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APhO 2013, Теория — Задача 1

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Problem · Conductors in Conducting Liquid

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14TH ASIAN PHYSICS OLYMPIAD
BOGOR, INDONESIA
Theory
Question 1
page 1 of 1

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14TH ASIAN PHYSICS OLYMPIAD
BOGOR, INDONESIA
Theoretical 1: Solution
Conductors in Conducting Liquid

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14TH ASIAN PHYSICS OLYMPIAD
BOGOR, INDONESIA
Theoretical 1: Solution
Conductors in Conducting Liquid

Бележка от източника

14TH ASIAN PHYSICS OLYMPIAD
BOGOR, INDONESIA
Theoretical 1: Solution
Conductors in Conducting Liquid

Бележка от източника

14TH ASIAN PHYSICS OLYMPIAD
BOGOR, INDONESIA
Theoretical 1: Solution
Conductors in Conducting Liquid

Бележка от източника

14TH ASIAN PHYSICS OLYMPIAD
BOGOR, INDONESIA
Theoretical 1: Solution
Conductors in Conducting Liquid

Бележка от източника

14TH ASIAN PHYSICS OLYMPIAD
BOGOR, INDONESIA
Theoretical 1: Solution
Conductors in Conducting Liquid

Условие

A system consisted of two conductor bodies is immersed in a uniform dielectric and weakly conducting liquid. When a constant voltage difference is applied between both conductors, the system has both electric and magnetic fields. In this problem we will investigate this system.

(0.4 pts) First consider an infinitely long line with charge per unit length $\lambda$ in vacuum. Calculate the electric field $\mathbf{E}(\mathbf{r})$ due to the line.
(0.4 pts) The potential due to the line charge could be written as

V(r) = f(r) + K,

where $K$ is a constant. Determine $f(r)$ .

(0.7 pts) Calculate the potential in all space $V(x,y,z)$ due to an infinitely long line with charge per unit length $\lambda$ at $x = -b, y = 0$ and another infinitely long line with charge per unit length $-\lambda$ at $x = b, y = 0$ . Both lines are parallel to the $z$ -axis. Take $V = 0$ at the origin. Sketch the equipotential surfaces.

For the following questions, ignore any edge effects.

(2.0 pts) Now consider two identical conducting cylinders, both with radius $R = 3a$ in vacuum. The length of each cylinders are the same and much larger than its radius ( $l \gg R$ ). The axis of both cylinders are on the $xz$ -plane and parallel to the $z$ -axis, one at $x = -5a$ , $y = 0$ and the other at $x = 5a$ , $y = 0$ . An electrical potential difference of $V_0$ is applied between the two cylinders (the cylinder at $x = -5a$ has the higher potential) by connecting them to a battery. Calculate the potential in all regions. Take $V = 0$ at the origin.
(0.5 pts) Calculate the capacitance $C$ of the system.
(1.0 pts) Now both cylinders are totally immersed in a weakly conducting liquid with conductivity $\sigma$ . Calculate the total current that flows between both cylinders. Assume the permittivity of the liquid is equal to that of vacuum, $\epsilon = \epsilon_0$ .
(0.5 pts) Calculate the resistance $R$ of the system. Calculate $RC$ of the system.
(1.5 pts) Calculate the magnetic field due to the current in question 6. Assume that the permeability of the liquid is equal to that of vacuum $\mu = \mu_0$ .

Решение

Using Gauss law

\oint \mathbf{E} \cdot d\mathbf{A} = \frac{q}{\epsilon_0}. \quad (1)

From symetry we know that the electric field only has radial component. Choose a cylinder (with a line charge as the axis) as the Gaussian surface, we obtain

E.2\pi r l = \frac{\lambda l}{\epsilon_0}.

Simplify to obtain

\mathbf{E} = \hat{r}\frac{\lambda}{2\pi\epsilon_0 r}. \quad (2)

The potential is given by

V = -\int_{ref}^{r} \mathbf{E} \cdot d\mathbf{l} = -\int_{ref}^{r} E.dr

V = -\frac{\lambda}{2\pi\epsilon_0}\ln r + K, \quad (3)

so $f(r) = -\frac{\lambda}{2\pi\epsilon_0}\ln r$ . where $K$ is a constant.

The potential from both line charges is a superposition of both potential

Diagram showing two line charges lambda at x=-b and -lambda at x=b on the x-axis, with vectors r1 and r2 drawn to a point on the y-axis, illustrating the geometry used to compute the potential.

