Question

An electron (charge \(-e\) ) is projected horizontally into the space between two oppositely charged parallel plates. The electric field between the plates is \(500.0 \mathrm{N} / \mathrm{C}\) upward. If the vertical deflection of the electron as it leaves the plates has magnitude \(3.0 \mathrm{mm},\) how much has its kinetic energy increased due to the electric field? [Hint: First find the potential difference through which the electron moves.]

Short answer

Answer: The increase in the kinetic energy of the electron due to the electric field is \(2.4 \times 10^{-19} \mathrm{J}\).

Step-by-step solution

1. Identify the given information

We are given the following information: - Electric field, \(E = 500.0 \mathrm{N} / \mathrm{C}\) - Vertical deflection, \(d = 3.0 \mathrm{mm} = 3.0\times 10^{-3}\mathrm{m}\) - electron charge, \(q = -e\) - Mass of electron, \(m_e = 9.11 \times 10^{-31} kg\)

2. Finding the potential difference

To find the potential difference, we will use the electric field formula: \(E = \frac{V}{d}\) Solving for potential difference (V): \(V = Ed\) \(V = (500.0 \mathrm{N} / \mathrm{C} ) \times (3.0\times 10^{-3}\mathrm{m})\) \(V = 1.5 \mathrm{V}\)

3. Apply conservation of energy

Now we apply the conservation of energy. The increase in kinetic energy is equal to the work done by the electric field on the electron. The potential energy gained is given by: \(\Delta PE = qV\) The change in potential energy is equal to the change in kinetic energy, so we have: \(\Delta KE = \Delta PE = qV\)

4. Calculate the kinetic energy increase

Now we can calculate the increase in kinetic energy due to the electric field: \(\Delta KE = ( - e) (1.5 \mathrm{V})\) \(\Delta KE = - (1.6\times 10^{-19}\mathrm{C})(1.5 \mathrm{V})\) \(\Delta KE = -2.4 \times 10^{-19} \mathrm{J}\) However, the electron gains kinetic energy, so the actual increase is the absolute value of the energy change: \(\Delta KE = 2.4 \times 10^{-19} \mathrm{J}\) So the increase in the kinetic energy of the electron due to the electric field is \(2.4 \times 10^{-19} \mathrm{J}\).