Laplace Equation, potential around cylinder

[unscientific]

Homework Statement

Part(a): State condition for laplace's to work. Find potential in space between electrodes.

Part (b): Find potential inside cylinder and outside.

Part (c): How would the answer change is one pair is kept at potential 0 while other V₀?

Homework Equations

The Attempt at a Solution

Part (a)

$$V = ax^n + by^n + cz^n = 0$$
$$(\frac{\partial ^2}{\partial x^2} + \frac{\partial ^2}{\partial y^2} + \frac{\partial ^2}{\partial z^2} )V = 0$$

$$ax^{n-2} + by^{n-2} + cz^{n-2} = 0$$

V must be finite as r→∞.

$$C = 0$$ as V doesn't depend on z.
$$V_0 = b(-d)^n = b(d)^n$$ (therefore n = 2m)
$$b = \frac {V_0}{d^{2m}}$$
Similarly,
$$a = -\frac {V_0}{d^{2m}}$$

$$∇^2V = 0$$,
$$2m(2m-1)(y^{2m-2} - x^{2m-2}) = 0,$$
$$m = \frac {1}{2}$$

$$V = \frac {V_0}{d}(y - x)$$

Part(b)

I'm not sure why the question wants us to use r^-2cos(2ø)? What I did was to do it the usual way, separation of variables:

$$\frac{1}{r} \frac {\partial}{\partial r}(r\frac{\partial V}{\partial r} + \frac{1}{r^2}\frac{\partial ^2 V}{\partial ø^2} = 0$$
$$V = \sum_{k=1}^{\infty} [A_k e^{kr} + B_k e^{-kr}][C_k cos (kø) + D_k sin (kø)]$$

Boundary Conditions
As r→ ∞, V = 0 (As the two electrodes touch each other to cancel out)
So, A = 0.

$$V_{(d,0)} = V_{(d,\pi)} = -V_0$$
$$V_{(d,\frac{\pi}{2})} = V_{(d,-\frac{\pi}{2})} = V_0$$
$$ε_r \vec {E_1}^{\bot} = ε_0 \vec {E_2}^{\bot}$$
$$\vec {E_1}^{||} = \vec {E_2}^{||}$$

Despite applying these boundary conditions, they don't help me in solving for the coefficients at all...

 

[TSny]

For part (a), you want to find conditions on $a, b, c$ and $n$ such that $V = ax^n + by^n + cz^n =0$ [Edit: the = 0 should not be here] is a solution to Laplace's equation in general. For this part, I don't think you are meant to assume anything about the presence of electrodes or the behavior at infinity.

$$V = ax^n + by^n + cz^n = 0$$
$$(\frac{\partial ^2}{\partial x^2} + \frac{\partial ^2}{\partial y^2} + \frac{\partial ^2}{\partial z^2} )V = 0$$

$$ax^{n-2} + by^{n-2} + cz^{n-2} = 0$$

You have canceled out some important factors in the last equation. Put those back in and then consider various possibilities for $n$.

 

[TSny]

I'm not sure why the question wants us to use r^-2cos(2ø)?

Did you check to see if this function satisfies Laplace's equation?

 

[unscientific]

For part (a), you want to find conditions on $a, b, c$ and $n$ such that $V = ax^n + by^n + cz^n = 0$ is a solution to Laplace's equation in general. For this part, I don't think you are meant to assume anything about the presence of electrodes or the behavior at infinity.



You have canceled out some important factors in the last equation. Put those back in and then consider various possibilities for $n$.

Possibilities are n = 0 or n = 1.

 

[unscientific]

Did you check to see if this function satisfies Laplace's equation?

Yes it does satisfy, so do i suggest a solution:

$$V = \frac {A}{r^2} cos(2ø)$$?

Then Solving for A, we get potential outside cylinder:

$$V = -V_0 (\frac {d}{r})^2 cos (2\phi)$$

But this form doesn't work for potential inside the cylinder, as r->0, the potential goes to infinity..

I propose using:
$$V = -B r^2 cos (2\phi)$$ instead. And it satisfies laplace as well.