Figure 1: System with two line charges

V = -\frac{\lambda}{2\pi\epsilon_0}\ln r_1 + \frac{\lambda}{2\pi\epsilon_0}\ln r_2 \quad (4)

= \frac{\lambda}{2\pi\epsilon_0}\ln\frac{\sqrt{(b-x)^2+y^2}}{\sqrt{(b+x)^2+y^2}}

V = \frac{\lambda}{4\pi\epsilon_0}\ln\frac{(b-x)^2+y^2}{(b+x)^2+y^2}. \quad (5)

We can rearrange eq.(5) in to:

\left(x - \left(\frac{1+\beta}{1-\beta}\right)\right)^2 + y^2 = b^2\left(\left(\frac{1+\beta}{1-\beta}\right)^2 - 1\right) \quad (6)

where $\beta = \exp\left(\frac{4\pi\epsilon_0 V}{\lambda}\right)$ . For an arbitrary potential $V$ , Eq. (6) is an equation of circle.

Two separate circular equipotential curves plotted for two values of beta, illustrating the cylindrical equipotential surfaces of the two line charges.

Figure 2: The equipotential surfaces with $b = 1$ , for $\beta = 12.35$ (left) and $\beta = \frac{1}{12.35}$ (right)

From eq.(5) and eq.(6), we see that for any arbitrary potential $V$ , the equipotential surfaces of these two equal but opposite lines charge, are cylindrical surfaces. From this observation, we can choose the specific position for each line charge in both cylinders so that the surface of each cylinder is an equipotential surface.

Consider the following figure

Diagram of two line charges lambda and -lambda with distances l1 and l2 to the center of a circle of radius R, and angle phi and distances r1, r2 to a point on the circle, used to derive the image-line-charge positions for the conducting cylinders.

Figure 3: Two line charges with its equipotential surfaces

We would like to find a cylindrical equipotential surface enclose one line charge, let say the $-\lambda$ (if we could find the surface, by symmetry, we surely can find the identical one that enclose the line $\lambda$ ). The potential is given by

V = -\frac{\lambda}{2\pi\epsilon_0}\ln r_1 + \frac{\lambda}{2\pi\epsilon_0}\ln r_2

= -\frac{\lambda}{4\pi\epsilon_0}\ln(l_1^2 + R^2 - 2l_1 R\cos\phi) + \frac{\lambda}{4\pi\epsilon_0}\ln(l_2^2 + R^2 - 2l_2 R\cos\phi). \quad (7)

Since the surface of the cylinder has to be the equipotential surface, so the potential should not depend on $\phi$ , i.e. $\frac{\partial V}{\partial \phi} = 0$ .

-\frac{\lambda}{4\pi\epsilon_0}\frac{2l_1 R\sin\phi}{l_1^2+R^2-2l_1 R\cos\phi} + \frac{\lambda}{4\pi\epsilon_0}\frac{2l_2 R\sin\phi}{l_2^2+R^2-2l_2 R\cos\phi} = 0 \quad (8)

\frac{l_1}{l_1^2+R^2-2l_1 R\cos\phi} = \frac{l_2}{l_2^2+R^2-2l_2 R\cos\phi}

l_1^2 l_2 + R^2 l_2 - 2l_1 l_2 R\cos\phi = l_1 l_2^2 + R^2 l_1 - 2l_1 l_2 R\cos\phi

l_1 l_2 (l_1 - l_2) = R^2 (l_1 - l_2)

l_1 l_2 = R^2. \quad (9)

From the data in the problem, we have

l_1 + l_2 = 10a, \quad (10)

l_1 l_2 = 9a^2. \quad (11)

Solve this quadratic equation to get

l_1 = 5a \pm 4a. \quad (12)

However, since $l_1 > l_2$ , we have

l_1 = 9a, \quad (13)

l_2 = a. \quad (14)

Using this results on eq.(5), we have

V = \frac{\lambda}{4\pi\epsilon_0}\ln\frac{(4a-x)^2+y^2}{(4a+x)^2+y^2}. \quad (15)

This is the potential in all region except inside both cylinders. For cylinders at $x = -5a$ , the potential is constant and equal to

V(x=-2a,y=0) = \frac{\lambda}{4\pi\epsilon_0}\ln\frac{(4a+2a)^2+0^2}{(4a-2a)^2+0^2} = \frac{\lambda}{2\pi\epsilon_0}\ln 3. \quad (16)

For cylinders at $x = 5a$ , the potential is constant and equal to

V(x=2a,y=0) = \frac{\lambda}{4\pi\epsilon_0}\ln\frac{(4a-2a)^2+0^2}{(4a+2a)^2+0^2} = -\frac{\lambda}{2\pi\epsilon_0}\ln 3. \quad (17)

