Oscillations of Dipole in Electric Field: Stability & SHM

[gracy]

In a uniform electric field if a dipole is slightly displaced from it's stable equilibrium position it executes angular SHM.
What if a dipole is slightly displaced from it's unstable equilibrium position ,will it execute angular SHM?

 

[blue_leaf77]

What if a dipole is slightly displaced from it's unstable equilibrium position ,will it execute angular SHM?

Do you mean like in a situation where the dipole is hold at an arbitrary angle from the field and then released?
The SHM resulting from a small initial displacement from the equilibrium position is actually an approximation. For significantly bigger starting angle, the solution of the equation of motion is no longer well approximated by SHM. For this kind of problem, the solution involves certain type of elliptic integral.

 

[gracy]

Do you mean like in a situation where the dipole is hold at an arbitrary angle from the field?

I mean if the dipole initially makes 180 degrees with electric field and then it is displaced from there will it execute SHM?

 

[blue_leaf77]

I mean if the dipole initially makes 180 degrees with electric field and then it is displaced from there will it execute SHM?

That example is also included in the situation I described above.

 

[gracy]

That example is also included in the situation I described above.

I mean if the dipole initially makes 180 degrees with electric field and then it is displaced from there will it execute SHM?

You mean it won't execute SHM?

 

[blue_leaf77]

It will not.

 

[gracy]

In a uniform electric field if a dipole is displaced for a large angle from it's stable equilibrium position will it execute angular SHM?I think no.

 

[blue_leaf77]

The only logical difference between "slightly displaced from it's unstable equilibrium position" and "displaced for a large angle from it's stable equilibrium position" is that the second statement being just a special case of the first. If you want to know which case is an approximate SHM which is not, you should go to the equation of motion and see whether any approximation can be assumed. On the whole, a pendulum-type motion under uniform force field is strictly speaking not SHM.

 

[gracy]

In a non uniform electric field if a dipole is displaced it won't execute angular SHM(no matter what it's starting angle is and whether it is displaced for a larger or smaller angle )
In this case motion of dipole is combination of translatory and rotatory motion .
Right?

 

[blue_leaf77]

In this case motion of dipole is combination of translatory and rotatory motion .
Right?

What does your trained intuition tell you? Try to trust yourself if you know your reasoning can be supported by the known physics formula.

 

[nasu]

What does your trained intuition tell you? Try to trust yourself if you know your reasoning can be supported by the known physics formula.

If the displacement from stable equilibrium is small enough you can model the motion as a SHO. This is valid for any system, not restricted to dipole in a specific type of field.

 

[SammyS]

If the displacement from stable equilibrium is small enough you can model the motion as a SHO. This is valid for any system, not restricted to dipole in a specific type of field.

Right. For many potential functions, particularly in the case of one dimensional motion, a small displacement from a stable equilibrium position results in SHM, Simple Harmonic Motion.

 

[gracy]

a small displacement from a stable equilibrium position results in SHM, Simple Harmonic Motion.

Even in non uniform electric field?

 

[SammyS]

Even in non uniform electric field?

What is the potential function in this case?

 

[gracy]

What is the potential function in this case?

I just know dipole in non uniform electric field experiences a net force given by
$F$=$Δ(P.E)$

where p is the dipole moment and E is the electric field.I don't know abut potential.

 

[SammyS]

I just know dipole in non uniform electric field experiences a net force given by
$F$=$Δ(P.E)$

where p is the dipole moment and E is the electric field.I don't know about potential.

(In Post #12, I was commenting on Post #11.)

That should be $\displaystyle \vec F = -\vec \nabla (\vec p \cdot \vec E) \$ .

Do you understand what all of those symbols mean ?For the case of a non-uniform field, whatever happens will depend on the details, details which are not specified here.

 

[gracy]

Please answer my post #13 .

 

[SammyS]

Please answer my post #13 .

I did.

Last sentence in Post #16.

 

[berkeman]

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