\( (-1,0) \)

The electric field points from large to small values of the electric potential. If we move a charge from the infinity in towards the dipole along the negative \( x \)-axis, all the arrows point toward the negative charge. The potential decreases as we move along this line. The mininal value is at the negative charge, where the potential approaches minus infinity.

\( (1,0) \)

The electric field points from large to small values of the electric potential. If we move a charge from the infinity in towards the dipole along the positive \( x \)-axis, all the arrows point away from the positive charge. The potential increases as we move along this line. The maximum value is at the positive charge, where the potential approaches infinity.

At infinity and at \( (0,y) \).

The electric potential is zero at infinity. However, if we move along the line \( (0,y) \), the direction of the field is normal to the direction of motion. All the points along this line will have zero potential. This line separates positive and negative values of the potential.

The electric field between your finger and the lamp depends on the potential difference \( V \) and the distance \( d \) between the two objects: \( E = - \partial V / \partial x \simeq -V/d \). As the distance becomes shorter, the electric field becomes larger. At some point the field increases beyond the critical strength of the air, producing a spark and a discharge of the charges on your finger to the lamp.

\( V_a = 100 \text{V} \), \( V_b = 300 \text{V} \), \( V_c = 300 \text{V} \).

The electric potential in point a is given as the integral: $$ \begin{equation} V_a = \int_a^{\text{ref}} \vec{E} \cdot d \vec{r} \; . \tag{3.1} \end{equation} $$ For point a this is the integral from a to the origin. In this region the electric field is uniform and directed along the \( x \)-axis, \( \vec{E} = -10^4 \text{N/C} \hat{x} \). The integral is therefore simply: $$ \begin{equation} V_a = \vec{E} \cdot \Delta \vec{r} \; , \tag{3.2} \end{equation} $$ where \( \Delta \vec{r} = (-1 \text{cm},-1\text{cm}) \). We see that only the component of \( \Delta \vec{r} \) in the \( x \)-direction will contribute, because \( E_y = 0 \), giving: \( V_a = \left( -10^4 \text{N/C}\right) \, \left( -1 \text{cm} \right) = 10^2 Nm/C = 10^2 V \). We check the sign by observing that the electric field point from high to low potential, as it should. Similarly, we find \( V_b = 3 \, 10^2 \text{V} \) and \( V_c = V_b = 3 \, 10^2 \text{V} \).

\( (F_x,F_y) = (0.5 \text{N},0) \)

For the velocity to be constant, the net force must be zero. The force from the electric field is \( q\vec{E} \), where \( \vec{E} \) is uniform along the line from a to b, \( \vec{E} = (-1,0) \, 10^4 \text{N/C} \). The net force is \( (F_x,F_y) + q \vec{E} = 0 \), which gives \( (F_x,F_y) = - q \vec{E} = 50 10^{-6} \text{C} \, 10^4 \text{N/C} \hat{x} = 0.5 \text{N} \hat{x} \).

\( 10^{-2} \text{Nm} \)

The work done by the force \( \vec{F} \) was \( W = \vec{F} \cdot \Delta \vec{r} = (0.5 \text{N},0) \cdot (2 \text{cm}, 0) = 10^{-2} \text{Nm} \).

\( 10^{-2} \text{Nm} \)

The potential energy difference is the negative work done by the electric field, which is the work done by the force \( \vec{F} \): \( U_b - U_a = 10^{-2} \text{Nm} \).

\( 200 \text{V} \)

The electric potential difference is \( V_b - V_a = 300 \text{V} - 100 \text{V} = 200 \text{V} \).

\( \Delta V = \Delta U/q \)

In general, \( V = U/q \), as long as the potential energy and the electric potential have the same reference points. Here, we see that \( (U_b-U_a)/a = 10^{-2} \text{Nm} / (50 \cdot 10^{-6}C) = 200 \text{V} \), which corresponds to the value we found from the electric field.

