No. From Gauss' law we know that the flux through any closed surface is just \( Q_{enc}/\epsilon_0 \), where \( Q_{enc} \) is the sum of all charges within the area. And no matter how much we inflate the balloon there will always be just 1 charge on the inside.

Gauss law on differential form states $$ \begin{equation} \nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}. \tag{4.1} \end{equation} $$ Since \( \rho > 0 \), \( \vec{E} \) cannot be constant and therefore not uniform.

From Gauss' law on differential form, (4.1), we get for \( \rho = 0 \) that the divergence of the electric field is zero. We cannot say anything about its value other than the fact than the divergence is zero. For example, the field from a point charge outside this region, will have zero divergence inside this region. We therefore cannot deduct that the field is zero in this region. In fact, that would be very unlikely given the positive charge density around it. However, if space was filled by a uniform volume charge density \( \rho \), and we cut a spherical hole in this infinite space, then the system would have spherical symmetry around the center of the sphere. In this particular case, the electric field would have to be zero inside the sphere.

Consider a spherical surface centered around a single point charge.

For a point charge we would have \( \mathbf{E}(r) = \frac{q}{4\pi\epsilon_0 r^3}\mathbf{\hat{r}} \). This field is spherically symmetric around a point charge. We take the flux out of a sphere \( S \) of radius \( R \) with the charge at the center: $$ \begin{eqnarray*} \Phi &=& \oint_S \mathbf{E}(R)\cdot \mathbf{dS} \\ &=& \frac{q}{4\pi\epsilon_0 R^3} \oint_S dS \\ &=& \frac{q}{4\pi\epsilon_0 R^3}\cdot 4\pi R^2 \\ &=& \frac{q}{\epsilon_0 R} \end{eqnarray*} $$ Gauss' law is no longer valid.

Alternative 1: If there is a field parallel to the wire and we rotate the wire by \( 180^{\circ} \) around an axis normal to the wire, the wire does not change, and the field should therefore also not change. This can only be the case if the field parallel to the wire is zero.

Alternative 2: The contribution to the electric field at a point will come from two small pieces of wire placed symmetrically around the normal from the point to the wire. The net contribution to the field from these two pieces will cancel in the direction parallel to the wire.

We can use the same argument as above: If we rotate the wire by \( 180^{\circ} \) around an axis normal to the wire, the wire does not change, and the field should therefore also not change. For this rotation, the tangential component will rotate onto its negative. This component can only be the same if it is zero.

Because for a cylinder with an axis along the wire, the electric field would be constant on the side cylinder surface.

\( q = L \lambda \)

The total charge on a small section of length \( L \) is \( q = L \lambda \).

\( \vec{E} = \frac{1}{2 \pi \epsilon_0} \frac{\lambda}{R} \hat{u} \), where \( \hat{u} = (x,y,0)/\sqrt{x^2 + y^2} \).

We use Gauss' law on a cylinder of length \( L \) and radius \( R \). The charge inside the cylinder is \( q = L \lambda \). The flux is \( E 2 \pi R L = q/\epsilon_0 \). The field is therefore \( E = \frac{1}{2 \pi \epsilon_0} \frac{\lambda}{R} \).

At the exact midpoint of the wire the symmetry argument would still hold. This is because our original arguments of symmetry come from claiming that an infinite wire at any point along the wire always has equal amount of wire at both sides of the point. At the midpoint of the wire segment this would still be true, although at any other point it would not, and in the latter case the symmetry arguments would not hold.

\( \nhat_A = (0,-1,0) \), \( \nhat_B = (0,0,-1) \), \( \nhat_C = (1,0,0) \), \( \nhat_D = (-1,0,0) \), and \( \nhat_E = (0,y,z)/\sqrt{y^2 + z^2} \).

The surface normal to surface \( A \) is \( (0,-1,0) \), to \( B \) is \( (0,0,-1) \), to \( C \) is \( (1,0,0) \), and to \( D \) is \( (-1,0,0) \). The normal vector for surface \( E \) points in the direction of the cylindrical radial vector with a cylinder axis along the \( x \)-axis, that is the vector pointing in the direction \( (0,y,z) \), which must be normalized by \( \sqrt{y^2 + z^2} \).

