D

A

D

The charge distribution will have cylindrical symmetry since the system has cylindrical symmetry. Since the electrical field must be zero inside the inner cylinder, the charges must be on the surface of the inner cylinder. They must be uniformly distributed due to the cylindrical symmetry. Inside the outer cylinder, the electric field must be zero. This can be achieved by having all the \( -Q \) charge uniformly distributed on the inner surface so that a cylindrical Gauss surface inside the outer cylinder would have zero net charge inside it.

Inside the inner cylinder, the electric field is zero, and the potential constant. In the region between the cylinders, the electric field can be found from Gauss law using a cylindrical Gauss surface. In the outer cylinder shell the electric field is zero, so potential is uniform. Outside the outer cylinder, we see that inside a cylindrical Gauss surface the net charge is zero, so that the electric field is zero. The potential is therefore constant from the inside of the outer cylinder and to infinity.

In this case, we can use the same argument as above. The charge \( +Q \) must be uniformly distributed on the outside of the inner cylinder. And there must be a charge \( -Q \) uniformly distributed on the inside of the outer cylinder. However, for the out cylinder to be neutral, a charge \( +Q \) must also be present. It can only be on the outside of the outer cylinder, and it must be uniformly distributed due to the cylindrical symmetry.

This problem is much more tricky. We can no longer assume cylindrical symmetry. However, we still know that the charges can only be on the surfaces and that they must be organized so that there is no electric field inside the condutors. This means that the electric field is zero inside the outer cylinder and the outer surface of the outer cylinder is an equipotential surface. This means that the charges on the surface of the inner cylinder will be non-uniform, probably with a higher charge density where it is closest to the outer cylinder, because the electric field will be the largest here since the distance is the smallest. (And the electric field depends on the difference in potential divided by the distance). Similarly, the charge density on the inner part of the outer cylinder will be largest where it is closest to the inner cylinder. Inside the outer cylinder the electric field is zero. Because the outer cylinder is neutral, there must be a corresponding positive charge on the outside of the conductor. However, this distribution must be uniform, otherwise it would induce an electric field inside the cylinder.

We can calculate the surface charge density using a short python program. The resulting distribution is shown in the figure.

This distribution was calculated by the following Python program (which you will need to find out how works for yourself. Notice how we have calculated the surface charge density as \( \vec{E} \cdot \rhat \), where \( \rhat \) is pointing outwards from the inner cylinder for the surface of the inner cylinder and inwards from outer cylinder for the inner surface of the outer cylinder.) This program only calculates the charges on the inner cylinder and on the inner part of the outer cylinder. We could use it to calculate the surface charge density on the outer part of the outer cylinder, but then we would need to make the boundaries far away (where the potential is zero) and to set the potential of the outer cylinder to a non-zero value. However, we know how the potential and the electric field will look like outside the outer cylinder: Similar to that of a cylinder with a net charge \( +Q \) on the outer surface, so we do not need to make this calculation.

# Definere randverdier
L = 200
b = np.zeros((L,L))
b[:] = np.float('nan')
# Make inner circle
a = 0.1*L
xca = L*0.5
yca = L*0.7
for ix in range(L):
    for iy in range(L):
        dx = ix - xca
        dy = iy - yca
        rr = np.sqrt(dx*dx+dy*dy)
        if (rr<a):
            b[ix,iy] = 1.0
# Make outer circle
rb = 0.4*L
rc = 0.45*L
xc = L/2
yc = L/2
for ix in range(L):
    for iy in range(L):
        dx = ix - xc
        dy = iy - yc
        rr = np.sqrt(dx*dx+dy*dy)
        if (rr>=rb and rr<=rc):
            b[ix,iy] = 0.0
V = solvepoissonvonneumann(b,10000)
Ey,Ex = np.gradient(-V)
# Visualize the surface charge distribution
rhos = Ex.copy()
rhos[:] = 0.0
# Inner cylinder
dr = 1.0
for ix in range(L):
    for iy in range(L):
        dx = ix - xca
        dy = iy - yca
        rr = np.sqrt(dx*dx+dy*dy)
        if (rr>=a and rr<=a+dr):
            ux = dx/rr
            uy = dy/rr
            rhos[ix,iy] = Ey[ix,iy]*ux + Ex[ix,iy]*uy
# Outer cylinder
for ix in range(L):
    for iy in range(L):
        dx = ix - xc
        dy = iy - yc
        rr = np.sqrt(dx*dx+dy*dy)
        if (rr>=rb and rr<=rb+dr):
            ux = dx/rr
            uy = dy/rr
            rhos[ix,iy] = -(Ey[ix,iy]*ux + Ex[ix,iy]*uy)
plt.figure(figsize=(8,8))
plt.imshow(rhos)
plt.colorbar()

Now, we are back to the symmetric case again, and we can use simpler arguments. We realize that all the charge \( +Q \) can be distributed uniformly on the outer surface of the outer cylinder. The electric field inside both cylinders (and in the region between them) will then be zero.