\( \d V = (Q/L)\d x'/(4 \pi \epsilon_0 ((x')^2 + y^2)^{1/2}) \)

The contribution is $$ \begin{equation} d V = \frac{dq}{4 \pi \epsilon_0 R} \tag{3.3} \end{equation} $$ First, let us note that we use the variable \( x' \) and \( dx' \) to describe the position of the small element to avoid confusion between the point \( (x',y',z') \) of the small element and the point \( (x,y,z) \) where we want to find the potential. For an element along the \( x \)-axis \( (x',y',z') = (x',0,0) \) and for a point on the \( y \)-axis we have \( \vec{r} = (0,y,0) \) so that \( \vec{R} = \vec{r} - \vec{r}' = (0,y,0) - (x',0,0) = (-x',y,0) \) and $ R^2= (x')^2 + y^2$. We also write \( dx' \) to make sure there is no confusion as to the choice of integration variable. We assume that the charge is uniformly distributed with a charge density \( \rho_l = Q/L \), so that \( dq = \rho_l dx' = (Q/L) dx' \). This gives: $$ \begin{equation} dV = \frac{(Q/L)dx'}{4 \pi \epsilon_0 ((x')^2 + y^2)^{1/2} } \; . \tag{3.4} \end{equation} $$

\( V = \int_{-L/2}^{L/2} (Q/L) / (4 \pi \epsilon_0 \left( (x')^2 + y^2 \right)^{1/2} ) dx' \)

The electric potential is found from the integral over all the charge contributions \( dq = \rho_l dx' \): $$ \begin{equation} V = \int_{-L/2}^{L/2} \frac{(Q/L) dx'}{4 \pi \epsilon_0 \left( (x')^2 + y^2 \right)^{1/2} } \; . \tag{3.5} \end{equation} $$ Notice that the integral is over \( dx' \) and not \( x \)!

No

No, she is not correct. We can take the partial derivative with respect to \( y \) inside the integral, but this will only generate a new integral that still needs to be solved.

\( V(x) = \frac{q}{4 \pi \epsilon_0} \left( \frac{1}{|x-a|} + \frac{1}{|x+a|} \right) \)

We use the superposition principle. The potential in a point \( x \) from a charge \( q \) at \( x_1 \) is $$ \begin{equation} V_1(x) = \frac{q}{4 \pi \epsilon_0 |x - x_1|} \; . \tag{3.6} \end{equation} $$ We add the contributions from \( x_1 = a \) and \( x_2 = -a \), getting: $$ \begin{equation} V(x) = \frac{q}{4 \pi \epsilon_0} \left( \frac{1}{|x-a|} + \frac{1}{|x+a|} \right) \; . \tag{3.7} \end{equation} $$

Equipotential points are point \( x \) where \( V(x) = V_0 \).

\( V(y) q/(2 \pi \epsilon_0 (y^2 + a^2 )^{1/2} ) \)

The potential along the \( y \)-axis is also found using the superposition principle. The contribution from each of the charges \( q \) are the same in this case: $$ \begin{equation} V(y) = \frac{q}{4 \pi \epsilon_0 (y^2 + a^2)^{1/2} } + \frac{q}{4 \pi \epsilon_0 (y^2 + a^2)^{1/2} } = \frac{q}{2 \pi \epsilon_0 (y^2 + a^2 )^{1/2} } \; , \tag{3.8} \end{equation} $$ where \( R = (y^2 + a^2 )^{1/2} \) is the distance from a charge (\( \pm q \)) to the point \( \vec{r} = (0,y,0) \).

We generate the electric potential using the code from the text. We introduce a function findpotential that finds the potential in the \( xy \)-plane with a given resolution.

import numpy as np
import matplotlib.pyplot as plt

def epotlist(r,Q,R):
    # Find V*4*pi*epsilon0 at r from a charges Q at positions R
    V = 0
    for i in range(len(R)):
        Ri = r - R[i]
        qi = Q[i]
        Rinorm = np.linalg.norm(Ri)
        V = V + qi/Rinorm
    return V

def findpot(R,Q,L,N):
    x = np.linspace(-L,L,N)
    y = np.linspace(-L,L,N)
    rx,ry = np.meshgrid(x,y)
    # Set up electric potential
    V = np.zeros((N,N),float)
    # Calculate the potential
    for i in range(len(rx.flat)):
        r = np.array([rx.flat[i],ry.flat[i]])
        V.flat[i] = epotlist(r,Q,R)
    return rx,ry,V

We setup the charges:

# Setup charges
R = []
Q = []
r0 = np.array([1,0])
q0 = 1
R.append(r0)
Q.append(q0)
r0 = np.array([-1,0])
q0 = 1
R.append(r0)
Q.append(q0)

And then we calculate and visualize the electric potential using contour lines:

# Calculate the potential
rx,ry,V = findpot(R,Q,3,81)
# Visualize the potential
plt.figure(figsize=(8,8))
plt.contour(rx,ry,V,200)
plt.axis('equal')

The resulting plot is shown in the figure:

The electric field is found using the gradient. The electric field is visualized using the vector field.

