Time period of small oscillations of the point dipole

[sid0123]

Homework Statement

In an infinite flat layer of thickness 2d, volume charge density is given according to the law: ρ=(ρ°)(x)/d and (-d≤x≤d). Here, x is the axis perpendicular to the plane. In the layer, there is a thin channel in which a point dipole of mass m and dipole moment p is placed. Calculate the time period of small oscillations of the dipole.

Homework Equations

Differential form of electric field:

div E = 4πρ and
div E = ∂E_x/∂x + ∂E_y/∂y + ∂E_z/∂z

The Attempt at a Solution

"Attempt to the solution has been made in CGS system"

div E = 4πρ
∂E_x/∂x + ∂E_y/∂y + ∂E_z/∂z = 4πρ

∂E_y/∂y and ∂E_z/∂z will be zero.
Therefore,
∂E_x/∂x = 4πρ =(4πρ°x)/d
E_x(x) = (4πρ°)/d ∫x dx = [(4πρ°)/d]*[x²]/2 = (2πρ°x²)/d + C

How will we find the constant C?
What exactly will be the electric field at point d from the midpoint of the layer? Will it be 2πρ°(2d) = 4πρ°d?

I am confused how to proceed from here.

 

[rude man]

Homework Statement

In an infinite flat layer of thickness 2d, volume charge density is given according to the law: ρ=(ρ°)(x)/d and (-d≤x≤d). Here, x is the axis perpendicular to the plane. In the layer, there is a thin channel in which a point dipole of mass m and dipole moment p is placed. Calculate the time period of small oscillations of the dipole.

Homework Equations

Differential form of electric field:

div E = 4πρ and

are you using some other, unfamiliar (at least to me) non-SI system of units?

 

[sid0123]

Yes, as I have already mentioned, I am not using SI system.
Here, I have used CGS system

 

[rude man]

OK, I might look at it from SI & will post if I'm successful, in which case maybe you can "translate" into cgs.

Funny, my introductory Physics course was also in cgs but I never mess with it anymore since it's totally at variance with electrical engineering.

 

[sid0123]

Wow! CGS?

OK, I might look at it from SI & will post if I'm successful, in which case maybe you can "translate" into cgs.

Funny, my introductory Physics course was also in cgs but I never mess with it anymore since it's totally at variance with electrical engineering.

That is very rare to hear. Which country did you graduate from?

 

[rude man]

Wow! CGS?
That is very rare to hear. Which country did you graduate from?

U.S. of A.
Cambridge, MA
1962
So it was a while ago & maybe even Harvard has "reformed"!

 

[sid0123]

Oh, that was long back ago. My university in Russia still uses CGS system for Electromagnetism course.

 

[rude man]

Oh, that was long back ago. My university in Russia still uses CGS system for Electromagnetism course.

I'm not sure but I think pure physics here is still taught using cgs but applied physicists and certainly engineers all use the Système Internationale.

 

[rude man]

How about a Gaussian right cylinder running from x=0 to x = ∞? What can you say about E_x at x = ∞?
EDIT: since the sheets are infinite in the y and z directions you don't have to put the far end of the Gaussian cylinder at infinity. What is the E field for all |x| > d?

 

[haruspex]

In the layer, there is a thin channel in which a point dipole

It is not clear to me what this means. Whereabouts in the layer is the dipole? Is it in the middle, at one extreme, or an unknown point between the two?

Anyway, consider just a thin infinite sheet of the layer at distance x from the middle, thickness dx. What is the field due to that?

 

[rude man]

I assume the dipole is located at x=0, is that right @sid0123?
(By means of the Gaussian surface, any other parallel layer within -d<x<d can also be easily chosen.)

 

[sid0123]

Sorry for the late reply.
I think the centre of the line joining the dipole is located at x=0 i.e. the middle of the plate.
There was a small hint to the problem given and there, they have solved for C and they finally came up with an equation:
E(x)=(2πρ°(x²-d²))/h and that means, they have taken C to be -h.

How I am trying to view this problem is that electric field at x=0 should be 0.
Let electric field at x=0 be E(0) and at x=x be E(x)

Now, solving the integral ∂Ex/∂x =(4πρ°x)/d i.e. ∫dE_x=(4πρ°/d) ∫x dx
Taking limits of LHS from E(0) to E(d) and of RHS from x=0 to x=d, we get,
E(x)-E(0)=((2πρ°)/d)[x²-0²] and we get E(x)-E(0)=((2πρ°x²)/d) and here, E(0) would be 0
So, the equation should finally become, E(x)=((2πρ°x²)/d)
Is it right till here?

 

[haruspex]

electric field at x=0 should be 0.

Certainly not.

that means, they have taken C to be -h.

I do not see how you conclude that. It's not what I get if I substitute -h for C in your expression for the integral.

