times 10^{-5} \text{ C} \)- Initial potential \( V_A = 5650 \text{ V} \)- Final potential \( V_B = 7850 \text{ V} \)
The change results in \( \Delta PE = 0.088 \text{ J} \).

The work done by the external force needs to overcome this change in potential energy, plus any additional work to increase the particle’s speed—in this way facilitating the overall transition between the surfaces.