nyuuzyou/wikis huggingface dataset:
Either-Or Topology is Locally Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is a locally connected space.
== Proof ==
:Either-Or Topology is Locally Path-Connected
:Locally Path-Connected Space is Locally Connected
== Sources ==
* : Part $\text {II}$: Counterexamples: $17$. Either-Or Topology: $4$
Category:Locally Connected Spaces
Category:Either-Or Topology
Either-Or Topology is Locally Path-Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is a locally path-connected space.
== Proof ==
Consider the set:
:$\BB := \set {\set x: x \in S, x \ne 0} \cup \openint {-1} 1$
Then by Basis for Either-Or Topology, $\BB$ is a basis for $T$.
From Point is Path-Connected to Itself, all $x \in S, x \ne 0$ are path-connected elements of $\BB$.
Finally we consider $\openint {-1} 1 \in \BB$.
Let $p \in \openint {-1} 1$ and consider the mapping $f: \closedint 0 1 \to \openint {-1} 1$:
:$\forall x \in \closedint 0 1: \map f x = \begin{cases}
0 & : x = 0 \\
p & : x \in \hointl 0 1 \end{cases}$
Then $f$ is a continuous mapping and so a path from $0$ to $p$.
So $\openint {-1} 1$ is a path-connected element of $\BB$.
Hence the result by definition of locally path-connected space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $17$. Either-Or Topology: $4$
Category:Locally Path-Connected Spaces
Category:Either-Or Topology
Either-Or Topology is Non-Meager

== Theorem ==
Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is non-meager.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $17$. Either-Or Topology: $3$
Category:Non-Meager Spaces
Category:Either-Or Topology
Category:Either-Or Topology is Non-Meager
Either-Or Topology is Non-Meager/Proof 1

== Theorem ==
== Proof ==
From the definition of the either-or space, we have that every point $x$ in $T$ (apart from $0$) forms an open set of $T$.
So every non-empty subset of $T$ (apart from $\left\{{0}\right\})$ contains at least one open set of $T$.
So no subset of $T$ is nowhere dense in $T$.
So $T$ is not a countable union of subsets of $S$ which are nowhere dense in $T$.
Hence the result by definition of non-meager.
Category:Either-Or Topology is Non-Meager
Either-Or Topology is Scattered

== Theorem ==
Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is a scattered space.
== Proof ==
By definition, $T$ is scattered every non-empty subset $H$ of $S$ contains at least one point which is isolated in $H$.
Let $H$ be a non-empty subset of $S$.
Let $x \in H$.
Let $x \ne 0$.
By definition of either-or space, $\set x$ is open in $T$.
So $\set x$ is an open set of $x$ containing only $x$.
Thus by definition $x$ is isolated.
Thus if $H$ is a non-empty subset of $S$ which contains any element other than $0$, $H$ contains at least one isolated point.
The other case is that $H$ contains no $x$ such that $x \ne 0$.
That is: $H = \set 0$.
But from Singleton Point is Isolated, $0$ is an isolated point in $H$.
Thus in all cases $H$ contains at least one point which is isolated in $H$.
So by definition $T$ is a scattered space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $17$. Either-Or Topology: $5$
Category:Scattered Spaces
Category:Either-Or Topology
Either-Or Topology is T0

== Theorem ==
Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is a $T_0$ (Kolmogorov) space.
== Proof ==
Let $x, y \in S$ such that $x \ne y$.
, let $x \ne 0$.
Then $U = \set x$ is open in $T$ from the definition of the either-or topology.
We have that $x$ and $y$ are arbitrary.
So:
:$\forall x, y \in S: \exists U \in \tau: x \in U, y \notin U$
and the result follows by definition of $T_0$ (Kolmogorov) space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $17$. Either-Or Topology: $1$
Category:T0 Spaces
Category:Either-Or Topology
Either-Or Topology is T4

== Theorem ==
Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is a $T_4$ space.
== Proof ==
Follows from:
:Either-Or Topology is $T_5$
:$T_5$ Space is $T_4$ Space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $17$. Either-Or Topology: $1$
Category:T4 Spaces
Category:Either-Or Topology
Either-Or Topology is T5