\( \Phi_A = -4 a^2 E_0 \), \( \Phi_B = \Phi_C = \Phi_D = 0 \).

We can find the flux either by taking the dot product of the surface \( \vec{A} = A\hat{n} \) and the field \( \vec{E} = E_0 \y \), or by realizing that it is only the projection of the surface onto a plane normal to \( \y \) that contributes to the flux. We use the last approach. Surfaces \( B \), \( C \) and \( D \) are normal to \( \vec{E} \) and hence the flux is zero. For surface \( A \), the area of the surface normal to \( \vec{E} \) is \( 2\times 2 \) (in units of \( a^2 \)), but the field points into the surface, so the flux is negative. The flux is therefore \( \Phi_A = -4 a^2 E_0 \).

\( \Phi_E = 4 a^2 E_0 \).

We notice that there are no charges inside the surface, hence the net flux through all the surfaces must be zero: $$ \begin{equation} \Phi_A + \Phi_B + \Phi_C + \Phi_D + \Phi_E = 0 \, \Rightarrow \, \Phi_E = - \left( \Phi_A + \Phi_B + \Phi_C + \Phi_D \right) \; , \tag{4.2} \end{equation} $$ which gives \( \Phi_E = 4 a^2 E_0 \).

We could have seen this simply be realizing that the projection of \( E \) onto the \( xz \) plane has an area \( 2 \times 2 a^2 \) and therefore that the flux is \( 4 a^2 E_0 \).

Make a sketch. Is there another surface you can easily find the flux through? How can you use Gauss' law?

\( \Phi = - (\pi a^2 /2) E(a) \)

To solve this problem, we can use Gauss' law. We construct a surface that consists of the triangle and a spherical surface with center at the origin and radius \( a \). To complete the surface, we need three smaller segments that are in the \( xy \), \( xz \) and \( yz \) planes - and therefore have no fluxes through them! Since there are no charges inside this surface, the net flux is zero. The flux through the triangle is negative and must be of the same magnitude as the flux through the part of the sphere in the first octant. The flux through this surface is simple to calculate, since the radius is constant and the field is normal to the surface: \( \Phi_S = (4 \pi a^2/8) E(a) \). The flux through the triangular surface is therefore \( \Phi_T = - (\pi/2) a^2 E(a) \).

\( \d S = a \d \phi \d z \), \( \hat{n} = (x,y,0)/\sqrt{x^2 + y^2} \).

This area is \( \d S = a \d \phi \d z \).

We assume cylindrical symmetry, which means that \( \vec{E} \) only can point in the radial direction normal to the \( z \)-axis.

We use a cylindrical Gauss surface of length \( L \) and radius \( r \). The field on this surface points in the radial (cylindrical) direction everywhere. The field is of the same magnitude on the surface. The flux of the field is therefore \( \Phi = 2 \pi r L E = Q/\epsilon_0 \), where \( Q \) is the charge inside the surface. When \( r < a \) there is no charge inside the surface, hence \( Q = 0 \) and \( E = 0 \). When \( r > a \), the charge inside is \( 2 \pi a L \rho_s \). The field is therefore $$ \begin{equation} 2 \pi r L E = \frac{2 \pi a L \rho_s}{\epsilon_0} \, \Rightarrow \, E = \left( \frac{a}{r} \right) \frac{\rho_s}{\epsilon_0} \; , \tag{4.4} \end{equation} $$ when \( r > a \).

We just concluded that the \( \vec{E} \)-field (due to the charge on the accelerator surface) is 0 inside the accelerator itself. This means that the charge accumulation would not affect the accelerated particles. However, there is a chance that the present \( \vec{E} \)-field on the outside of the accelerator might damage any bounding equipment, and the physicists should proceed with caution.

\( E(z) = \frac{\sigma}{2 \epsilon_0} \text{sign}(z) \).

