Capacitor, Dielectrics and height movement

[TFM]

Homework Statement

Two long coaxial cylindrical metal tubes (inner radius a, outer radius b) stand vertically in a tank of (dielectric) oil (suceptibility X-e, mass density rho). the inner one is maintained at potential V, the outer one is grounded. to what height (h) does the oil rise between the tubes.

Homework Equations

F = (1/2) V squared dc/dx

x = (v squared C)/(2F)

C = epsilon-r epsilon-0 A/d

Q = CV

The Attempt at a Solution

I'm not quite sure where to start - I've put a few equations that I believe may be relevant.

TFM

 

[alphysicist]

Hi TFM,

Homework Statement

Two long coaxial cylindrical metal tubes (inner radius a, outer radius b) stand vertically in a tank of (dielectric) oil (suceptibility X-e, mass density rho). the inner one is maintained at potential V, the outer one is grounded. to what height (h) does the oil rise between the tubes.

Homework Equations

F = (1/2) V squared dc/dx

x = (v squared C)/(2F)

I'm not sure if this equation would be correct here. Where did this come from?

C = epsilon-r epsilon-0 A/d

This would not be true. This is the formula for the capacitance of a parallel plate capacitor; you need the capacitance formula for a cylindrical capacitor.

Q = CV

The Attempt at a Solution

I'm not quite sure where to start - I've put a few equations that I believe may be relevant.

TFM

The oil will rise until it is in equlibrium. By setting the sum of the forces equal to zero, I think you should be able to get the height.

 

[TFM]

Hi TFM,

F = (1/2) V squared dc/dx

x = (v squared C)/(2F)

I'm not sure if this equation would be correct here. Where did this come from?

I rearranged

F = (1/2) V squared dc/dx

to give

Fdx = (1/2) V squared dc

which I presumed would give

F*x = (1/2) V squared C

and then rearranged?

TFM

 

[TFM]

This is the formula for the capacitance of a parallel plate capacitor; you need the capacitance formula for a cylindrical capacitor.

Is the formula for a cylindrical capacitor:

C/L = (2*pi*k*epsilon-0)/(ln(b/a))

$$\frac{C}{L} = \frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}$$

where L is the length of the capacitor?

TFM

 

[alphysicist]

I rearranged

F = (1/2) V squared dc/dx

to give

Fdx = (1/2) V squared dc

which I presumed would give

F*x = (1/2) V squared C

and then rearranged?

TFM

I don't believe this is valid. Intead, I think you need to leave it in the original form, find an expression for the capacitance, and once you take the deriviative you will have the force.

Is the formula for a cylindrical capacitor:

C/L = (2*pi*k*epsilon-0)/(ln(b/a))

$$\frac{C}{L} = \frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}$$

where L is the length of the capacitor?

TFM

The bottom equation looks fine to me:

$$\frac{C}{L} = \frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}$$

for calculating the capacitance where there is no oil.

The top one:

C/L = (2*pi*k*epsilon-0)/(ln(b/a))

is right and you'll need it to calculate the capacitance in the area where there is oil. However, you have k (the dielectric constant) in the equation, but in the problem statement they give the susceptibility $\chi_e$. How are these related? You probably want to rewrite this with the susceptibility instead of the dielectric constant?

 

[TFM]

So I need to use:

$$\frac{C}{L} = \frac{2*\pi*k*\epsilon_0}{ln\frac{b}{a}}$$

and

$$X_e = \epsilon_r - 1$$

I'm assuming k = $$\epsilon_r$$

Where

 

[alphysicist]

So I need to use:

$$\frac{C}{L} = \frac{2*\pi*k*\epsilon_0}{ln\frac{b}{a}}$$

and

$$X_e = \epsilon_r - 1$$

I'm assuming k = $$\epsilon_r$$

Yes, that's true in this case, and so the capacitance is:

$$\frac{C}{L} = (1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}}$$

 

[TFM]

So the Capacitance is:

$$C= L(1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}}$$


So do I now somehow insert it into:

$$F = \frac{1}{2}V^2 \frac{dc}{dx}$$

?

TFM

 

[alphysicist]

So the Capacitance is:

$$C= L(1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}}$$


So do I now somehow insert it into:

$$F = \frac{1}{2}V^2 \frac{dc}{dx}$$

?