0

Because the electric field inside both conductors and inside the region between the conductors are zero, the potential is constant in this region and the potential difference \( \Delta V = 0 \).

No

The surface charge density on the inner surface is \( Q/(2 \pi a L) \) and on the outer surface \( -Q/(2 \pi b L) \).

Choose a Gaussian surface and use Gauss' law.

\( \vec{E} = (Q/L)/(2 \pi \epsilon r) \rhat \) for \( a < r < b \) and zero otherwise.

We notice that we expect the electric field to have cylindrical symmetry and that it only will have a radial component. We then find the electric field by using Gauss' law on a cylindrical surface with radius \( r \), getting $$ \begin{equation} \oint_S \vec{E} \cdot d \vec{S} = E A = E 2 \pi r L = Q_{in}/\epsilon\; , \tag{6.1} \end{equation} $$ where \( r \) is the radius, \( L \) is the length of the cylinder, $ Q_{in} $ is the net, free charge inside the cylinder, and \( \epsilon \) is the dielectric constant for the material inside the cylinder.

For \( r < a \) $Q_{in} = 0$ and \( E = 0 \).

For \( a < r < b \), \( Q_{in} = Q \) and \( E = (Q/L)/(2 \pi \epsilon r) \).

For \( b < r \), \( Q_{in} = 0 \) and \( E = 0 \).

\( V(a) = (Q/L)/(2 \pi \epsilon) \ln (b/a) \)

We find the potential \( V(r) \) by integrating the line intergral from \( r \) to \( r=b \) where \( V=0 \): $$ \begin{equation} V(r) - V(b) = \int_{r}^{b} \vec{E} \cdot \d \vec{l} = \int_{r}^{b} \frac{(Q/L)}{2 \pi \epsilon r}\d r = \frac{(Q/L)}{2 \pi \epsilon}\left( \ln b - \ln r \right) = \frac{(Q/L)}{2 \pi \epsilon} \ln\left(\frac{b}{r}\right) \; . \tag{6.2} \end{equation} $$ and $$ \begin{equation} V(a) = \frac{(Q/L)}{2 \pi \epsilon} \ln\left(\frac{b}{a}\right) \; . \tag{6.3} \end{equation} $$

\( C = (2 \pi \epsilon L)/(\ln(b/a)) \)

The capacitance is given as \( C=Q/V \), which gives $$ \begin{equation} C = \frac{Q}{V} = \frac{2 \pi \epsilon L}{\ln(b/a)} \tag{6.4} \end{equation} $$

\( Q_1/(2 A_1) \)

The total charge of \( Q_1 \) will be distributed uniformly on both sides of the plate, that is, over an area \( 2A_1 \). The charge density is therefore \( \rho_a = Q_1/(2A_1) \).

We can find the contribution from one surface charge density by applying Gauss' law on a small cylinder of area \( dA \). We place one end outside the surface and one end inside the conductor. We can make the sides infinitely small. Thus, the only contribution is from the surfaces and not from sides. The flux through the top and bottom surfaces are \( EdA + EdA = 2EdA = Q/\epsilon = dA \rho_a / \epsilon \), and thus \( E = \rho_a/(2\epsilon) \) is a constant. However, the contribution from each surface will cancel, making the field zero inside the conductor.

\( E = Q_1/(A_1\epsilon) \)

We use the result from above. The field will be the field from the charge density on each side, giving a total field of \( E = 2 \rho_a/(2\epsilon) = Q_1/(A_1 \epsilon) \).

\( \rho_a'=Q_1/A_1 \), \( E = Q_1/(A_1 \epsilon) \).

The charge density is \( \rho_a' = Q_1/A_1 \). The electric field will be \( E = \rho_a'/\epsilon = Q_1/(A_1 \epsilon) \).