First, we notice that due to symmetry the field must be perpendicular to the surface. We draw a cube with area \( A \) extending a distance \( z \) on each side of the plane. Applying Gauss' law to this cube, the total flux is \( EA + EA \) with contributions from each side. (Ensure that you understand the signs here). The total charge in this volume is \( A\sigma \), thus Gauss law gives that \( 2EA = A \sigma/\epsilon_0 \), and therefore \( E = \frac{\sigma}{2 \epsilon_0} \). Including the directions gives, \( E = \frac{\sigma}{2 \epsilon_0} \text{sign}(z) \), pointing in positive \( z \)-direction when \( z > 0 \) and in the negative direction when \( z < 0 \).

\( \sigma_+ = \rho_0 \Delta z \), \( \sigma_- = - \rho_0 \Delta z \)

In a region of area \( V = A \Delta z \) the charge is \( Q_{+} = A \Delta z \rho_0 \) on the positive side (large \( z \)) and \( Q_{-} = - A\Delta z \rho_0 \) on the negative side (small \( z \)). Hence, the surface charge density is \( \sigma_{+} = Q_{+}/A = \rho_0 \Delta z \) and \( \sigma_{-} = Q_{-}/A = - \rho_0 \Delta z \).

\( E_i = - \rho_0 \Delta z / \epsilon_0 \)

We find the electrical field inside the slab from the formula we found for the electric field and by superimposing the field from the top and the bottom surface charges. The field from the top charges is \( E_+ = -\rho_0 \Delta z/(2 \epsilon_0) \) and the field from the bottom charges is \( E_- = - \rho_0 \Delta z/(2 \epsilon_) \). The net induced field is therefore \( E_i = - \rho_0 \Delta z /\epsilon_0 \).

\( E_i = - \rho_0 c E_0/\epsilon_0 \)

We insert \( \Delta z = c E_{tot} \), getting: \( E_i = - \rho_0 c E_{tot} / \epsilon_0 \).

\( E_{tot} = E_0 /(1 + \rho_0 c/\epsilon_0) \)

\( E_{tot} = E_0 + E_i = E_0 - \rho_0 c E_{tot}/ \epsilon_0 \), which gives \( E_{tot} = E_0 /(1 + \rho_0 c/\epsilon_0) \).

Notice that this is a self-consistent equation for \( E_{tot} \).

The effective field is indeed smaller than the applied field inside the plastic slab.

\( \vec{E}(r) = \frac{Q}{4\pi\epsilon_0 r^2} \rhat \)

I alle deloppgavene har vi kulesymmetri, slik at feltet må være \( \mathbf{E} = E(r) \rhat \). Vi velger oss derfor Gaussflater som er kuleflater med radius \( r \). slik at \( \d \mathbf{S} = \nhat \d S = \rhat dS \). I kulekoordinater vet vi også at \( \d S = r^2\sin\theta\d\phi\d\theta \).

I denne oppgaven er innesluttet ladning \( Q \), slik at vi får $$ \begin{align} \oint_S \vec{E} \cdot \d\vec{S} &= 4\pi r^2 E(r) = \frac{Q}{\epsilon_0} \tag{4.5}\\ \vec{E}(r) &= \frac{Q}{4\pi\epsilon_0 r^2} \rhat \tag{4.6} \end{align} $$

Venstresiden av Gauss' lov blir som i forrige deloppgave (pga kulesymmetri). På høyresiden trenger vi nå å integrere opp ladningsfordelingen for å finne den innesluttede ladningen. Vi begynner med tilfellet \( r < a \). $$ \begin{align} Q_{in} &= \int_0^{\pi}\int_0^{2\pi}\int_0^{r} \rho_v \ r^2 \sin \theta \d r \d\phi\d\theta \tag{4.8}\\ &= 4\pi\int_0^r \rho_v \ r^2 \d r = 4\pi\int_0^r \frac{Q}{4\pi a^3/3} \ r^2 \d r \tag{4.9}\\ &= \frac{3Q}{a^3}\left[ \frac{1}{3} r^3]_0^r \right] = \frac{Qr^3}{a^3} \tag{4.10} \end{align} $$ Dette utrykket gjelder for \( r \leq a \). For \( r > a \) for vi \( Q_{encl} = Q \). Dermed har vi $$ \begin{align} 4\pi r^2 E(r) &= \frac{Qr^3}{a^3} \tag{4.11}\\  \vec{E}(r) = \frac{Q}{4\pi\epsilon_0 a^3}r\rhat \tag{4.12} \end{align} $$ for \( r < a \). For \( r > a \) blir situasjonen lik som for punktladningen slik at \( \vec{E}(r) = \frac{Q}{4\pi\epsilon_0 r^2} \rhat \).