Applying continuity at surface,
$$V_{in} = V_{out}$$
$$Bd^2 = -V_0$$

So potential inside:
$$V_{in} = -V_0 (\frac{r}{d})^2 cos (2\phi)$$

 

[TSny]

Possibilities are n = 0 or n = 1.

Are there any restrictions on $a, b$ and $c$ for these two possibilities?

You have overlooked one other possibility for $n$.

 

[TSny]

OK, $V = \frac{A}{r^2} \cos2\phi$ satisfies Laplace's equation everywhere except at the origin $r = 0$.

You have also noted that $V = Br^2\cos2\phi$ satisfies Laplace's equation.

Before introducing the dielectric cylinder, you are asked to solve for $V$ with just the electrodes in place. So, you need to do that before going on to the problem with the cylinder. It will help if you find the other value of $n$ besides 0 and 1 that makes $ax^n+by^n+cz^n$ a solution of Laplace's equation (with maybe some restrictions on $a, b$ and $c$).

 

[unscientific]

OK, $V = \frac{A}{r^2} \cos2\phi$ satisfies Laplace's equation everywhere except at the origin $r = 0$.

You have also noted that $V = Br^2\cos2\phi$ satisfies Laplace's equation.

Before introducing the dielectric cylinder, you are asked to solve for $V$ with just the electrodes in place. So, you need to do that before going on to the problem with the cylinder. It will help if you find the other value of $n$ besides 0 and 1 that makes $ax^n+by^n+cz^n$ a solution of Laplace's equation (with maybe some restrictions on $a, b$ and $c$).

Before the cylinder was introduced, the potential is:

$$V = \frac{V_0}{d}(y-x)$$

The only values I have found are: n = 2m, m = 0 or 1/2 so this implies n = 0 or 1.

 

[TSny]

Before the cylinder was introduced, the potential is:

$$V = \frac{V_0}{d}(y-x)$$

Does this satisfy the boundary condition at the surface of the electrodes?

The only values I have found are: n = 2m, m = 0 or 1/2 so this implies n = 0 or 1.

Can you make n = 2 work if you add some conditions on $a, b$ and $c$?

 

[unscientific]

Does this satisfy the boundary condition at the surface of the electrodes?

Yup it satisfies the boundary condition, at (0,d) potential is $V_0$, at (d,0) the potential is $-V_0$ etc.

Can you make n = 2 work if you add some conditions on $a, b$ and $c$?

The equation of a circle touches tangentially at all 4 closest points of the electrodes, coincidentally.

I suppose $V = \frac{-V_0}{d^2}x^2 + \frac{V_0}{d^2}y^2$ satisfies the conditions. But I never got this answer when considering the conditions for laplace to work, by differentiating twice..

These means that either this or $V = \frac{V_0}{d}(y-x)$ satisfies the conditions!

I'm not sure which to choose..

 

[TSny]

Yup it satisfies the boundary condition, at (0,d) potential is $V_0$, at (d,0) the potential is $-V_0$ etc.

But does it satisfy the boundary condition at all points of the electrodes? The surfaces of the pair of electrodes where $V=V_0$ are points which satisfy $x^2-y^2 = d^2$. The surfaces of the pair of electrode where $V=-V_0$ are points which satisfy $x^2-y^2 = -d^2$. Although (0, d) and (d, 0) are isolated points of the electrodes, you need your solution to satisfy the boundary condition at all points where $x^2-y^2 = \pm d^2$

I suppose $V = \frac{-V_0}{d^2}x^2 + \frac{V_0}{d^2}y^2$ satisfies the conditions.

That's very close to a solution of Laplace's equation which will satisfy the electrode boundary conditions. I think there's a sign error in your expression. Check to see if the potential is $+V_0$ on the electrodes $x^2-y^2 = d^2$ and $-V_0$ on the electrodes $x^2-y^2 = -d^2$ .

But I never got this answer when considering the conditions for laplace to work, by differentiating twice..

OK. Go back to the general trial function $ax^n+by^n+cz^n$. What is the most general condition that $a, b$ and $c$ must satisfy in order for this function to satisfy Laplace's equation when $n=2$?

