Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 24: Problem 76

A proton traveling along the \(x\) -axis at a speed of \(1.0 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) enters the gap between the plates of a \(2.0-\mathrm{cm}-\) wide parallel plate capacitor. The surface charge distributions on the plates are given by \(\sigma=\pm 1.0 \cdot 10^{-6} \mathrm{C} / \mathrm{m}^{2}\). How far has the proton been deflected sideways \((\Delta y)\) when it reaches the far edge of the capacitor? Assume that the electric field is uniform inside the capacitor and zero outside.

Short Answer

Expert verified
Answer: The sideways deflection of the proton when it reaches the far edge of the capacitor is 4.32 x 10^-4 m.

Step by step solution

01

Calculate the electric field inside the capacitor

Using the relation between electric field (E) and surface charge density (σ), we can find the electric field inside the capacitor: \(E = \frac{\sigma}{\epsilon_0}\) where \(ε_0\) is the vacuum permittivity (\(8.85 \cdot 10^{-12} \mathrm{C}^2 / \mathrm{N}\mathrm{m}^2\)). \(\sigma = 10^{-6} \mathrm{C}/\mathrm{m}^2\) \(E = \frac{10^{-6} \mathrm{C}/\mathrm{m}^2}{8.85 \cdot 10^{-12} \mathrm{C}^2 / \mathrm{N}\mathrm{m}^2} = 1.13 \cdot 10^5 \mathrm{N}/\mathrm{C}\)
02

Determine the force acting on the proton

The force acting on the proton due to the electric field can be calculated using: \(F = qE\) where \(q\) is the charge of the proton (\(1.6 \cdot 10^{-19} \mathrm{C}\)). \(F = (1.6 \cdot 10^{-19} \mathrm{C})(1.13 \cdot 10^5 \mathrm{N}/\mathrm{C}) = 1.81 \cdot 10^{-14} \mathrm{N}\)
03

Calculate the acceleration of the proton in the y-direction

Using Newton's second law, we can find the acceleration of the proton in the y-direction: \(F = ma\) \(a = \frac{F}{m}\) where \(m\) is the mass of the proton (\(1.67 \cdot 10^{-27} \mathrm{kg}\)). \(a = \frac{1.81 \cdot 10^{-14} \mathrm{N}}{1.67 \cdot 10^{-27} \mathrm{kg}} = 1.08 \cdot 10^{13} \mathrm{m}/\mathrm{s}^2\)
04

Calculate the time taken for the proton to travel across the width of the capacitor

Using the initial speed \(v_x = 1.0 \cdot 10^6 \mathrm{m}/\mathrm{s}\) and the width of the capacitor \(d = 2.0 \mathrm{cm} = 0.02 \mathrm{m}\), we can find the time taken for the proton to travel across the width: \(t = \frac{d}{v_x} = \frac{0.02 \mathrm{m}}{1.0 \cdot 10^6 \mathrm{m}/\mathrm{s}} = 2.0 \cdot 10^{-8} \mathrm{s}\)
05

Calculate the sideways deflection (Δy) of the proton

Using the time \(t\) and the acceleration \(a\) found in the previous steps, we can calculate the sideways deflection (Δy) of the proton: Δy = \(\frac{1}{2}at^2\) Δy = \(\frac{1}{2}(1.08 \cdot 10^{13} \mathrm{m}/\mathrm{s}^2)(2.0 \cdot 10^{-8} \mathrm{s})^2 = 4.32 \cdot 10^{-4} \mathrm{m}\) Therefore, the sideways deflection of the proton when it reaches the far edge of the capacitor is \(4.32 \cdot 10^{-4} \mathrm{m}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two concentric metal spheres are found to have a potential difference of \(900 . \mathrm{V}\) when a charge of \(6.726 \cdot 10^{-8} \mathrm{C}\) is applied to them. The radius of the outer sphere is \(0.210 \mathrm{~m}\). What is the radius of the inner sphere?

Which of the following capacitors has the largest charge? a) a parallel plate capacitor with an area of \(10 \mathrm{~cm}^{2}\) and a plate separation of \(2 \mathrm{~mm}\) connected to a \(10-\mathrm{V}\) battery b) a parallel plate capacitor with an area of \(5 \mathrm{~cm}^{2}\) and a plate separation of \(1 \mathrm{~mm}\) connected to a \(10-\mathrm{V}\) battery c) a parallel plate capacitor with an area of \(10 \mathrm{~cm}^{2}\) and a plate separation of \(4 \mathrm{~mm}\) connected to a \(5-\mathrm{V}\) battery d) a parallel plate capacitor with an area of \(20 \mathrm{~cm}^{2}\) and a plate separation of \(2 \mathrm{~mm}\) connected to a \(20-\mathrm{V}\) battery e) All of the capacitors have the same charge.

A parallel plate capacitor is charged with a battery and then disconnected from the battery, leaving a certain amount of energy stored in the capacitor. The separation between the plates is then increased. What happens to the energy stored in the capacitor? Discuss your answer in terms of energy conservation.

A parallel plate capacitor has a capacitance of \(120 .\) pF and a plate area of \(100 . \mathrm{cm}^{2}\). The space between the plates is filled with mica whose dielectric constant is \(5.40 .\) The plates of the capacitor are kept at \(50.0 \mathrm{~V}\) a) What is the strength of the electric field in the mica? b) What is the amount of free charge on the plates? c) What is the amount of charge induced on the mica?

A large parallel plate capacitor with plates that are square with side length \(1.00 \mathrm{~cm}\) and are separated by a distance of \(1.00 \mathrm{~mm}\) is dropped and damaged. Half of the areas of the two plates are pushed closer together to a distance of \(0.500 \mathrm{~mm}\). What is the capacitance of the damaged capacitor?
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Geometrical and Physical Optics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Translational Dynamics

Read Explanation

Waves Physics

Read Explanation

Physical Quantities And Units

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free