Venstresiden i Gauss' lov blir som de forrige deloppgavene (pga. kulesymmetri). På høyresiden har vi nå en situasjon der \( Q_encl \) kommer til å være en step-funksjon i \( r \). Altså kommer \( Q_encl \) til å oke fra \( 0 \) til \( 4\pi a^2 \cdot \rho_s = Q \) når vi flytter Guassflaten fra innsiden til utsiden av det ladde kuleskallet. Dermed får vi $$ \begin{equation} \vec{E}(r) = \begin{cases} \frac{Q}{4\pi\epsilon_0 r^2}\rhat &\text{ for } r > a \\ \vec{0} &\text{ for } r < a \end{cases} \tag{4.14} \end{equation} $$

Bestem \( k \) ved å regne ut \( Q = \int_v \rho \mathrm{d}v \). Geometrien i denne oppgaven gjør at det er lurt å jobbe i kulekoordinater. Husk da på å få inn rett volumelement i integralet.

Vi velger oss et kuleskall som Gaussflate og begynner med å finne den innesluttede ladningen innenfor radius \( a \), uttrykt ved \( k \), slik at vi kan finne konstanten \( k \). $$ \begin{equation} Q = \int_v \rho_v \d v = 4\pi k \int_0^a r^3 \d r = \pi k a^4 \tag{4.15} \end{equation} $$ Dermed ser vi at dersom den totale ladningen skal være \( Q \), så må \( k=\frac{Q}{\pi a^4} \) Så finner vi den innesluttede ladningen innenfor en radius \( r \): $$ \begin{equation} Q(r) = \int_v \rho_v \d v = \pi k r^4 \tag{4.16} \end{equation} $$ Vi har fortsatt kulekymmetri slik at \( \vec{E} = E(r) \rhat \). Dermed kan vi skrive $$ \begin{align} 4\pi r^2 E(r) = \frac{\pi k r^4}{\epsilon_0} = \frac{\pi Q r^4}{\pi a^4 \epsilon_0} \tag{4.17}\\  \vec{E}(r) = \frac{Qr^2}{4\pi \epsilon_0 a^4} \rhat \tag{4.18} \end{align} $$

Et system med translasjonssymmetri har den egenskapen at systemet ser helt likt ut etter å ha blitt translatert. Et uendelig plan vil se helt likt ut dersom det translateres i en retning som ligger i samme plan som planet selv. Sagt på en annen måte. Om man observerer planet fra et punkt \( z \) over planet, så vil man ikke se forskjell dersom planet blir flyttet på i \( x \)- eller \( y \)-retning.

Et system med rotasjonssymmetri ser helt likt ut etter en rotasjon om en eller annen akse, for eksempel gjennom origo. Vi sier da at systemet er rotasjonssymmetrisk om origo. Det uendelige planet over er rotasjonssymmetrisk om \( z \)-aksen og alle andre rotasjoner som er parallelle med \( z \)-aksen. Om vi observerer planet fra punktet \( z \), kan vi ikke se om det har blitt rotert om \( z \)-aksen.

I tillegg har vi speilsymmetri om \( xy \)-planet. Det er det samme som rotasjonssymmetri med vinkel \( \pi \) om alle akser som ligger i \( xy \)-planet.

Dersom det fysiske systemet (ladningsfordelingen) ser lik ut etter en rotasjon eller translasjon, må også feltet se likt ut.