 

[unscientific]

But does it satisfy the boundary condition at all points of the electrodes? The surfaces of the pair of electrodes where $V=V_0$ are points which satisfy $x^2-y^2 = d^2$. The surfaces of the pair of electrode where $V=-V_0$ are points which satisfy $x^2-y^2 = -d^2$. Although (0, d) and (d, 0) are isolated points of the electrodes, you need your solution to satisfy the boundary condition at all points where $x^2-y^2 = \pm d^2$



That's very close to a solution of Laplace's equation which will satisfy the electrode boundary conditions. I think there's a sign error in your expression. Check to see if the potential is $+V_0$ on the electrodes $x^2-y^2 = d^2$ and $-V_0$ on the electrodes $x^2-y^2 = -d^2$ .



OK. Go back to the general trial function $ax^n+by^n+cz^n$. What is the most general condition that $a, b$ and $c$ must satisfy in order for this function to satisfy Laplace's equation when $n=2$?

At (0,d) and (0,-d) potential is $V_0$ and at (d,0) and (-d,0) potential is $-V_0$. So I think my expression for potential is right.

And the condition is a = -b.

I do not see where this question is going, and I do not see where the $r^{-2} cos 2 phi$ comes in ...This is terrible.

 

[TSny]

At (0,d) and (0,-d) potential is $V_0$ and at (d,0) and (-d,0) potential is $-V_0$. So I think my expression for potential is right.

Suppose you pick the point $x = \sqrt{3}d$, $y = \sqrt{2}d$, and $z = 5$. Does that point lie on one of the electrodes? If so, what is the potential at that point? Does your solution give the right potential at that point?

And the condition is a = -b.
I do not see where this question is going

That is not the most general condition on $a$, $b$ and $c$. See the first sentence in the image of the statement of the problem in your first post. That's what we are trying to answer right now. That question has nothing to do with the electrodes, it's just a general question about Laplace's equation. The electrodes and the cylinder are introduced after this question.

In your first post you had $ax^{n-2} + by^{n-2} + cz^{n-2} = 0$. What does this give you for $n=2$?

After answering this, we can then move on and see how it applies to the problem with the electrodes.

and I do not see where the $r^{-2} cos 2 phi$ comes in ...This is terrible.

You will see where this comes in after the cylinder is introduced later.

 

[unscientific]

Suppose you pick the point $x = \sqrt{3}d$, $y = \sqrt{2}d$, and $z = 5$. Does that point lie on one of the electrodes? If so, what is the potential at that point? Does your solution give the right potential at that point?



That is not the most general condition on $a$, $b$ and $c$. See the first sentence in the image of the statement of the problem in your first post. That's what we are trying to answer right now. That question has nothing to do with the electrodes, it's just a general question about Laplace's equation. The electrodes and the cylinder are introduced after this question.

In your first post you had $ax^{n-2} + by^{n-2} + cz^{n-2} = 0$. What does this give you for $n=2$?

After answering this, we can then move on and see how it applies to the problem with the electrodes.



You will see where this comes in after the cylinder is introduced later.

Okay, my circle equation does not include the points beyond the circle, so it's wrong. (Then why are we even considering n = 2 below?)

When n = 2, it means that a + b + c = 0

 

[TSny]

Right. So let's summarize the answer to the first question. $V = ax^n+by^n+cz^n$ is a solution to Laplace's equation for n = 1 and n= 2 without any restriction on a, b, c; while n = 2 yields a solution as long as a + b + c = 0.

Now we can move onto the next paragraph of the question and introduce the electrodes. You want to find the solution of Laplace's equation that satisfies the boundary conditions on the surfaces of the electrodes. The cross sections of the electrodes are hyperbolas $x^2-y^2 = \pm d^2$. Since the electrodes extend from -∞ to ∞ in the z direction, we should be able find a solution where V does not depend on z. To fit the electrodes, it might be worth trying to find a solution V which is also quadratic in x and y.

So, from the results of your previous post, how would you choose V such that it is quadratic in x and y, does not depend on z, satisfies Laplace's equation everywhere in the space between the electrodes, and also satisfies the boundary conditions on the electrodes?