Ey,Ex = np.gradient(-V)
# Calculate field magnitude and unit vectors
Emag = np.sqrt(Ex**2 + Ey**2)
minlogEmag = min(np.log10(Emag.flat))
scaleE = np.log10(Emag) - minlogEmag
uEx = Ex / Emag
uEy = Ey / Emag
# Visualize using both arrows and colors
levels = np.arange(0, 3.5+0.2, 0.2)
cmap = plt.cm.get_cmap('plasma')
plt.figure(figsize=(16,8))
ax1 = plt.subplot(1,2,1)
plt.contourf(rx,ry,V,10, cmap=cmap, levels=levels, extend='both')
skip = (slice(None, None, 4), slice(None, None, 4))
plt.quiver(rx[skip],ry[skip],uEx[skip]*scaleE[skip],uEy[skip]*scaleE[skip])
ax1.set_aspect('equal', 'box')
ax2 = plt.subplot(1,2,2)
plt.contourf(rx,ry,V,10, cmap=cmap, levels=levels, extend='both')
plt.streamplot(rx,ry, Ex, Ey)
ax2.set_aspect('equal', 'box')

The figure above shows the streamline plots. These provide insight into the direction of the field, but do not visualize the strength of the field. In order to visualize the strength we would need to show the field lines (see description in text).

It may be useful to work in cylinder coordinates. Remember that the area element in the \( r\phi \)-plane when we integrate in cylinder coordinates is \( r\vec{d}r\vec{d}\phi \)

We integrate the contributions to the electric field in the point \( (0,0,z) \). The distance from a point \( (r,\phi, 0) \) on the disk and to the observation point \( (0,0,z) \) is \( R = \sqrt{r^2 + z^2} \). Consequently, we get $$ \begin{equation} \vec{d}E = \frac{\rho_s}{4\pi\epsilon_0 R^2} \vec{d}S = \frac{\rho_s}{4\pi\epsilon_0 (r^2+z^2)} \vec{d}S \tag{3.10} \end{equation} $$ These contributions have components both in the \( z \)-direction and a component normal to the \( z \)-axis, that is, in the direction \( \hat{\mathbf{r}} \) in cylinder coordinates. We call the angle between the \( z \)-axis and the line from \( (0,0,z) \) to \( (r, \phi, 0) \) $\alpha$. We know that the electric field points in the same direction as the vector from the charge to the observation point. This gives $$ \begin{align} \vec{d}E_z = \vec{d}E\cos\alpha \tag{3.11}\\ \vec{d}E_r = \vec{d}E\sin\alpha \tag{3.12} \end{align} $$ We find that \( \cos\alpha = \frac{z}{\sqrt{r^2+z^2}} \) and \( \sin\alpha = \frac{r}{\sqrt{r^2+z^2}} \) so that $$ \begin{equation} \vec{d}\mathbf{E} = \frac{\rho_s}{4\pi\epsilon_0}\frac{-r\hat{\mathbf{r}} + z\hat{\mathbf{z}}}{(r^2+z^2)^{3/2}} \ \vec{d}S \tag{3.13} \end{equation} $$

Integrate over the disk \( \vec{d}q = \rho_s\vec{d}S \). It may help to start by introducing a symmetry argument so that you only need to find one of the components of the \( \vec{E} \)-field.