\( \mathbf{E} = E(z)\z \)

Dersom systemet ser helt likt ut som tidligere, kan ikke konsekvensene av systemet endre seg. Altså må det elektriske feltet fra systemet ha de samme symmetriene som systemet selv. Translasjonssymmetrien gir oss at feltet må være uniformt i ethvert plan parallellt med ladningsfordelingen. Rotasjonssymmetrien om en hvilken som helst akse parallell med \( z \)-aksen fører til at feltet må stå normalt på ladningsfordelingen. Dermed kan vi skrive at $$ \begin{equation} \mathbf{E} = E(z)\z \tag{4.19} \end{equation} $$

Om vi i tillegg legger på speilsymmetrien, får vi at \( E(z) = -E(-z) \) siden feltet må se likt ut om vi snur hele platen opp ned.

Vi ønsker å velge oss en Gaussflate slik at vi har et uniformt felt eller felt parallellt med flaten over klart definerbare deler av flaten. Det kan vi oppnå for eksempel med et prisme med høyde \( 2z \) som strekker seg fra \( -z \) til \( z \) og sidekanter \( a \) i \( x \)- og \( y \)-retning. Da har vi uniformt felt lik \( E(z)\z \) på endeflatene øverst og nederst i \( z \)-retning, og felt som ligger parallellt med flaten for alle andre flater på prismet (dem med normalvektor \( \pm\x \) og \( \pm\y \)). Vi setter så opp Gauss' lov: $$ \begin{equation} \Phi_S = \oint_S \mathbf{E} \cdot \d \mathbf{S} = \frac{Q_{encl}}{\epsilon_0} \tag{4.20} \end{equation} $$ På venstresiden skal vi integrere over overflaten til prismet. Vi vet at \( \mathbf{E}\cdot \d \mathbf{S} \) kommer til å være \( 0 \) for flater med normalvektor \( \pm\x \) eller \( \pm\y \). For flater med normalvektor \( \z \) og \( -\z \) får vi henholdsvis \( \Phi_{z} = a^2E(z) \) og \( \Phi_{-z} = -a^2E(-z) \).

På høyresiden har vi innesluttet ladning. Ladningen vi har innesluttet er den som ligger mellom sidekanter med lengde \( a \), altså \( Q_{encl} = a^2\rho_s \).

Til sammen har vi dermed $$ \begin{align} a^2E(z)-a^2E(-z) &= \frac{a^2\rho_s}{\epsilon_0} \tag{4.21}\\ 2E(z) &= \frac{\rho_s}{\epsilon_0} \tag{4.22}\\ E(z) = \frac{\rho_s}{2\epsilon_0} \tag{4.23} \end{align} $$

In this case we expect the electric field to have spherical symmetry.

The field can only depend on the distance \( r \) and not on any of the rotational angles (\( \phi \), \( \theta \)) because the sphere is the same when it is rotated and therefore the field must also be the same in all directions.

Similarly, the electric field can only have a radial component pointing directly outward or inward from the center. Why? If we assume that the field has a component that is in any other direction, we can rotate the sphere half a turn, and the field will point in the opposite direction. This is only possible if the field is zero in these directions.

The field therefore has the form \( \vec{E}(\vec{r}) = E_r(r) \rhat \).

Due to the spherical symmetry, we select a spherical Gauss surface of radius \( r \) centered in the center of the sphere. On this surface, the magnitude of the field is constant and the field is normal to the surface.

We expect the \( \vec{D} \)-field to have the same symmetry as the \( \vec{E} \)-field. (Strictly, this is because the dielectric material is isotropic --- that is the same in all directions): \( \vec{D} = D_r(r) \rhat \).