 

[unscientific]

Right. So let's summarize the answer to the first question. $V = ax^n+by^n+cz^n$ is a solution to Laplace's equation for n = 1 and n= 2 without any restriction on a, b, c; while n = 2 yields a solution as long as a + b + c = 0.

Now we can move onto the next paragraph of the question and introduce the electrodes. You want to find the solution of Laplace's equation that satisfies the boundary conditions on the surfaces of the electrodes. The cross sections of the electrodes are hyperbolas $x^2-y^2 = \pm d^2$. Since the electrodes extend from -∞ to ∞ in the z direction, we should be able find a solution where V does not depend on z. To fit the electrodes, it might be worth trying to find a solution V which is also quadratic in x and y.

So, from the results of your previous post, how would you choose V such that it is quadratic in x and y, does not depend on z, satisfies Laplace's equation everywhere in the space between the electrodes, and also satisfies the boundary conditions on the electrodes?

$$V = \frac{V_0}{d^2}(y^2 - x^2)$$

It appears quite similar to the equation of the electrodes, I think that is where I got my hint from.

 

[TSny]

$$V = \frac{V_0}{d^2}(y^2 - x^2)$$

OK, up to an overall sign, that looks good.

So the idea was that we know $V=ax^2+by^2+cz^2$ satisfies Laplace's equation as long as $a+b+c=0$. Since we want a solution that doesn't depend on $z$, we choose $c=0$. That means we then need to choose $a$ and $b$ such that $a+b=0$. Hence, $b=-a$ and we now have $V=a(x^2-y^2)$ where the constant $a$ is yet to be chosen.

Now, we are very lucky that the shape of the electrodes are defined by $x^2-y^2 = \pm d^2$. Thus, by a suitable choice of $a$ you can make $V$ satisfy the boundary conditions on the electrodes. Just make sure to choose $a$ so that $V = +V_0$ on the electrodes that satisfy $x^2-y^2 = + d^2$ and then check that this choice will also satisfy the boundary condition $V = -V_0$ on the electrodes defined by $x^2-y^2 = - d^2$.

 

[unscientific]

OK, up to an overall sign, that looks good.

So the idea was that we know $V=ax^2+by^2+cz^2$ satisfies Laplace's equation as long as $a+b+c=0$. Since we want a solution that doesn't depend on $z$, we choose $c=0$. That means we then need to choose $a$ and $b$ such that $a+b=0$. Hence, $b=-a$ and we now have $V=a(x^2-y^2)$ where the constant $a$ is yet to be chosen.

Now, we are very lucky that the shape of the electrodes are defined by $x^2-y^2 = \pm d^2$. Thus, by a suitable choice of $a$ you can make $V$ satisfy the boundary conditions on the electrodes. Just make sure to choose $a$ so that $V = +V_0$ on the electrodes that satisfy $x^2-y^2 = + d^2$ and then check that this choice will also satisfy the boundary condition $V = -V_0$ on the electrodes defined by $x^2-y^2 = - d^2$.

Yes, it does.

Now for the second part, where does $r^{-2} cos (2\phi)$ come in?? I know it does satisfy laplace's equation.

Do I add both of them, by superposition principle? So general solution is:

$$V = \frac{A}{r^2}cos(2\phi) + V_0(x^2 - y^2)$$

But $r^{-2} cos (2\phi)$ doesn't work inside the cylinder, so:

$$V_{in} = V_0(x^2 - y^2)$$ (since polarization of cylinder = 0, potential inside is just as without the cylinder)
$$V_{out} = \frac{A}{r^2}cos(2\phi) + V_0(x^2 - y^2)$$

 

[unscientific]

bumpp

 

[TSny]

Yes, it does.

Now for the second part, where does $r^{-2} cos (2\phi)$ come in?? I know it does satisfy laplace's equation.

Do I add both of them, by superposition principle? So general solution is:

$$V = \frac{A}{r^2}cos(2\phi) + V_0(x^2 - y^2)$$

OK. But I think you left out $d^2$ in the second term. Note that adding the term $\frac{A}{r^2}cos(2\phi)$ prevents $V$ from satisfying the boundary conditions on the electrodes! But we are only looking for an approximate solution with the cylinder in place. So, if it turns out that $A$ is small, the first term will be small out at the electrodes and we will still be approximately satisfying the boundary conditions on the electrodes. And we will be satisfying Laplace's equation everywhere between the cylinder and the electrodes.