Because the system has rotation symmetry around the \( z \)-axis (it is identical if we rotate the disk around the \( z \)-axis), the field contributions in the \( x \) and \( y \)-directions must cancel. This means we only need to find the \( z \)-component. We need to solve: $$ \begin{equation} E_z = \int_S \d E_z = \int_S \frac{\rho_s z}{4\pi\epsilon_0 (r^2+z^2)^{3/2}} \d S \tag{3.14} \end{equation} $$ We choose to integrate the surface \( S \) in concentric circles with perimeter \( 2 \pi r \) and thickness \( d r \), so that \( d S = 2 \pi r d r \): $$ \begin{eqnarray} E_z &=& \int_0^a \frac{\rho_s z}{4\pi\epsilon_0 (r^2+z^2)^{3/2}} 2\pi r \d r \\ &=& \frac{\rho_s}{2\epsilon_0} \int_0^a \frac{zr}{(r^2 + z^2)^{3/2}} \d r \\ &=& \frac{\rho_s}{2\epsilon_0} \left[ -\frac{z}{\sqrt{z^2 + r^2}} \right]_0^a \\ &=& \frac{\rho_s}{2\epsilon_0} \left(1 - \frac{z}{\sqrt{z^2+a^2}}\right) \end{eqnarray} $$ The field is therefore $$ \begin{equation} \vec{E} = \frac{\rho_s}{2\epsilon_0} \left(1 - \frac{z}{\sqrt{z^2+a^2}}\right) \hat{z} \tag{3.15} \end{equation} $$

We integrate to find the electric potential. Since we know the field along the \( z \)-axis, we choose to integrate from infinity along the \( z \)-axis. $$ \begin{eqnarray} V &=& - \int_\infty^z E_z(z') \vec{d} z' = -\int_\infty^z \frac{\rho_s}{2\epsilon_0} \left(1 - \frac{z'}{\sqrt{z'^2+a^2}}\right) \d z' \\ &=& -\frac{\rho_s}{2\epsilon_0} \left[z' - \sqrt{z'^2 + a^2}\right]_\infty^z \\  &=& \frac{\rho_s}{2\epsilon_0} \left(\sqrt{z^2+a^2} - z\right) \end{eqnarray} $$ In the last line we have used the limit \( \lim_{x\to \infty} x - \sqrt{x^2 + 1} = 0 \).

\( V_{z\to 0} = \frac{\rho_s}{2\epsilon_0} \)

We find the behavior in the limit $$ \begin{equation} \lim_{x\to 0} 1-\frac{z}{\sqrt{z^2+a^2}} = \lim_{x\to 0} 1-\frac{1}{\sqrt{1+\frac{a^2}{z^2}}} \\ = 1 \tag{3.17} \end{equation} $$ The field for small \( z \) therefore becomes \( \vec{E} = \frac{\rho_s}{2\epsilon_0}\hat{z} \). This is the same as the field from an infintely large plane, which is reasonable because the plan relatively becomes larger and larger the closer we get to it.

We plot the field using the following code:

import numpy as np
import matplotlib.pyplot as plt
z = np.linspace(0,3,100)
E = 1-z/np.sqrt(z**2+1)
plt.plot(z,E)
plt.xlabel(r"$z/a$")
plt.ylabel(r"$2\epsilon_0 E/\rho_s$")

Use the series expansion \( (1+x)^a \approx 1 + ax \) when \( x \ll 1 \).

We use the approximation given in the hint: $$ \begin{equation} \frac{z}{\sqrt{z^2+a^2}} = \left(1 + \frac{a^2}{z^2}\right)^{-1/2} \approx 1 - \frac{1}{2} \frac{a^2}{z^2} \tag{3.18} \end{equation} $$ when \( a/z \ll 1 \), that is \( z \gg a \). In addition, we use that the total charge of the disk is \( Q_S = \rho_s \pi a^2 \) and get: $$ \begin{eqnarray} \mathbf{E} &=& \frac{\rho_s}{2\epsilon_0} \left[ 1- \left( 1-\frac{1}{2} \frac{a^2}{z^2} \right) \right]\z \\ &=& \frac{\rho_s}{4\epsilon_0}\frac{a^2}{z^2} \z \\ &=& \frac{Q_S}{4\pi\epsilon_0 z^2} \z \end{eqnarray} $$ which is the same expression as Coulomb's law, that is, the field from a point charge. This is because from far away the disk is approximately a point charge.

Superposition

The field over an infinite plate without a hole is the same as the field from an infinite plate in the limit \( z \to 0 \).