 

[sid0123]

I do not see how you conclude that. It's not what I get if I substitute -h for C in your expression for the integral.

Yes, we don't get -h. I skipped 2πρ°
Yeah, so we get -2πρ°h
But I didn't understand how?

 

[sid0123]

Certainly not.

But why? The volume charge density is related according to the relation ρ=(ρ°)(x)/d
At x=0, won't the volume charge density at the centre become zero?
And if it would be zero, how would there be any electric field at that point?

 

[rude man]

But why? The volume charge density is related according to the relation ρ=(ρ°)(x)/d
At x=0, won't the volume charge density at the centre become zero?
And if it would be zero, how would there be any electric field at that point?

Because there is + charge above x=0 and - charge below x=0 so a test charge at x=0 will be repelled by the + charges and attracted by the - charges, giving a negative, non-zero E field at x=0.

Why are you using h instead of d all of a sudden?
The derived expression E(x)=(2πρ°(x²-d²))/h with a - sign in front of it and the condition -h <= x <= h looks right.

 

[sid0123]

Why are you using h instead of d all of a sudden?

My apologies for this. The hint that I was referring to had used h and I forgot to change it to d.

The derived expression E(x)=(2πρ°(x2-d2))/h with a - sign in front of it and the condition -h <= x <= h looks right.

But there is no negative sign in front of this expression and also, I failed to understand how did you reach even close to this expression. Can you please throw some more light on this?

 

[rude man]

There is another problem with the statement of the question: in order to solve for frequency of oscillations it is insufficient to specify just the dipole moment. The separation distance of the dipole charges is also needed.

 

[sid0123]

Because there is + charge above x=0 and - charge below x=0 so a test charge at x=0 will be repelled by the + charges and attracted by the - charges, giving a negative, non-zero E field at x=0.

And if there were no point charges, i.e. no point dipole inside the infinite plate, would the electric field at x=0 then be 0?

 

[sid0123]

There is another problem with the statement of the question: in order to solve for frequency of oscillations it is insufficient to specify just the dipole moment. The separation distance of the dipole charges is also needed

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

 

[rude man]

But there is no negative sign in front of this expressio

The E field points along the -x direction! So the - sign is needed.

and also, I failed to understand how did you reach even close to this expression. Can you please throw some more light on this?

You can proceed with your integration with haruspex's help but I recommend invoking Gauss's theorem which I find makes the problem easier. Remember? ∫D⋅dA = Q (combining ∇⋅D = ρ with the Divergence Theorem?

Of course, to get the period a simple ordinary differential equation must be solved unless you have been told its solution beforehand.

 

[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)]

Are you sure "d" does not refer to the separation distance between the dipole charges? "d" cannot be negative! I think you're mixing up d and h.

 

[sid0123]

Are you sure "d" does not refer to the separation distance between the dipole charges? "d" cannot be negative! I think you're mixing up d and h

Yes, I am sure. I reread the question and 2d is the thickness of the infinite flat plate and not the distance between the dipole.

 

[sid0123]

"d" cannot be negative

If we take the midpoint of the thickness of the infinite flat layer at x=0, then the right wall of the flat layer is at distance d in the positive x direction and the left layer is at a distance d in the negative x direction. That is what is meant by -d≤x≤d

 

[sid0123]

be

What about this one?

 

[rude man]

And if there were no point charges, i.e. no point dipole inside the infinite plate, would the electric field at x=0 then be 0?

No.
Your dipole does not affect the E field used to compute the torque on the dipole. I have exlained why the E field at x=0 is not zero in post 16.

 

[sid0123]

I have exlained why the E field at x=0 is not zero in post 16.

In post #16, you've explained that in reference to the dipole.
"Because there is + charge above x=0 and - charge below x=0 so a test charge at x=0 will be repelled by the + charges and attracted by the - charges, giving a negative, non-zero E field at x=0."

+ charge above x and - charge below x : Aren't these the charges of the dipole you're talking about?

 

[rude man]

If we take the midpoint of the thickness of the infinite flat layer at x=0, then the right wall of the flat layer is at distance d in the positive x direction and the left layer is at a distance d in the negative x direction. That is what is meant by -d≤x≤d

OK.
The E field is negaive but the magnitude of the E field, which is what you will need to compute the period, is of course positive.

 

[rude man]

In post #16, you've explained that in reference to the dipole.
"Because there is + charge above x=0 and - charge below x=0 so a test charge at x=0 will be repelled by the + charges and attracted by the - charges, giving a negative, non-zero E field at x=0."

+ charge above x and - charge below x : Aren't these the charges of the dipole you're talking about?

No, they are the charges per ρ = ρ⁰x/h. That is what acts on the dipole.

 

[sid0123]

No, they are the charges per ρ = ρ0x/h. That is what acts on the dipole.

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?

 

[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.