Three charged particles (their sign are shown) move along the path in a uniform electric field
as shown. In each of three cases, state whether the potential energy increases or decreases?
![](https://hi-static.z-dn.net/files/d13/213112a1d002337719a6c82cd88c8134.jpg)
Answers
Answer:
The electric field E is uniform inside the parallel plates.
The bending (deviation) from a straight line path depends on the Electric field perpendicular to velocity.
For the same velocity and electric field, the bending depends on charge per mass ratio.
The charged particle 3 bends most. So its charge/mass ratio is the highest.The particle 3 moves towards negative charges.
So it is a positive charge.Particles 1 and 2, move towards positive charges. So they are negative.
hope these will help you...
![](https://hi-static.z-dn.net/files/df5/e14f8c6abaad0590a1aa898049969ab4.jpg)
explanation
for particle A:- (-)
=>>>>>>>>>>≥> direction of electric field
<<<<<<<<<≤=. direction of negatively charged A
thus it's velocity decreases => kinetic energy decreases=> potential energy increases
for particle B:- (+)
=>>>>>>≥>>> direction of electric field.
<<<<<<≤<<<= direction of positively charged B
thus it's velocity increases => kinetic energy increases => potential energy decreases
for particle C:- (+)
=>>>>>>>>>> direction of electric field
=>>>>>>>>>> direction of positively charged C
thus it's velocity decreases => kinetic energy decreases => potential energy increases