We apply Gauss' law to this Gauss surface \( S \): $$ \begin{equation*} \int_S \vec{D} \cdot \d \vec{S} = \int_S D_r \d S = Q \; . \end{equation*} $$ In this case, \( D_r \) is constant over the surface and can be placed outside the integral: $$ \begin{equation*} \int_S D_r \d S = D_r \int_S \d S = D_r 4 \pi r^2 = Q \; , \end{equation*} $$ which gives $$ \begin{equation*} D_r = \frac{Q}{4 \pi r^2} \; . \end{equation*} $$ We find \( E_r \) from \( \epsilon(r) E_r = D_r \) and therefore \( E_r = \frac{D_r}{\epsilon(r)} \). We know that \( \epsilon(r) = \epsilon \) when \( r\le a \) and \( \epsilon(r) = \epsilon_0 \) when \( r > a \). The electric field is therefore $$ \begin{equation*} E_r = \left\{ \begin{array}{cc} \frac{Q}{4 \pi \epsilon r^2} & \quad r\le a \\ \frac{Q}{4 \pi \epsilon_0 r^2} & \quad r >a \end{array} \right. \; . \end{equation*} $$

This is because there are bound charges on the surface of the sphere, but no free charges. The \( \vec{D} \) field only depends on the free charges, so it does not change. But the \( \vec{E} \)-field depends on all the charges, including the bound charges, so the \( \vec{E} \) field must change across the surface of the sphere due to the bound surface charges here.

The polarization is found from $$ \begin{equation*} \vec{D} = \epsilon_0 \vec{E} + \vec{P} \quad \Rightarrow \quad \vec{P} = \vec{D} - \epsilon_0 \vec{E} = \left( \epsilon(r) - \epsilon_0\right) \vec{E} \; . \end{equation*} $$ Thus the polarization has the same symmtry as \( \vec{E} \) and therefore only has a radial component, \( P_r(r) \). In addition, we see that the polarization is zero in vacuum, where \( \epsilon(r) = \epsilon_0 \). This is as it should be since there is no polarizable material here. The polarization is therefore only non-zero when \( r\le a \): $$ \begin{equation*} P_r = \left( \epsilon - \epsilon_0\right) E_r = \left( \epsilon - \epsilon_0\right)\frac{Q}{4 \pi \epsilon r^2} \; , \end{equation*} $$ when \( r \le a \).

The bound surface charge density on the surface of a dielctric material is \( \rho_{s,b} = \vec{P} \cdot \nhat \). Here, the normal vector to the surface points in the \( \rhat \) direction, so that \( \rho_{s,b} = P_r \). The surface is at \( r =a \) so the surface charge density is $$ \begin{equation*} \rho_{s,b} = P_r = \left( \epsilon - \epsilon_0\right)\frac{Q}{4 \pi \epsilon a^2} \; . \end{equation*} $$

Vi lager som alltid først en skisse av systemet.

Dette systemet har kulesymmetri. Det er derfor naturlig å beskrive det elektriske feltet i kulekoordinater: $$ \begin{equation} \vec{E} = E_r(r,\phi,\theta)\rhat + E_{\phi}(r,\phi,\theta)\phihat + E_{\theta}(r,\phi,\theta)\thetahat \; . \tag{4.24} \end{equation} $$ Siden systemet har kulesymmetri, kan ikke feltet avhenge av \( \phi \) eller \( \theta \). Dessuten kan hverken \( \phi \) eller \( \theta \)-komponenten av feltet være forskjellig fra null. Det elektriske feltet og dermed også \( \vec{D} \)-feltet har derfor formen $$ \begin{equation} \vec{E} = E_r(r) \rhat \; , \tag{4.25} \end{equation} $$ og $$ \begin{equation} \vec{D} = D_r(r) \rhat \; . \tag{4.26} \end{equation} $$

Siden systemet har kulesymmetri kan vi anvende Gauss lov. Som Gauss-flate \( S \) velger vi en kuleflate med samme sentrum som kuleskallet og radius \( r \). I dette tilfellet gir Gauss lov på integralform at $$ \begin{equation} \oint_S \vec{D} \cdot \d \vec{S} = \oint_S D_r(r) \rhat \cdot \d S \rhat = D_r(r) \oint_S \d S = D_r(r) 4 \pi r^2 = Q_{\text{in}}\; . \tag{4.27} \end{equation} $$ Dette fluksintegralet skal være lik den frie ladningen innenfor overflaten. Hvis \( r < a \) er det ikke noe fri ladning innenfor overflaten og feltet \( D_r = 0 \). Hvis \( r>b \) er netto ladning innenfor flaten \( (+Q) + (-Q) = 0 \), og dermed \( D_r(r) = 0 \). Hvis \( a < r < b \) er den frie ladningen innenfor overflaten lik \( +Q \) og vi finner at: $$ \begin{equation} D_r 4 \pi r^2 = +Q \quad \Rightarrow \quad D_r = \frac{Q}{4 \pi r^2} \; . \tag{4.28} \end{equation} $$ I området \( a < r < b \) er systemet dielektriske med dielektrisitetskonstant \( \epsilon \) slik at \( D_r = \epsilon E_r \) og dermed er \( E_r = D_r/\epsilon \): $$ \begin{equation} E_r = \frac{Q}{4 \pi \epsilon r^2} \; . \tag{4.29} \end{equation} $$

