Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 1556

A charged particle of mass \(1 \mathrm{~kg}\) and charge \(2 \mu \mathrm{c}\) is thrown from a horizontal ground at an angle \(\theta=45^{\circ}\) with speed $20 \mathrm{~m} / \mathrm{s}\(. In space a horizontal electric field \)\mathrm{E}=2 \times 10^{7} \mathrm{~V} / \mathrm{m}$ exist. The range on horizontal ground of the projectile thrown is $\ldots \ldots \ldots$ (A) \(100 \mathrm{~m}\) (B) \(50 \mathrm{~m}\) (C) \(200 \mathrm{~m}\) (D) \(0 \mathrm{~m}\)

Short Answer

Expert verified
The range of the projectile for the charged particle in the presence of the electric field is \(100 \mathrm{m}\).

Step by step solution

01

Find the Vertical and Horizontal Components of Initial Velocity

We have the initial speed \(v_{0} = 20 \mathrm{m/s}\) and the angle of projection \(\theta = 45^{\circ}\). The components of initial velocity in the horizontal and vertical directions are: \[v_{0x} = v_{0} \cos \theta \] \[v_{0y} = v_{0} \sin \theta \] Find the values of \(v_{0x}\) and \(v_{0y}\).
02

Calculate Time of Flight

The time of flight for a projectile is the time taken for it to reach the ground again. Since the vertical component of velocity and acceleration are only affected by gravity, we can calculate the time of flight using this equation: \[t_{f}=\frac{2v_{0y}}{g}\] where \(t_f\) is the time of flight and \(g\) is the acceleration due to gravity (\(g = 9.81 \mathrm{m/s^2}\)). Calculate the time of flight.
03

Find Vertical Force and Acceleration Due to Electric Field

The electric field influences the charged particle with a force in the vertical direction: \[F_{y} = Eq\] where \(E\) is the electric field, \(q\) is the charge of the particle, and \(F_{y}\) is the vertical force. In this case, \(E = 2 \times 10^{7} \mathrm{V/m}\) and \(q = 2 \times 10^{-6} \mathrm{C}\). Next, calculate the acceleration in the vertical direction due to this force: \[a_{y} = \frac{F_{y}}{m}\] where \(a_y\) is the vertical acceleration and \(m = 1\mathrm{kg}\) is the mass of the particle. Calculate the value of \(a_{y}\).
04

Adjust the Time of Flight in the Presence of Electric Field

Since there is an upward acceleration due to the electric field, the time of flight will be affected. The adjusted time of flight can be calculated by finding the effective gravitational acceleration in the presence of the electric field: \[g_{eff} = g - a_{y}\] Now compute the new time of flight with the effective gravitational acceleration: \[t_{f_{eff}} = \frac{2v_{0y}}{g_{eff}}\] Calculate the adjusted time of flight \(t_{f_{eff}}\).
05

Calculate Range of the Projectile

The range of the projectile is the horizontal distance traversed during the time of flight. In the presence of the electric field, it can be calculated using: \[R_{eff} = v_{0x} \cdot t_{f_{eff}}\] Determine the range \(R_{eff}\) and compare it to the given choices. The solution should provide the correct range for the charged particle in the presence of the electric field.

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 uniformaly charged spherical conductors \(\mathrm{A}\) and \(\mathrm{B}\) having radius \(1 \mathrm{~mm}\) and \(2 \mathrm{~mm}\) are separated by a distance of \(5 \mathrm{~cm} .\) If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres \(\mathrm{A}\) and \(\mathrm{B}\) is..... (A) \(4: 1\) (B) \(1: 2\) (C) \(2: 1\) (D) \(1: 4\)

A small sphere whose mass is \(0.1 \mathrm{gm}\) carries a charge of $3 \times 10^{-10} \mathrm{C}\( and is tie up to one end of a silk fiber \)5 \mathrm{~cm}$ long. The other end of the fiber is attached to a large vertical conducting plate which has a surface charge of \(25 \times 10^{-6} \mathrm{Cm}^{-2}\), on each side. When system is freely hanging the angle fiber makes with vertical is \(\ldots \ldots \ldots\) (A) \(41.8^{\circ}\) (B) \(45^{\circ}\) (C) \(40.2^{\circ}\) (D) \(45.8^{\circ}\)

Point charges \(q_{1}=2 \mu c\) and \(q_{2}=-1 \mu c\) care kept at points \(\mathrm{x}=0\) and \(\mathrm{x}=6\) respectively. Electrical potential will be zero at points ..... (A) \(\mathrm{x}=-2, \mathrm{x}=2\) (B) \(\mathrm{x}=1, \mathrm{x}=5\) (C) \(\mathrm{x}=4, \mathrm{x}=12\) (D) \(\mathrm{x}=2, \mathrm{x}=9\)

An electric circuit requires a total capacitance of \(2 \mu \mathrm{F}\) across a potential of \(1000 \mathrm{~V}\). Large number of \(1 \mu \mathrm{F}\) capacitances are available each of which would breakdown if the potential is more then \(350 \mathrm{~V}\). How many capacitances are required to make the circuit? (A) 24 (B) 12 (C) 20 (D) 18

Two point charges of \(+16 \mathrm{c}\) and \(-9 \mathrm{c}\) are placed $8 \mathrm{~cm}\( apart in air \)\ldots \ldots\(.. distance of a point from \)-9$ c charge at which the resultant electric field is zero. (A) \(24 \mathrm{~cm}\) (B) \(9 \mathrm{~cm}\) (C) \(16 \mathrm{~cm}\) (D) \(35 \mathrm{~cm}\)
See all solutions

Recommended explanations on English Textbooks

English Language Study

Read Explanation

Semiotics

Read Explanation

Single Paragraph Essay

Read Explanation

Textual Analysis

Read Explanation

Language Acquisition

Read Explanation

Pragmatics

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