Time period of oscillation of charge in front of infinite charged plane

Click For Summary
The discussion focuses on calculating the time period of oscillation for a charged particle near an infinite charged plane. The initial calculation for the time to reach the plane is expressed as t = root(4εmd/σq). It is debated whether the oscillation period is double this time, with a consensus emerging that it should actually be four times t. The final agreement confirms that the correct time period of oscillation is 4t. The conversation highlights the importance of careful analysis in deriving physical equations.
Saptarshi Sarkar
Messages
98
Reaction score
13
Homework Statement
A negetive point charge -q with mass m is held at rest at a distance d from an infinite charged plane with charge density σ and released. Find time period of oscillation (assuming that charge can freely move through plane without disturbing the charge density).
Relevant Equations
F = -qσ/2ε
a = -σq/2εm
I tried to calculate the time the charged particle will take to reach the plane using the a and using d=1/2at² and found the t to be equal to root(4εmd/σq).

I guess the time period of oscillation will be double of t (by symmetry), i.e. 2root(4εmd/σq). I don't know if this is correct.
 
Physics news on Phys.org
I should have looked at your work more closely. You have the picture quite correct.
Only small thing - if t is the time to fall to the plane, is the period really 2t?
 
  • Like
Likes Saptarshi Sarkar
rude man said:
I should have looked at your work more closely. You have the picture quite correct.
Only small thing - if t is the time to fall to the plane, is the period really 2t?

Thanks, got it. It should be 4t.
 
Saptarshi Sarkar said:
Homework Statement:: A negetive point charge -q with mass m is held at rest at a distance d from an infinite charged plane with charge density σ and released. Find time period of oscillation (assuming that charge can freely move through plane without disturbing the charge density).
Homework Equations:: F = -qσ/2ε
OK
a = -σq/2εm

I tried to calculate the time the charged particle will take to reach the plane using the a and using d=1/2at² and found the t to be equal to root(4εmd/σq).

I guess the time period of oscillation will be double of t (by symmetry), i.e. 2root(4εmd/σq). I don't know if this is correct.
[/QUOTE]
Saptarshi Sarkar said:
Thanks, got it. It should be 4t.
Yes. Good going.
 
  • Like
Likes Saptarshi Sarkar
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
9
Views
690
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
8
Views
2K
Replies
9
Views
2K
Replies
4
Views
5K
Replies
26
Views
1K
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K