\( \vec{E} = V_0/(r \ln (b/a)) \rhat \), \( C' = 2 \pi \epsilon/\ln (b/a) \)

We can choose to use either method 1 or method 2. Here, since we have given the potential, we attempt to use method 2 where we first find the potential and then find the electric field. Laplace equation in the region between the two conductors in cylindrical coordinates is $$ \begin{equation} \frac{1}{r}\frac{\partial}{\partial r} \left( r \frac{\partial V}{\partial r} \right) = 0 \; , \tag{6.5} \end{equation} $$ where we only have included the radial component due to the cylindrical symmetry of the system. This means that $$ \begin{equation} r \frac{\partial V}{\partial r} = c_1 \; , \tag{6.6} \end{equation} $$ where \( c_1 \) is a constant. This gives us: $$ \begin{equation} \frac{\partial V}{\partial r} = \frac{c_1}{r} \; , \tag{6.7} \end{equation} $$ and therefore \( V = c_1 \ln r + c_2 \). We know that \( V(b) = 0 \), which means that \( V = c_1 \ln (r/b) \) and \( V(a) = c_1 \ln (a/b) = V_0 \), and \( c_1 = V_0/\ln (a/b) \). Therefore $$ \begin{equation} V(r) = V_0 \ln (r/b)/\ln (a/b) \; . \tag{6.8} \end{equation} $$ We now need to find the electric field, which is $$ \begin{equation} \vec{E} = -\nabla V = \frac{\partial V}{\partial r} \rhat = -\frac{V_0}{r \ln (a/b)} \rhat \; , \tag{6.9} \end{equation} $$ which points outward since \( \ln(a/b) \) is negative. And we find the surface charge density on the inner cylinder to be $$ \begin{equation} \rho_s = \epsilon \vec{E} \cdot \rhat = -\frac{V_0}{a \ln (a/b)} \; . \tag{6.10} \end{equation} $$ The total charge on a cylinder of radius \( a \) and length \( L \) is \( Q = \rho_s A \), where \( A = 2 \pi a L \), and therefore: $$ \begin{equation} Q = A \rho_s = - \frac{V_0 \epsilon}{a \ln (a/b)} 2 \pi a L = - \frac{2 \pi V_0 L \epsilon}{ \ln (a/b)} \; . \tag{6.11} \end{equation} $$ We change the sign by realizing that \( \ln(a/b) = - \ln(b/a) \) and that \( \ln(b/a) \) is positive. The capacitance of a length \( L \) of the cylinder is then: $$ \begin{equation} C = Q/V_0 = \frac{2 \pi L \epsilon}{\ln(b/a)} \;. \tag{6.12} \end{equation} $$ and the capacitance per unit length, \( C' \), is $$ \begin{equation} C' = \frac{2 \pi \epsilon}{\ln (b/a)} \; . \tag{6.13} \end{equation} $$ (You could try to solve the same problem using method 1).

\( C' = 8.58 \times 10^{-11}\text{F/m} \)

We find that \( C' = 2 \pi 3 \epsilon_0/ \ln (b/a) = 2 \pi 3 \, 8.85 \times 10^{−12}/\ln(7) = 8.58 \times 10^{-11}\text{F/m} \).

\( U' = \frac{\pi \epsilon}{\ln \frac{b}{a}}V_0^2 \)

We apply the formula for the energy stored in a capacitor: $$ \begin{equation} U = \frac{1}{2} CV^2 \tag{6.14} \end{equation} $$ This means that the energy density (per unit length) must be: $$ \begin{eqnarray} U' &=& \frac{1}{2} C'V^2 \\ &=& \frac{\pi \epsilon}{\ln \frac{b}{a}}V_0^2 \end{eqnarray} $$

Zero.

Due to symmetry, the total force must be zero.

Let us address this a bit closer. There cannot be any force along the cable, because the system has mirror symmetric along the \( z \)-axis: \( z \rightarrow -z \). Only a zero force has this symmetry. There cannot be any force in an azimuthal direction because the cable has rotational symmetry around the cable. There might by a force acting radially, but because the system has rotational symmetry, there would always be another equal force on the opposite site of the cable. The net radial force must therefore also be zero.

Speilladningsmetoden

La oss si at det ledende planet ligger i \( xy \)-planet.

Potensialet over \( xy \)-planet må være likt som potensialet ville vært dersom vi i stedet for å ha et ledende plan, speilet ladningsfordelingen om \( xy \)-planet og satte motsatt ladning på ladningene på nedsiden.

Det skyldes at når vi gjør dette får vi konstant potensial akkurat i \( xy \)-planet. Det er flere måter vi kan se det på. En av dem er å innse at alle komponenter av det elektriske feltet i planet må kansellere mellom ladningene over og ladningene under \( xy \)-planet: For enhver ladning på oversiden finnes det en tilsvarende plassert ladning med motsatt fortegn på undersiden, og denne kansellerer komponenten i planet. Men den adderer til feltet normalt på planet, altså i \( z \)-retning.

