/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 An electron (charge \(-e\) ) is ... [FREE SOLUTION] | 91影视

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Chapter 17: Problem 46

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

Expert verified
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

01

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, \(m_e = 9.11 \times 10^{-31} kg\)
02

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): \(V = Ed\) \(V = (500.0 \mathrm{N} / \mathrm{C} ) \times (3.0\times 10^{-3}\mathrm{m})\) \(V = 1.5 \mathrm{V}\)
03

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: \(\Delta KE = \Delta PE = qV\)
04

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 \(2.4 \times 10^{-19} \mathrm{J}\).

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Most popular questions from this chapter

(a) Calculate the capacitance per unit length of an axon of radius $5.0 \mu \mathrm{m} \text { (see Fig. } 17.14) .$ The membrane acts as an insulator between the conducting fluids inside and outside the neuron. The membrane is 6.0 nm thick and has a dielectric constant of \(7.0 .\) (Note: The membrane is thin compared with the radius of the axon, so the axon can be treated as a parallel plate capacitor.) (b) In its resting state (no signal being transmitted), the potential of the fluid inside is about \(85 \mathrm{mV}\) lower than the outside. Therefore, there must be small net charges \(\pm Q\) on either side of the membrane. Which side has positive charge? What is the magnitude of the charge density on the surfaces of the membrane?
A parallel plate capacitor has a charge of \(0.020 \mu \mathrm{C}\) on each plate with a potential difference of \(240 \mathrm{V}\). The parallel plates are separated by 0.40 mm of air. What energy is stored in this capacitor?
The inside of a cell membrane is at a potential of \(90.0 \mathrm{mV}\) lower than the outside. How much work does the electric field do when a sodium ion (Na') with a charge of \(+e\) moves through the membrane from outside to inside?
Before a lightning strike can occur, the breakdown limit for damp air must be reached. If this occurs for an electric field of $3.33 \times 10^{5} \mathrm{V} / \mathrm{m},$ what is the maximum possible height above the Earth for the bottom of a thundercloud, which is at a potential $1.00 \times 10^{8} \mathrm{V}$ below Earth's surface potential, if there is to be a lightning strike?
A parallel plate capacitor is charged by connecting it to a \(12-V\) battery. The battery is then disconnected from the capacitor. The plates are then pulled apart so the spacing between the plates is increased. What is the effect (a) on the electric field between the plates? (b) on the potential difference between the plates?
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