TFM

You need to find the formula for the capacitance of this capacitor (for example, say it has length L, and the oil has risen to a height x), and then the derivative dC/dx is what goes in the force formula.

To find the capacitance, remember that oil is only in part of the capacitor, so you'll need to treat them separately and then combine them.

Once you find this force, this will be the electric force pulling upwards. What does this have to equal, for the oil to be at an equilibrium?

 

[TFM]

So the capacitance for the part above the oil is

$$C = l_1\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}$$

and the part for the capacitor in the oil is:


$$C= L_2(1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}}$$

TFM

 

[TFM]

So if you combine the two, do you get:

$$C = (l_1\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) + (L_2(1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}})$$

??

TFM

 

[alphysicist]

So if you combine the two, do you get:

$$C = (l_1\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) + (L_2(1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}})$$

??

TFM

That's the right idea; but remember that you will have to take the derivative of this with respect to the height (h in the problem statement, x in your formula), so you need to write L₁ and L₂ in terms of the height. (Hint: One of them is just equal to the height.)

 

[TFM]

So it should be:

$$C = (hx\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) + (h(1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}})$$

Since I assume that the part in oil is the height h, and the part out of oil is some unknwon part of h, which I have called x?

??

TFM

 

[alphysicist]

So it should be:

$$C = (hx\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) + (h(1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}})$$

Since I assume that the part in oil is the height h, and the part out of oil is some unknwon part of h, which I have called x?

??

TFM

Instead of using hx as the part out of oil, call then entire length of the capacitor something like L. Then if the length of the capacitor is L, and the length in oil is h, what is the length out of the oil?

 

[TFM]

So say the capacitor is length L,

Capcitor in Oil is h

Part of Capacitor out of water is L-h

so:

$$C = ((L-h)\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) + (h(1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}})$$

TFM

 

[alphysicist]

So say the capacitor is length L,

Capcitor in Oil is h

Part of Capacitor out of water is L-h

so:

$$C = ((L-h)\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) + (h(1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}})$$

TFM

That looks right to me. The derivative of that with respect to h is what appears in the force formula, so you can find the force.

Then this force has to be canceled by what other force at equilibrium? Find out what the value of this other force is for the oil at height h, and set the forces equal to each other, and you can find h.

 

[TFM]

Right then, so we have the formula:

$$F = \frac{1}{2} V squared dc/dx$$

and we have

$$C = ((L-h)\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) + (h(1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}})$$

I assume we put them together like so,

$$F = \frac{1}{2} V squared \frac{d\left(((L-h)\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) + (h(1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}})\right)}{dh}$$

and I need to find another force which will equal this?

TFM

 

[TFM]

Also, would the other force be mg?

TFM

 

[alphysicist]

Right then, so we have the formula:

$$F = \frac{1}{2} V squared dc/dx$$

and we have

$$C = ((L-h)\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) + (h(1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}})$$

I assume we put them together like so,

$$F = \frac{1}{2} V squared \frac{d\left(((L-h)\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) + (h(1+\chi_e) \frac{2*\pi*\epsilon_0}{ln\frac{b}{a}})\right)}{dh}$$

and I need to find another force which will equal this?

TFM

That's right (of course when you do the derivative it will be a lot cleaner).

Also, would the other force be mg?

TFM

Yes, but they don't give the mass, they give the density. So you'll need to rewrite m in terms of the density.

 

[TFM]

I am assuing that I can break up the derivative:

$$F = \frac{1}{2} V^2 (\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}(\frac{d(l-h)}{dh}))+(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})(\frac{d(h(1+\chi_e))}{dh}))$$

Does this look right?

Also, the mass of the oil will be:

$$\rho * \left(\pi*b^2*h - \pi*a^2*h\right)$$

TFM

 

[alphysicist]

I am assuing that I can break up the derivative:

$$F = \frac{1}{2} V^2 (\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}(\frac{d(l-h)}{dh}))+(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})(\frac{d(h(1+\chi_e))}{dh}))$$

Does this look right?

Also, the mass of the oil will be:

$$\rho * \left(\pi*b^2*h - \pi*a^2*h\right)$$

TFM

Yes, looks correct so far. Go ahead and perform the deriviatives, simplify, and see if you can get the final answer.

