An electron and a proton are in any electric field E=5x10^2 V/m. What is their speed 1.0 cm after being released?

F=Eq= 5x10^2 x 1.6x10^-19 = 8x10^-17 N

Kinetic energy theory: deltaEc= Fxd =1/2mv^2

thus v= squareroot(2xFxd/m)

v(proton) = squareroot(2x8x10^-17x1x10^-2/9.1x10^-31)=1.3x 10^6 m/s

v(electron) = squareroot(2x8x10^-17x1x10^-2/1.67x10^-27)= 3.1x10^6 m/s

EH
Answered by Emma H. Physics tutor

1822 Views

See similar Physics A Level tutors

Related Physics A Level answers

All answers ▸

Derive I = nAVe


Calculate the length of a 120m (as measured by the astronaut) spaceship travelling at 0.85c as measured by a stationary observer


A positively charged particle enters a magnetic field oriented perpendicular to its direction of motion. Does the particle: A) Change its velocity, B) Change its speed, C) Accelerate in the direction of the magnetic field.


Explain what simple harmonic motion means?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences