Potential Energy of Electron & Proton in an Electric Field

AI Thread Summary
When an electron moves closer to the source of an electric field, its potential energy decreases due to the attractive force exerted by the field. Conversely, if a proton approaches the same field source, its potential energy increases because it is repelled by the field. The kinetic energy of both particles is lowest at infinity and increases as they approach the field source. The direction of the electric field significantly influences the force acting on charged particles, which is determined by the equation F = qE. Understanding these principles is essential for analyzing the behavior of charged particles in electric fields.
da_coolest
Messages
15
Reaction score
0
when an electron moves closer to the electric field source, does the potential energy get higher or lower?

if it is being accelerated from the infinity, then at the infinity it should have the lowest kinetic energy? then when it is closer to the source of the field, it should have the highest kinetic energy??

ok then what about a proton?
 
Physics news on Phys.org


da_coolest said:
when an electron moves closer to the electric field source, does the potential energy get higher or lower?

if it is being accelerated from the infinity, then at the infinity it should have the lowest kinetic energy? then when it is closer to the source of the field, it should have the highest kinetic energy??

ok then what about a proton?

Welcome to the PF. Your question is a bit incomplete, but maybe that is what is confusing you. You need to specify what the Electric field's direction is. Electric field is a vector field, having magnitude and direction. The force on a charged particle is

F = qE

Where q is the value of the charge in Coulombs, and E is the Electric field value in Volts/meter. The direction of the force is determined by both the charge polarity and the direction of the vector E field.
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top