 

[TFM]

I may be missing something here, but, isn't:

$$\frac{dh}{dh} = 1$$

giving:

$$F = \frac{1}{2} V^2 (\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}l)+(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})((1+\chi_e))$$

Which would the makes:


$$\rho * \left(\pi*b^2*h - \pi*a^2*h\right) = \frac{1}{2} V^2 (\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}l)+(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})((1+\chi_e))$$

?

TFM

 

[alphysicist]

I may be missing something here, but, isn't:

$$\frac{dh}{dh} = 1$$

Yes.

giving:

$$F = \frac{1}{2} V^2 (\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}l)+(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})((1+\chi_e))$$

No, this is not quite right. What is:

$$(\frac{d( \ell -h)}{dh})=?$$

You seem to have it equal to $\ell$, which is not correct.

 

[TFM]

No, this is not quite right. What is:

$$(\frac{d( \ell -h)}{dh})=?$$

You seem to have it equal to $\ell$, which is not correct.

It should be l - 1

So:

$$F = \frac{1}{2} V^2 (\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}(l-1))+(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})((1+\chi_e))$$

TFM

 

[alphysicist]

It should be l - 1

So:

$$F = \frac{1}{2} V^2 (\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}(l-1))+(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})((1+\chi_e))$$

TFM

No, $\ell$ is a constant. So:

$$\frac{d}{dh}(\ell-h) = \frac{d\ell}{dh} - \frac{dh}{dh}=?$$

What's the derivative of a constant?

 

[TFM]

if you derive a constant you get 0

So should it be:

$$F = \frac{1}{2} V^2 (\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}(-1))+(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})((1+ \chi_e))$$

??

TFM

 

[Vuldoraq]

Thats right, although the V^2 should be multiplying everything in the brackets. Now try simplifying it down further (you have two equal and oppisite terms in your equation). Then equate and see what you get.

 

[TFM]

So

$$F = \frac{1}{2} V^2(-(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})+(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})((1+ \chi_e)))$$

Which cancels down to:

$$\frac{1}{2} V^2 (1 + \chi_e)$$

TFM

 

[Vuldoraq]

So

$$F = \frac{1}{2} V^2(-(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})+(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})((1+ \chi_e)))$$

Which cancels down to:

$$\frac{1}{2} V^2 (1 + \chi_e)$$

TFM

No! It might be easier if you multiply out your brackets for this part first;

$$(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})*(1+ \chi_e)$$

and then simplify down.

 

[TFM]

I though it seemed to small...

So:

$$(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})*(1+\chi_e)$$

$$(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})*(1+\chi_e)$$

Goes to:

$$\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})} + \chi_e(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})$$

so

$$F = \frac{1}{2}V^2(\chi_e(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) )$$

?

TFM

 

[alphysicist]

I though it seemed to small...

So:

$$(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})*(1+\chi_e)$$

$$(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})*(1+\chi_e)$$

Goes to:

$$\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})} + \chi_e(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})})$$

so

$$F = \frac{1}{2}V^2(\chi_e(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) )$$

?

TFM

Looks good to me. What do you get for the final value of h?

 

[TFM]

F also = mg, so

$$\rho * \left(\pi*b^2*h - \pi*a^2*h\right) = \frac{1}{2}V^2(\chi_e(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) )$$

take the h out the brackets:

$$\rho *h \left(\pi*b^2 - \pi*a^2\right) = \frac{1}{2}V^2(\chi_e(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) )$$

giving:

$$h = \frac{(\frac{1}{2}V^2(\chi_e(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) ))}{\rho*\left(\pi*b^2 - \pi*a^2\right)}$$

 

[alphysicist]

F also = mg, so

$$\rho * \left(\pi*b^2*h - \pi*a^2*h\right) = \frac{1}{2}V^2(\chi_e(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) )$$

take the h out the brackets:

$$\rho *h \left(\pi*b^2 - \pi*a^2\right) = \frac{1}{2}V^2(\chi_e(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) )$$

giving:

$$h = \frac{(\frac{1}{2}V^2(\chi_e(\frac{2*\pi*\epsilon_0}{ln(\frac{b}{a})}) ))}{\rho*\left(\pi*b^2 - \pi*a^2\right)}$$

You are missing the factor of g. (You set the force equal to the mass, but it needs to be set equal to the weight.)