The potential difference between both cylinders are

\Delta V = \frac{\lambda}{\pi\epsilon_0}\ln 3 \equiv V_0. \quad (18)

Substituting this results in the potential equation, the potential outside the two cylinders are:

V = \frac{V_0}{4\ln 3}\ln\frac{(4a-x)^2+y^2}{(4a+x)^2+y^2}. \quad (19)

And the potential inside the cylinders are:
The potential inside the cylinder centered at $(x=5a, y=0)$ is $V = -V_0/2$ .
The potential inside the cylinder centered at $(x=-5a, y=0)$ is $V = V_0/2$ .

From eq.(18), we have

V_0 = \frac{q}{l\pi\epsilon_0}\ln 3, \quad (20)

so we get

C = \frac{q}{V_0} = \frac{l\pi\epsilon_0}{\ln 3} \quad (21)

The electric field produces by both cylinders are

E_x = \frac{V_0}{2\ln 3}\left(\frac{4a+x}{(4a+x)^2+y^2} + \frac{4a-x}{(4a-x)^2+y^2}\right). \quad (22)

E_y = \frac{V_0}{2\ln 3}\left(\frac{y}{(4a+x)^2+y^2} - \frac{y}{(4a-x)^2+y^2}\right). \quad (23)

The volume current density is given by

\mathbf{J} = \sigma\mathbf{E} \quad (24)

To calculate the total current, we may choose to calculate the current that flow through the $x = 0$ plane. On this plane, there is no current in the $y$ direction. The total current is given by

I = \int \mathbf{J} \cdot d\mathbf{A} \quad (25)

= \int \sigma E_x l\,dy

= \sigma l \frac{8aV_0}{2\ln 3}\int_{\infty}^{\infty} \frac{dy}{(4a)^2+y^2}

I = \frac{V_0 \pi \sigma l}{\ln 3} \quad (26)

The resistance is given by

R = \frac{V_0}{I} = \frac{\ln 3}{\pi \sigma l} \quad (27)

and therefore

RC = \frac{\epsilon_0}{\sigma} \quad (28)

Since the system has a high symmetry, we may use Ampere's law. The magnetic field should not have any $z$ dependence, since the current has no $z$ dependence.

Figure 4 shows the current density $\mathbf{J}$ flow from one cylinder to the other cylinder. Choose an Ampere loop on a constant $x$ plane in a symmetrical way, so that the first path is pointing in the positive $z$ direction with constant $y$ coordinate, the second path is pointing to the negative $y$ direction with constant $z$ coordinate. The third path is pointing to the negative $z$ direction, but with constant $-y$ coordinate. The fourth path is pointing in the positive $y$ direction with constant $-z$ coordinate.

Having this path, we need to calculate the current that flow through the loop

I = \int \mathbf{J} \cdot d\mathbf{A}

= \int J_x l\,dy

= \frac{V_0 \sigma l}{2\ln 3}\int_{-y}^{y}\left(\frac{4a+x}{(4a+x)^2+y^2} + \frac{4a-x}{(4a-x)^2+y^2}\right)dy

Three-dimensional sketch of the two cylinders immersed in the conducting liquid with current density J flowing between them, and a rectangular Ampere loop (paths 1,2,3,4) on a constant-x plane used to compute the magnetic field.

Figure 4: The Ampere loop

I = \frac{V_0 \sigma l}{\ln 3}\left(\arctan\frac{y}{4a+x} + \arctan\frac{y}{4a-x}\right) \quad (29)

Using the Ampere's law

\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I \quad (30)

2B_z l = \frac{\mu_0 V_0 \sigma l}{\ln 3}\left(\arctan\frac{y}{4a+x} + \arctan\frac{y}{4a-x}\right)

B_z = \mu_0 \frac{V_0 \sigma}{2\ln 3}\left(\arctan\frac{y}{4a+x} + \arctan\frac{y}{4a-x}\right) \quad (31)

therefore

\mathbf{B} = \hat{z}\frac{\mu_0 V_0 \sigma}{2\ln 3}\left(\arctan\frac{y}{4a+x} + \arctan\frac{y}{4a-x}\right) \quad (32)

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Notes $\int \frac{\alpha\,dx}{\alpha^2+x^2} = \arctan\frac{x}{a} + const$

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Conductors in Conducting Liquid Page 1 of 6

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Теми и трудност

Трудност: неоценена.

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