But $r^{-2} cos (2\phi)$ doesn't work inside the cylinder, so:

$$V_{in} = V_0(x^2 - y^2)$$ (since polarization of cylinder = 0, potential inside is just as without the cylinder)
$$V_{out} = \frac{A}{r^2}cos(2\phi) + V_0(x^2 - y^2)$$

Good. But there will be polarization of the cylinder that will produce surface charge on the cylinder. Nevertheless, Laplace's equation still holds within the cylinder. Your trial solution for inside looks good except you should express it more generally as $V_{in} = B(x^2 - y^2)$ where $B$ is a constant that will be determined from the boundary conditions on the surface of the cylinder.

To deal with the boundary conditions on the surface of the cylinder, I think it would be a good idea to express $x^2-y^2$ in terms of the cylindrical coordinates $r$ and $\phi$.

 

[unscientific]

OK. But I think you left out $d^2$ in the second term. Note that adding the term $\frac{A}{r^2}cos(2\phi)$ prevents $V$ from satisfying the boundary conditions on the electrodes! But we are only looking for an approximate solution with the cylinder in place. So, if it turns out that $A$ is small, the first term will be small out at the electrodes and we will still be approximately satisfying the boundary conditions on the electrodes. And we will be satisfying Laplace's equation everywhere between the cylinder and the electrodes.
Good. But there will be polarization of the cylinder that will produce surface charge on the cylinder. Nevertheless, Laplace's equation still holds within the cylinder. Your trial solution for inside looks good except you should express it more generally as $V_{in} = B(x^2 - y^2)$ where $B$ is a constant that will be determined from the boundary conditions on the surface of the cylinder.

To deal with the boundary conditions on the surface of the cylinder, I think it would be a good idea to express $x^2-y^2$ in terms of the cylindrical coordinates $r$ and $\phi$.

$$V_{in} = \frac{V_0}{d^2}(x^2-y^2) = \frac {2V_0r^2}{d^2}cos(2\phi)$$
$$V_{out} = \frac{V_0}{d^2}(x^2 - y^2) + \frac{A}{r^2}cos(2\phi) = (2\frac{V_0}{d^2} + \frac{A}{r^2})cos(2\phi)$$

Boundary Conditions
$V_{in} = V_{out}$ at r = R

Not satisfied unless A ≈ 0.

When r = d, $\phi = 0, \pi$ $V = V_0$. This means that $A = -V_0 d^2$

$$V_{out} = (2(\frac{r}{d})^2 - (\frac{d}{r})^2) V_0 cos(2\phi)$$

can be simplified to give:

$$V_{out} = V_0(\frac{r}{d})^2 cos (2\phi)$$

This is exactly half of the answer in the first part..

 

[TSny]

1.) Looks like you are saying $x^2-y^2 = 2r^2cos2\phi$. Check to make sure that's right.

2.) As I mentioned in my last post, for inside the cylinder you want to start with the general solution $V = B(x^2-y^2)$ and let the boundary conditions at the surface of the cylinder tell you the value of $B$. Also, the boundary conditions will give you $A$, which will not be zero.

You might need to review the boundary conditions on the electric field at the surface between two dielectrics. There's more to it than just $V_{in} = V_{out}$.

 

[unscientific]

1.) Looks like you are saying $x^2-y^2 = 2r^2cos2\phi$. Check to make sure that's right.

2.) As I mentioned in my last post, for inside the cylinder you want to start with the general solution $V = B(x^2-y^2)$ and let the boundary conditions at the surface of the cylinder tell you the value of $B$. Also, the boundary conditions will give you $A$, which will not be zero.

You might need to review the boundary conditions on the electric field at the surface between two dielectrics. There's more to it than just $V_{in} = V_{out}$.