Vi finner potensialforskjellen mellom den ytre og den indre overflaten gjennom integralet: $$ \begin{equation} V(b) - V(a) = \int_b^a \vec{E} \cdot \d \vec{l} = \int_b^a E_r \rhat \cdot \d r \rhat = \int_b^a E_r \d r \; . \tag{4.30} \end{equation} $$ Vi setter inn resultatet for \( E_r \) og finner at $$ \begin{equation} V(b) - V(a) = \int_b^a \frac{Q}{4 \pi \epsilon r^2} \d r = \frac{Q}{4 \pi \epsilon} \int_b^a \frac{\d r}{r^2} = \frac{Q}{4 \pi\epsion} \left( \frac{1}{b} - \frac{1}{a} \right) \; . \tag{4.31} \end{equation} $$

Vi ser at i dette tilfellet er \( b = a + d \). Vi bruker Python til å regne ut tallverdiene:

import numpy
from scipy.constants import epsilon_0
from scipy.constants import e
dV = -70e-3 #V
a = 1e-6 # m
d = 5e-9 # m
epsilon = 80*epsilon_0
Qdive = dV*4*numpy.pi*epsilon/((1/(a+d)-1/a)*e)
print(Qdive)

Out[] : 781685.9182907249

Som forteller oss at det er omtrent 1 million ioner på eller nær overflaten til en celle og som danner den potensialforskjellen vi kan måle i en nervecelle.

(Ekstra --- ikke forventet som en del av svaret): Men hvor stort er dette tallet? For å få en ide om hvor stort dette tallet er, la oss sammenlikne med en størrelse vi kan forholde oss til. Vi kan måle hvor mange ioner det er per kvadratnanometer på overflaten. Vi vet at et vannmolekyl er ca. 0.27 nanometer i størrelse, så det gir oss noe å sammenlikne med. Vi ser at \( a = 1 \mu \text{m} = 10^3 \text{nm} \) slik at overflatearealet målt i kvadratnanometer er \( S = 4 \pi a^2 = 4 \pi (10^3 \text{nm})^2 = 4 \pi 10^6 \text{nm} \). Antall ioner per kvadratnanometer blir da:

S = 4*np.pi*(1e3)**2 # i nm^2
print(Qdive/S)

0.06220458892064814

Som et overslag tenker oss disse ionene befinner seg i et 1 nm tykt vannlag nær celleoverflaten. Hvis det er litt mer enn 3 vannmolekyler per nanometer er det ca 30 vannmolekyler i et 1 nm tykt vannlag. Det betyr at det er ca. 2 ioner per tusen vannmolekyler i dette volumet.

Re-construct the situation by placing \( q \) in the center of a cubic box of sides \( 2a \).

We construct a cube of sides \( 2a \) and place \( q \) in the center. To construct this cube it would require 4 cubes of sides \( a \). The shaded area would then only account for 1/24 of the entire surface area of the large cube. By using Gauss' law on the large cube, we find that the total flux through all the walls is \( \Phi = q/\epsilon_0 \). The flux through the shaded area is then \( \Phi/24 = q/24\epsilon_0 \).

We use Gauss law. The field can only depend on the distance \( z \) to the middle of the sheet: \( \vec{E} = \vec{E}(z) \) because the sheet is infinitely large. Also the field, cannot have any component in the \( xy \)-plane, because the sheet is the same when it is rotated around the \( z \)-axis. Therefore, the field must have the form \( \vec{E} = E_z \z \). We use cylindrical Gauss surface which is symmetric around the middle of the sheet at \( z=0 \).

