An electron is accelerated by a constant electric field of magnitude 270 N/C.(a) Find the acceleration of the electron.m/s2(b) Use the equations of motion with constant acceleration to find the electron's speed after 9.50 × 10 s, assuming it starts fromrest.m/s

An electron is accelerated by a constant electric field of magnitude 270 N/C. (a) Find the acceleration of the electron. m/s? (b) Use the equations of motion with constant acceleration to find the electron's speed after 9.50 x 10-9 s, assuming it starts from rest. m/s

An electron is accelerated by a constant electric field of magnitude 270 N/C. (a) Find the acceleration of the electron. m/s? (b) Use the equations of motion with constant acceleration to find the electron's speed after 9.50 x 10-9 s, assuming it starts from rest. m/s

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter7: Electric Potential

Section: Chapter Questions

Problem 105AP: An electron has an initial velocity of 5.00106m/s in a uniform 2.0010m/s electric field. The field...

See similar textbooks

Related questions

Q: To study the properties of various particles, you can accelerate the particles with electric fields.…

A: Electric field magnitude (E) = 296 N/CTime (t) = 8.75 x 10-9sMass of positron (m) = 9.109 x…

Q: Electric force in the Hydrogen atom. In a hydrogen atom the separation between the proton and the…

A: Here we have to Compute the magnitude of the attractive force of attraction on the electron by the…

Q: Given an electron that is released from rest in a uniform electric field, it accelerates upward at…

Q: In the ever popular game from the future, Electrodisc, players can score double points by pushing a…

A: Given: q1=1.3 mC = 1.3*10-3 C q2=1.7 mC = 1.7*10-3 C q3 = 0 r12 = 3.7m r23=10m

Q: kg. Q.1. The mass of an electron is 9.11 × 10*°T kg, that of a proton is 1.67 x 1072' Find the ratio…

Q: To study the properties of various particles, you can accelerate the particles with electric fields.…

A: .

Q: Shown are the configuration of a gel electrophoresis experiment. Charged plates establish an…

A: The electric field is directed from the positive charge to the negative charge. Let q denote the…

Q: A charged particle moves under the influence of an electric field. Its x-position and y-position,…

A: (a) Let vx and vy be defined as the velocity of a particle in x-direction and y-direction…

Q: An electron with an initial velocity of (3X105 m/s)j enters a region; in which the electric field is…

Q: An electron is accelerated by a constant electric field of magnitude 295 N/C. (a) Find the…

A: Given:- The magnitude of the constant electric field is E = 295 N/C. An electron is accelerated in…

Q: In an experiment in space, one proton is held fixed and another proton is released from rest a…

Q: An electron travels with = 4.00 x 106 m/s through a point in space where E=(2.10 x 105 - 2.10 x 105)…

A: Given: Velocity Electric field Magnetic field Charge on electron Required: Force on the elctron .

Q: An electron is released from rest in a uniform electric field. The electron accelerates vertically…

A: Given that:u=0 m/ss=4.5 mt=3 ms

Q: To study the properties of various particles, you can accelerate the particles with electric fields.…

Q: centripetal acceleration

A: Coulomb force provides centripetal force

Q: An electron is accelerated by a constant electric field of magnitude 295 N/C. (a) Find the…

A: (a)Electric field of an electron,(E)=(-)295 N/C -----(1) [Given] [Negative sign just indicates the…

Q: To study the properties of various particles, you can accelerate the particles with electric fields.…

A: Given that, Mass of the positron, m=mass of electron=9.1×10-31 kg Charge on positron, q=+e=1.6×10-19…

Q: Shown are the configuration of a gel electrophoresis experiment. Charged plates establish an…

A: The filled fragments transfer to the positively charged plate.There is higher potential for a…

Q: An electron is accelerated by a constant electric field of magnitude 300 N/C. (a) Find the…

A: (a) acceleration a = 5.274 × 1013 m/s2 (b) Electrons speed V = 5.274 × 105 m/s

Q: An electron is accelerated by a constant electric field of magnitude 295 N/C. (a) Find the…

A: Given data The electric field is E=295 N/C (a) The force on the electron is, F1=eE here e is the…

Q: An electron is accelerated by a constant electric field of magnitude 290 N/C. (a) Find the…

Q: A positively charged particle of mass 1.92 x 10-27 kg initially moves left to right long the x axis…

A: Diagram, Given, Mass of +ve charge particle, m = 1.92 x 10-27 kg Particle is coming to…

Q: The electron and proton of a hydrogen atom are separated by a distance of approximately 5.3*10-11 m.…

A: We are given a hydrogen atom. There is a proton and electron in hydrogen atom. We are given charge…

Q: Find the magnitude of the acceleration in SI Unit experienced by an electron in an electric field of…

A: Given data The magnitude of the electric field is given as E = 750 N/C. The force due to the…

