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Chapter 17: Problem 94

Two parallel plates are \(4.0 \mathrm{cm}\) apart. The bottom plate is charged positively and the top plate is charged negatively, producing a uniform electric field of \(5.0 \times 10^{4} \mathrm{N} / \mathrm{C}\) in the region between the plates. What is the time required for an electron, which starts at rest at the upper plate, to reach the lower plate? (Assume a vacuum exists between the plates.)

Short Answer

Expert verified
Answer: It takes approximately \(9.56 \times 10^{-9}\text{ s}\) for an electron to travel from the upper plate to the lower plate.

Step by step solution

01

Identify the Known Parameters

We're given: 1. Distance between plates (d): \(4.0\text{ cm}\) (converting to meters, \(d = 0.04\text{ m}\)) 2. Electric field (E): \(5.0 \times 10^{4} \text{ N/C}\) 3. Electron's initial speed (v_0): \(0\text{ m/s}\) (starts at rest) 4. Charge of the electron (e): \(-1.6 \times 10^{-19}\text{ C}\) 5. Mass of electron (m): \(9.11 \times 10^{-31}\text{ kg}\) We need to find the time (t) it takes for the electron to reach the lower plate.
02

Calculate Acceleration Due to Electric Field

The force acting on an electron due to an electric field is given by: $$F = e \times E$$ Where F is the force, e is the electron's charge, and E is the electric field strength. Since F = m × a, where m is the mass of the electron and a is the acceleration, we can solve for acceleration: $$a = \frac{e \times E}{m}$$ Substitute the given values: $$a = \frac{-1.6 \times 10^{-19}\text{ C} \times 5.0 \times 10^{4} \text{ N/C}}{9.11 \times 10^{-31}\text{ kg}}$$ Calculate the value of a: $$a = -8.77 \times 10^{13} \text{ m/s}^2$$ The negative sign indicates the acceleration is towards the positively charged bottom plate.
03

Use Kinematic Equations to Find Time

Using the equation \(d = v_0 t + \frac{1}{2}at^2\), where d is the distance, v_0 is the initial velocity, a is the acceleration, and t is the time, we can solve for the time t: $$0.04\text{ m} = 0\text{ m/s} \times t -\frac{1}{2}(8.77 \times 10^{13}\text{ m/s}^2) \times t^2$$ Simplify the equation: $$0.04\text{ m} = -4.385 \times 10^{13}\text{ m/s}^2 \times t^2$$ Solve for t: $$t^2 = \frac{0.04\text{ m}}{-4.385 \times 10^{13}\text{ m/s}^2}$$ $$t^2 = 9.124 \times 10^{-16}\text{ s}^2$$ $$t = \sqrt{9.124 \times 10^{-16}\text{ s}^2}$$ $$t = 9.56 \times 10^{-9}\text{ s}$$ The time required for the electron to reach the lower plate is approximately \(9.56 \times 10^{-9}\text{ s}\).

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Most popular questions from this chapter

A parallel plate capacitor has a capacitance of \(1.20 \mathrm{nF}\) There is a charge of \(0.80 \mu \mathrm{C}\) on each plate. How much work must be done by an external agent to double the plate separation while keeping the charge constant?
A parallel plate capacitor has a capacitance of \(2.0 \mu \mathrm{F}\) and plate separation of \(1.0 \mathrm{mm} .\) (a) How much potential difference can be placed across the capacitor before dielectric breakdown of air occurs \(\left(E_{\max }=3 \times 10^{6} \mathrm{V} / \mathrm{m}\right) ?\) (b) What is the magnitude of the greatest charge the capacitor can store before breakdown?
A shark is able to detect the presence of electric fields as small as $1.0 \mu \mathrm{V} / \mathrm{m} .$ To get an idea of the magnitude of this field, suppose you have a parallel plate capacitor connected to a 1.5 - \(V\) battery. How far apart must the parallel plates be to have an electric field of $1.0 \mu \mathrm{V} / \mathrm{m}$ between the plates?
Suppose you were to wrap the Moon in aluminum foil and place a charge \(Q\) on it. What is the capacitance of the Moon in this case? [Hint: It is not necessary to have two oppositely charged conductors to have a capacitor. Use the definition of potential for a spherical conductor and the definition of capacitance to get your answer. \(]\)
A van de Graaff generator has a metal sphere of radius \(15 \mathrm{cm} .\) To what potential can it be charged before the electric field at its surface exceeds \(3.0 \times 10^{6} \mathrm{N} / \mathrm{C}\) (which is sufficient to break down dry air and initiate a spark)?
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