Siden entydighetsteoremet sier oss at det kun skal være én løsning for det elektrostatiske potensialet gitt randbetingelser og ladningsfordeling, og vi kan sette randbetingelsen til \( V= \) konstant i \( xy \)-planet uansett om det er pga. et ledende plan eller en motsatt ladning speilet, kan det ikke ha noe å si om vi ser på situasjonen med et ladet plan, eller situasjonen med en speilet ladning. Så lenge vi ser på feltet over \( xy \)-planet.

Dermed ser vi egentlig bare på potensialet fra en dipol, langs dipolaksen: $$ \begin{equation} V = \frac{Q}{4\pi\epsilon_0 r_1} - \frac{Q}{4\pi\epsilon_0 r_2} \tag{6.15} \end{equation} $$

Vi skulle finne potensialet i området \( [0, h) \), og der får vi $$ \begin{equation} V = \frac{Q}{4\pi\epsilon_0} \left( \frac{1}{h-z} - \frac{1}{h+z} \right). \tag{6.16} \end{equation} $$

Vi finner så det elektriske feltet. Av symmetri ser vi at feltet må virke kun i \( z \)-retning, slik at vi kun ser på \( z \)-komponenten: $$ \begin{eqnarray} E_z &=& -\pz \frac{Q}{4\pi\epsilon_0} \left( \frac{1}{h-z} - \frac{1}{h+z} \right) \\ &=& -\frac{Q}{4\pi\epsilon_0} \left( \frac{1}{(h-z)^2} + \frac{1}{(h+z)^2} \right) \\ \mathbf{E} &=& -\frac{Q}{4\pi\epsilon_0} \left( \frac{1}{(h-z)^2} + \frac{1}{(h+z)^2} \right) \z \text{ for } 0 \leq z < h \end{eqnarray} $$

Bruk \( \rho_s = \epsilon_0 E \), der \( E \) evalueres rett ovenfor lederen.

\( \rho_s = -\frac{Qh}{2\pi(r^2+h^2)^\frac{3}{2}} \)

Vi kjenner til at feltet fra en dipol normalt på dipolens akse er $$ \begin{equation} E = -\frac{Q h }{2\pi \epsilon_0 (r^2+h^2)^\frac{3}{2}} \tag{6.17} \end{equation} $$ der \( r \) er avstanden over dipolen (her er det til side for), og \( h \) er halve avstanden mellom ladningene.

Videre vet vi at i grensen mellom en leder og vakuum, så er ladningstettheten gitt som \( \rho_s = \epsilon_0 E \). Dermed er ladningsteheten gitt som $$ \begin{equation} \rho_s = -\frac{Qh}{2\pi(r^2+h^2)^\frac{3}{2}}. \tag{6.18} \end{equation} $$

Vi integrerer opp ladningsfordelingen. Her må vi huske på at vi skal integrere over hele \( xy \)-planet. Vi vekger å gjøre det i polarkoordinater, siden vi allerede har ladningsfordelingen uttrykt ved \( r \), avstanden fra \( z \)-aksen. $$ \begin{eqnarray} Q_{\text{indusert}} &=& \int_S \rho_s \d S \\ &=& -\frac{Qh}{2\pi} \int_0^\infty \frac{2\pi r \d r }{(r^2 + h^2)^\frac{3}{2}} \\ &=& \left[\frac{Qh}{\sqrt{r^2 + h^2}}\right]_0^\infty \\ &=& -Q. \end{eqnarray} $$

Vi bruker Coulombs lov for å finne kraften. Dette er kraften mellom to motsatt ladde partikler med ladning \( \pm Q \) og avstand \( 2h \): $$ \begin{eqnarray} \mathbf{F} &=& \frac{1}{4\pi\epsilon_0}\frac{Q(-Q)}{(2h)^2} \z \\ &=& -\frac{Q^2}{16 \pi \epsilon_0 h^2}\z \end{eqnarray} $$

Se først på ett enkelt linjestykke og bruk deretter superposisjonsprinsippet.

Du kan få bruk for integralet \( \int \frac{\d x}{\sqrt{C^2 + x^2}} = \textrm{arcsinh} \frac{x}{C} \).

Visualiser det analytiske og det numeriske uttrykket \( V(z) \) i det samme plottet. (Hvis de ikke faller oppå hverandre har du gjort noe galt).

Det holder å se på et av kvadratene og så bruke superposisjonsprinsippet for å finne potensialforskjellen mellom de to kvadratene.