Q: Older televisions and computer monitors, as well as some modern electronic instruments, use a…

A: (a) The initial velocity of the electron is zero. Now when the electron will move in the electric…

Q: Para el circuito que se muestra en la siguiente figura, use el método que prefiera para encontrar la…

Q: Charges q₁ q2 = 93 = 10.0 µC have positions ₁ = 1.60 m î+ 1.20 m ĵ ₂ = 1.20 m ĵ, and 73 = (0.00,…

A: Solution:-Given thatq1=q2=q3=10 μC=10×10-6 C=10-5 Cr1→=(1.6 i^+1.2 j^) mr2→=1.2 j^ mr3→=(0, 0)

Concept explainers

Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

An electron is accelerated by a constant electric field of magnitude 270 N/C.
(a) Find the acceleration of the electron.
m/s2
(b) Use the equations of motion with constant acceleration to find the electron's speed after 9.50 × 10 s, assuming it starts from
rest.
m/s

Expert Solution

Part (a)

Given Data:

Electric field = E = 270 N/C

Charge of an electron = e = 1.6 x 10^-19 C

Mass of an electron = m_e = 9.11 x 10^-31 kg

Solution:

The electric force is

$F = q E$ . . . . . (i)

Using Newton's second law

$F = m a$ . . . . . (ii)

Equating (i) & (ii)

$a = \frac{q E}{m}$

$a = \frac{e E}{m_{e}}$

$a = \frac{1.6 \times 10^{- 19} \times 270}{9.11 \times 10^{- 31}}$

$a =$ 4.74 x 10¹³ m/s²

Trending now

This is a popular solution!

Step by step

Solved in 3 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions

An electron and a proton, each starting from rest, are accelerated by the same uniform electric field of 200 N/C. Determine the distance and time for each particle to acquire a kinetic energy of 3.21016 J.
An electron has an initial velocity of 5.00106m/s in a uniform 2.0010m/s electric field. The field accelerates the election in the direction opposite to its initial velocity, (a) What is the direction of the electric field? (b) How far does the electron travel before coming to rest? (c) How long does it take the electron to come to rest? (d) What is the electron's velocity when it returns to its starting point?
An electron is accelerated by a constant electric field of magnitude 295 N/C. (a) Find the acceleration of the electron. m/s2 (b) Use the equations of motion with constant acceleration to find the electron's speed after 1.25 x 108 s, assuming it starts from rest. m/s
On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 110 N/C. Compare the gravitational and electric forces on a small dust particle of mass 3,1 x 10 g that carries a single electron charge. 1.1134 What is the acceleration (both magnitude and direction) of the dust particle? (Enter the magnitude in m/s.) magnitude m/s direction downward v
To study the properties of various particles, you can accelerate the particles with electric fields. A positron is a particle with the same mass as an electron but the opposite charge (+e). If a positron is accelerated by a constant electric field of magnitude 271 N/C, find the following. (a) Find the acceleration of the positron. m/s² (b) Find the positron's speed after 6.50 x 10-9 s. Assume that the positron started from rest. m/s
To study the properties of various particles, you can accelerate the particles with electric fields. A positron is a particle with the same mass as an electron but the opposite charge (+e). If a positron is accelerated by a constant electric field of magnitude 340 N/C, find the following. (a) Find the acceleration of the positron. m/s2(b) Find the positron's speed after 7.75 10-9 s. Assume that the positron started from rest. m/s
An electron is accelerated by a constant electric feld of magnitude 290 N/C. (a) Find the acceleration of the electron. m/s (b) Use the equations of motion with constant acceleration to find the electron's speed after 115 x 10s, assuming it starts from rest. m/s Need Helpz E Read It Wetch
A =7, B= 17, C=117, D=8.
Q5) An electron enters a region of uniform electric field (E = 50 N/C) with an initial velocity of 4.0x10ʻm/s in the same as the electric field. What is the speed of the electron 1.5 ns after entering this region? a. 5.30x10°m/s b. 2.70x10°m/s c. 1.80x10°m/s d. 6.20x10°m/s e. 4.2 0x10°m/s
B4
Given an electron that is released from rest in a uniform electric field, it accelerates upward at 3.5 m in the first 8 x 10^-6 s after it is released. What is the magnitude and acceleration of the electron. The mass of an electron is = 9.109 x 10-31 kg e = 1.602 x 10-19 C
2- A particle with charge (q= 2µC ) and mass (m= 4x10 Kg) moves with initial velocity (v = 2x10°î m/s) into a region of uniform electric field (Ë = 6x10° } N/C) of length (L=0.4m),as shown in the figure. Determine the displacement of the particle along the y-axis when it leaves the region of the electric field. L A- Ay = 6x10²m B- Ay = 6x10³m C- Ay = 6x10**m D- Ay = 6µm