There will be no flux through the side surfaces. The top and bottom surface will contribute with the same flux. So that Gauss law gives the following inside the sheet: $$ \begin{equation*} \int_S \vec{D} \cdot \d\vec{S} = A D(z) + AD(-z) = 2 A D(z) = (2z)A \rho \quad \Rightarrow \quad D(z) = \rho z \; , \end{equation*} $$ when \( |z| \le d/2 \). Here, we have used that \( D(z) = D(-z) \), that is, the magnitudes are the same. The fields will point in opposite directions. When \( z>d/2 \) the expression becomes $$ \begin{equation*} 2 A D(z) = 2 (d/2) A \rho \quad \Rightarrow \quad D(z) = \rho \frac{d}{2} \; . \end{equation*} $$

No

No. Because the electric field is constant outside the sheet.

Uniform.

It is still uniform. Since the distribution is symmetric around \( z=0 \) we can use the same argument as above, and the field outside the sheet only depends on the total charge inside the cylindrical Gauss surface.

To find the field in a point \( z \) outside, we need to integrate over the charge distribution: $$ \begin{equation*} \int_S D_z \d S = AD(z) + AD(-z) = 2 AD(z) = \int{-d/2}^{d/2} A \rho(z) \d z \; , \end{equation*} $$ where the right-hand side is a number that depends on the details of the charge distribution.

No.

No. The electric field is then not the same on the top and bottom surfaces, and we cannot use Gauss law to find the field.

\( \d D_z = \frac{1}{2}\rho(z) \d z \)

This is essentially the same problem as above, and we can use the same result. The contribution \( \d D_z \) to the electric field from a thin sheet of thickness \( \d z \) at \( z \) is $$ \begin{equation*} \d D_z = \frac{1}{2}\rho(z) \d z \; . \end{equation*} $$

No. The contribution to the field does not depend on the distance \( z \) for any infinite sheet. The total field is the sum of all of these fields. The total field will therefore not depend on the distance, but only on the integral of the charge distribution across the sheet. $$ \begin{equation} D_z = \int_{-d/2}^{d/2} \frac{1}{2}\rho(z) \d z \; . \tag{4.32} \end{equation} $$ And we also see that the field is trivially symmetric in the \( z \)-direction outside the sheet because the field is uniform.

No.

\( D_z(z') = \frac{\rho_0}{d}\left( (z')^2 - (d/2)^2 \right) \)

Since the field outside each thin sheet is uniform, we only need to add up the positive and negative contributions from the thin sheets. At a position \( z' \), the contribution from a sheet above it, points in the negative direction (if the charge is positive). The contribution from a sheet below it, points in the positive direction (if the charge is positive). The contribution to the field is therefore $$ \begin{equation} D_z(z') = \int_{-d/2}^{z'} \frac{1}{2} \rho(z) \d z - \int_{z'}^{d/2} \frac{1}{2} \rho(z) \d z \; , \tag{4.33} \end{equation} $$ We insert the expression for \( \rho(z) \) and find that $$ \begin{equation} \int_{-d/2}^{z'} \frac{1}{2} \rho(z) \d z = \frac{\rho_0}{d}\int_{-d/2}^{z'} z \d z = \frac{\rho_0}{2d} \left( (z')^2 - (d/2)^2 \right) \; , \tag{4.34} \end{equation} $$ and $$ \begin{equation} \int_{z'}^{z} \frac{1}{2} \rho(z) \d z = \frac{\rho_0}{d}\int_{z'}^{d/2} z \d z = \frac{\rho_0}{2d} \left( (d/2)^2 - (z)^2 \right) \; , \tag{4.35} \end{equation} $$ The total is $$ \begin{equation} D_z(z') = \frac{\rho_0}{d}\left( (z')^2 - (d/2)^2 \right) \tag{4.36} \end{equation} $$ This is negative with a maximum at \( z' = 0 \), which seems reasonable.