Vi tar utgangspunkt i at vi kjenner feltet over et uendelig stort ladd \( xy \)-plan: \( \mathbf{E} = \frac{\rho_s}{2\epsilon_0} \z \). Vi kjenner også, fra en tidligere oppgave, at feltet over en ladd, sirkulær disk i \( xy \)-planet er \( \mathbf{E} = \frac{\rho_s}{2\epsilon_0} \left(1 - \frac{z}{\sqrt{z^2+a^2}}\right) \z \). Vi bruker så superposisjon ved at vi tar feltet fra et ladd plan og trekker fra feltet fra en disk, slik at vi får feltet fra en plate med et siruklært hull. $$ \begin{eqnarray} \vec{E}_{\text{hullplan}} &= \vec{E}_{\text{plan}}-\vec{E}_{\text{disk}} \\ &=& \frac{\rho_s}{2\epsilon_0} \z - \frac{\rho_s}{2\epsilon_0} \left(1 - \frac{z}{\sqrt{z^2+a^2}}\right) \z \\ &=& \frac{\rho_s}{2\epsilon_0}\frac{z}{\sqrt{z^2+a^2}} \z \end{eqnarray} $$

Vi ser at \( \vec{R} \)-vektoren er \( \vec{R} = \vec{r} - \vec{r}' = (x,0) - (x',0) = (x-x',0) \) og \( R = |x-x'| \). Vi setter dette inn i utrykket for det elektriske potensialet: $$ \begin{equation} \d V = \frac{\rho_l \d x'}{4 \pi \epsilon_0 R} = \frac{\rho_l \d x'}{4 \pi \epsilon_0 |x-x'|} \; . \tag{3.20} \end{equation} $$

Vi finner det elektriske potensialet ved å integrere \( \d V \) fra \( x'=-L \) til \( x' = 0 \). Fordi \( x>0 \) vil \( (x-x')>0 \) for hele integralet og vi kan sette \( |x-x'| = (x-x') \) i integralet: $$ \begin{equation} V(x) = \int_{-L}^{0} \frac{\rho_l \d x'}{4 \pi \epsilon_0 (x-x')} = \frac{\rho_l }{4 \pi \epsilon_0} \int_{-L}^{0} \frac{\d x'}{x-x'} = \frac{\rho_l }{4 \pi \epsilon_0} \left( \ln (x+L) - \ln (x) \right) \; . \tag{3.21} \end{equation} $$ Vi har valgt å bruke et utrykk for det elektriske potensialet som er basert på Coulombs lov. For dette uttrykket er potensialet satt til null uendelig langt vekk. Vi ser også at potensialet faktisk går mot null når \( x \) blir veldig stor fordi forskjellen mellom \( \ln (x+L) \) og \( \ln x \) relativt sett blir liten når \( x \) går mot uendelig.

Det elektriske feltet er relatert til det elektriske potensialet gjennom \( \vec{E} = - \nabla V \). Det betyr at \( x \)-komponenten av det elektriske feltet er \( E_x = - \partial V/\partial x \): $$ \begin{equation} E_x = - \frac{\partial}{\partial x} \frac{\rho_l }{4 \pi \epsilon_0} \left( \ln (x+L) - \ln (x) \right) = - \frac{\rho_l }{4 \pi \epsilon_0} \left( \frac{1}{x+L} - \frac{1}{x} \right) \; . \tag{3.22} \end{equation} $$ Vi ser at \( E_x \) går mot uendelig når vi nærmer oss tuppen av linjen, både for \( x \rightarrow 0^{+} \) og for \( x \rightarrow -L^{-} \).

For å finne det elektriske feltet \( E_y \) fra \( V(x,y) \) må vi beregne \( E_y = - \partial V(x,y) / \partial y \). Det kan vi ikke gjøre, fordi vi kun kjenner \( V(x,0) \) og ikke \( V(x,y) \) som vi ville trenge for å finne \( E_y \).

(Dette kreves ikke som svar på oppgaven): På den annen side, kan vi av symmetri se at \( E_y(x,0) \) må være null. Fordi hvis vi antar at \( E_y(x,0) >0 \), så kan vi speile systemet om \( x \)-aksen. Da får vi det samme fysiske system, men vi ville da få at \( E_y(x,0) < 0 \). Fordi \( E_y(x,0) \) ikke både kan være større og mindre enn null, ser vi at antagelsen om at \( E_y(x,0)>0 \) må være gal og at \( E_y(x,0) = 0 \).