Ok, so here are the potential inside and outside:$$V_{in} = B(x^2-y^2) = Br^2(cos^2(\phi) - sin^2(\phi)) = B'r^2cos(2\phi)$$

$$V_{out} = \frac{V_0}{d^2}(x^2-y^2) + \frac{A}{r^2}cos(2\phi) = (2\frac{V_0}{d^2} + \frac{A}{r^2})cos(2\phi)$$

Boundary Conditions
1. $$\vec {E}_{in}^{||} = \vec {E}_{out}^{||}$$

2. $$\epsilon _1 E^{\bot} = \epsilon _2 E^{\bot}$$Boundary condition 1 can be read as $\frac{\partial V_{in}}{\partial \theta} = \frac{\partial V_{out}}{\partial \theta}$ which is the same as $V_{in} = V_{out}$

Boundary condition 2 can be read as $\epsilon _0 \epsilon_r \frac{\partial V_{in}}{\partial r} = \epsilon_0 \frac{\partial V_{out}}{\partial r}$Putting Together, for boundary condition 1 we get:

$$B'R^2 = 2\frac{V_0}{d^2} + \frac{A}{R^2}$$

For boundary condition 2 we get:

$$2\epsilon_r R B' = -2\frac{A}{R^3}$$

Solving, we find
$$A = -2 V_0 (\frac{R}{d})^2 \frac{\epsilon_r}{\epsilon_r + 1}$$
$$B = -2 V_0 (\frac{1}{Rd})^2 \frac{1}{1 + \epsilon_r}$$

Substituting, we obtain the final expression for $V_{in}$ and $V_{out}$:
$$V_{in} = -2 \frac{V_0}{d^2} (\frac{r}{R})^2 \frac{1}{\epsilon_r + 1} cos (2\phi)$$
$$V_{out} = 2\frac{V_0}{d^2}\left [ 1 - (\frac{R}{r})^2(\frac{\epsilon_r}{\epsilon_r + 1})\right ] cos (2\phi)$$

 

[TSny]

Ok, so here are the potential inside and outside:


$$V_{in} = B(x^2-y^2) = Br^2(cos^2(\phi) - sin^2(\phi)) = B'r^2cos(2\phi)$$

$$V_{out} = \frac{V_0}{d^2}(x^2-y^2) + \frac{A}{r^2}cos(2\phi) = (2\frac{V_0}{d^2} + \frac{A}{r^2})cos(2\phi)$$

There's a mistake in the far right hand side of the second equation. The 2 should be replaced by something else. What does $x^2-y^2$ reduce to in cylindrical coordinates $r, \phi$?

Boundary Conditions
1. $$\vec {E}_{in}^{||} = \vec {E}_{out}^{||}$$
2. $$\epsilon _1 E^{\bot} = \epsilon _2 E^{\bot}$$
Boundary condition 1 can be read as $\frac{\partial V_{in}}{\partial \theta} = \frac{\partial V_{out}}{\partial \theta}$ which is the same as $V_{in} = V_{out}$

Boundary condition 2 can be read as $\epsilon _0 \epsilon_r \frac{\partial V_{in}}{\partial r} = \epsilon_0 \frac{\partial V_{out}}{\partial r}$

Yes, the boundary conditions look good. Your method of applying the boundary conditions also looks good, but because of the mistake pointed out above, your final result is not yet correct.

As a partial check on your answer, note that for the special case of $\epsilon_r = 1$, the cylinder acts like the vacuum. So, if you substitute $\epsilon_r = 1$ into the result for $V$, it should reduce to the same result as for no cylinder.

 

[unscientific]

There's a mistake in the far right hand side of the second equation. The 2 should be replaced by something else. What does $x^2-y^2$ reduce to in cylindrical coordinates $r, \phi$?



Yes, the boundary conditions look good. Your method of applying the boundary conditions also looks good, but because of the mistake pointed out above, your final result is not yet correct.

As a partial check on your answer, note that for the special case of $\epsilon_r = 1$, the cylinder acts like the vacuum. So, if you substitute $\epsilon_r = 1$ into the result for $V$, it should reduce to the same result as for no cylinder.

Yup, there should be an additional factor of 'R²'. I think I got this now, thanks alot!