Time period of small oscillations of the point dipole

[rude man]

So. you mean to say that the flat layer will be divided into two parts, one to the left of x=0 and one to the right and to the left of x=0, - charge will be present and to the right of x=0, + charge would be present. Is it like that?

Yes.
You still haven't told us the location of the dipole. Is the center at x=0?

 

[sid0123]

I already mentioned the position of the dipole. Let me attach a picture to make things more understandable. The charges of the dipole are on the opposite sides of the imaginary vertical line passing through x=0.

(The distance between the charges of the dipole was not given in the question and it was assumed here. Because in the end, q*l will give p and p was given in the question)

 

[rude man]

No, the answer is independent of the distance between the charges of the dipole.
The answer to this problem is √[(πmd)/(ρ°p)]

What is dipole moment p in cgs? (In SI it's simply ql).

 

[sid0123]

What is dipole moment p in cgs? (In SI it's simply ql).

In CGS also it is the same ql

 

[rude man]

I already mentioned the position of the dipole. Let me attach a picture to make things more understandable. The charges of the dipole are on the opposite sides of the imaginary vertical line passing through x=0.

(The distance l between the charges of the dipole was not given in the question and it was assumed here. Because in the end, q*l will give p and p was given in the question)

That's fine if we assume l → 0 so both dipole charges see the same E field. If this is not assumed then the problem becomes more complicated. If l is finite then in order for a uniform E field for the dipole, the dipole must be centered at x=0.

Did you check the given answer for T for dimensional correctness? I can't do it in cgs but I still believe l is needed (where p = ql).
EDIT: Never mind, I checked dimensions & they are OK. Still trying to figure out the answer though.

 

[rude man]

Note that p has to be in same direction as E (both in -x direction) except for the small initial angular displacement of course.

 

[sid0123]

Note that p has to be in same direction as E (both in -x direction) except for the small initial angular displacement of course.

Yes, and in our assumption also they are in the same direction. I still am not able to get this expression E(x)=(2πρ°(x²-d²))/d
What are the limits that we took for the integral? I tried to make a Gaussian cylinder.
On the right hand side, the direction of electric field will be towards right and ds will also be on the right, so flux through right face will be Eds
On the left hand side, the direction of electric field will be towards right but ds will be towards left and the flux becomes E.ds=-Eds
And when we add them up, the total flux becomes zero which shouldn't be the case I feel. Kindly tell me where am I going wrong.

I just need to get through this and after that I will try to solve for the time period by my own.

 

[rude man]

Look at

 

[rude man]

Yes, and in our assumption also they are in the same direction. I still am not able to get this expression E(x)=(2πρ°(x²-d²))/d
What are the limits that we took for the integral? I tried to make a Gaussian cylinder.
On the right hand side, the direction of electric field will be towards right and ds will also be on the right, so flux through right face will be Eds
On the left hand side, the direction of electric field will be towards right but ds will be towards left and the flux becomes E.ds=-Eds
And when we add them up, the total flux becomes zero which shouldn't be the case I feel. Kindly tell me where am I going wrong.

I just need to get through this and after that I will try to solve for the time period by my own.

You picked the wrong Gaussian cylinder.
One end is at x > d and the other end is at x, -d < x < d.

 

[rude man]

I still am not able to get this expression E(x)=(2πρ°(x²-d²))/d

BTW the expression for E is correct, there should be no minus sign in front of it. It is always negative within -d < x < d.

 

[sid0123]

Is this the gaussian surface you were talking about?

 

[sid0123]

You picked the wrong Gaussian cylinder.
One end is at x > d and the other end is at x, -d < x < d.

Is this the gaussian surface you were talking about?

 

[sid0123]

The magnitude of forces on both the charges of the dipole will be equal as they're at an equal distance from the centre. I am sure about this.
Talking about the gaussian surface,
What will be the gaussian surface, we use the formula ∫E.ds = 4πq or q/ε
Here, the what will be E through the right face of the cylinder? Would the + charge that is on the right hand side of the plate, be distributed on the right outside surface of the plate too? If that is the case, then Φ₁ would be Eds (+x direction)
and through the left surface also it would be the same.
The net flux will then become 2E.ds
Now, we are given volume charge density which is uniform. What value of q will we take on the right hand side?
Because q is the total charge contained inside our gaussian cylinder. And charge distribution is non uniform.

 

[rude man]

The magnitude of forces on both the charges of the dipole will be equal as they're at an equal distance from the centre. I am sure about this.

OK. Very important & I did not see this until now.

Now, we are given volume charge density which is uniform

?? It is not uniform. It varies with x.

. What value of q will we take on the right hand side?
Because q is the total charge contained inside our gaussian cylinder. And charge distribution is non uniform.

You need to calculate the total charge within the gaussian surface. You also need to know E at the right-hand end at x > d. Then you get E at the left-hand end which is anywhere within -d < x < d. So then you know E(x) for all x, -d < x < d.

 

[rude man]

Не забудьте зайти в постель!

 

[sid0123]

It is not uniform. It varies with x.

Yeah, non-uniform. That is what I intended to write, was a typo.If we keep left face of the gaussian cylinder inside the positive half of the plate, the flux through that surface would be E.ds and would be positive.
If we keep left face of the gaussian cylinder inside the negative half of the plate (i.e. to the left of x=0), then the flux through that surface would be negative as E and ds would be in opposite directions.
So, how to decide where will the left face of our gaussian cylinder would lie?

 

[haruspex]

Yeah, non-uniform. That is what I intended to write, was a typo.If we keep left face of the gaussian cylinder inside the positive half of the plate, the flux through that surface would be E.ds and would be positive.
If we keep left face of the gaussian cylinder inside the negative half of the plate (i.e. to the left of x=0), then the flux through that surface would be negative as E and ds would be in opposite directions.
So, how to decide where will the left face of our gaussian cylinder would lie?

If the cylinder extends from x to +∞, what is the total charge inside it?

 

[rude man]

Yeah, non-uniform. That is what I intended to write, was a typo.If we keep left face of the gaussian cylinder inside the positive half of the plate, the flux through that surface would be E.ds and would be positive.
If we keep left face of the gaussian cylinder inside the negative half of the plate (i.e. to the left of x=0), then the flux through that surface would be negative as E and ds would be in opposite directions.
So, how to decide where will the left face of our gaussian cylinder would lie?

E and ds are in the same direction in the entire region -d < x < d. So E⋅ds is positive everywhere in the region. The E field is in the -x direction everywhere in the region. This may not be intuitive but it's true because the plates are infinite in extent.

The net charge in the gaussian cylinder you depict is either + or 0, never - .

 

[TSny]

I don't think this problem is talking about rotational oscillations. You will get the answer in post #20 if you interpret the problem as shown below. The dipole oscillates in linear simple harmonic motion.

 

[rude man]

I don't think this problem is talking about rotational oscillations. You will get the answer in post #20 if you interpret the problem as shown below. The dipole oscillates in linear simple harmonic motion.

View attachment 207797

Fabulous perception T! So the dipole jumps up and down about x=0 and yes the force is proportional to p and of course x.
Thanks!
EDIT: I got T = √(πmd/kρ⁰p) in SI units where k is our usual 1/4πε₀ ~ 1e9.

 

[rude man]

@sid0123, when you have derived the E field to your satisfaction, look at post #49!
(Using gaussian cylinder method is still applicable).

 

[sid0123]

Thank you very much, everyone. I understood it but took a little time to reply back. And yes, they were linear oscillations.

 

[sid0123]

Thank you, everyone for making me understand this. Sorry for taking a little long to reply. And yes, they were linear oscillations