== Theorem ==
Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is a $T_5$ space.
== Proof ==
Let $A, B \subseteq S$ such that $A \cap B \ne \O$.
Suppose neither $A$ nor $B$ contain $0$.
Then $A$ and $B$ are both open in $T$.
From Limit Points of Either-Or Topology, the only limit point that either $A$ or $B$ may have is $0$.
So either $A^- = A \cup \set 0$ or $A^- = A$, where $A^-$ is the closure of $A$
Similarly for $B$.
So $A^- \cap B = \O$, and so are separated, and such that:
:$\exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$
where $U = A$ and $V = B$, fulfilling the condition for a $T_5$ space.
Suppose one of $A$ and $B$ contains $0$.
, suppose $0 \in A$.
Then for $A$ and $B$ to be separated, $0 \notin B^-$ and so $B^-$ has to be $\set {-1}$ or $\set 1$ or their union.
In any of these cases we have that $B$ and $S \setminus B$ are disjoint open sets containing $B$ and $A$ respectively.
So in all cases we see that if $A$ and $B$ are separated, then:
:$\exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$
where $U = A$ and $V = B$.
and so by definition $T$ is a $T_5$ space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $17$. Either-Or Topology: $1$
Category:T5 Spaces
Category:Either-Or Topology
Either-Or Topology is Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be the either-or topology.
Then $\tau$ is a topology on $T$.
== Proof ==
=== ===
Let $\UU \subseteq \tau$.
Then either:
:$\forall U \in \UU: \set 0 \nsubseteq U$ in which case $\set 0 \nsubseteq \bigcup \UU$
or:
:$\exists U \in \UU: \openint {-1} 1 \subseteq U$ in which case $\openint {-1} 1 \subseteq \bigcup \UU$
In both cases:
:$\ds \bigcup \UU \in \tau$
Note that:
:$\set 0 \nsubseteq U \implies \openint {-1} 1 \nsubseteq U$
so there does not exist the confusion of what happens if the conditions are contradictory.
=== ===
Let $U_1, U_2 \in \tau$.
By definition of either-or topology, either:
:$\openint {-1} 1 \subseteq U_1$ and $\openint {-1} 1 \subseteq U_2$
or:
:$\set 0 \nsubseteq U_1$ or $\set 0 \nsubseteq U_2$
Suppose:
:$\openint {-1} 1 \subseteq U_1$ and $\openint {-1} 1 \subseteq U_2$
From Intersection is Largest Subset:
:$\openint {-1} 1 \subseteq U_1 \cap U_2$
By definition of either-or topology:
:$U_1 \cap U_2 \in \tau$
Suppose:
:$\set 0 \nsubseteq U_1$ or $\set 0 \nsubseteq U_2$
Then:
By definition of either-or topology:
:$U_1 \cap U_2 \in \tau$
Thus by Proof by Cases:
:$\forall U_1, U_2 \in \tau: U_1 \cap U_2 \in \tau$
=== ===
From Open Real Interval is Subset of Closed Real Interval:
:$\openint {-1} 1 \subseteq \closedint {-1} 1 = S$
By definition of either-or topology:
:$S \in \tau$
Hence the result, from the definition of topology.
== Sources ==
* : Part $\text {II}$: Counterexamples: $17$. Either-Or Topology
Category:Either-Or Topology
Either-Or Topology is not Locally Arc-Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is not a locally arc-connected space.
== Proof ==
Let $\BB$ be a basis for $\tau$.
Suppose that $0 \in B$, where $B \in \BB$.
Then by definition of the either-or topology, $\openint {-1} 1 \subseteq B$.
In particular, $\dfrac 1 2 \in B$.
Let $f: \closedint 0 1 \to B$ be any injection such that:
:$\map f 0 = 0$
:$\map f 1 = \dfrac 1 2$
Since $0 \notin \set {\dfrac 1 2}$, it follows that $\set {\dfrac 1 2}$ is open in $\tau$.
As $f$ is an injection, it follows that:
:$f^{-1} \sqbrk {\set {\dfrac 1 2} } = \set 1$
By Closed Real Interval is not Open Set, $\set 1 = \closedint 1 1$ is not open in $\closedint 0 1$.
Thus, $f$ cannot be continuous.
Hence, no arc from $0$ to $\dfrac 1 2$ can exist.
It follows that $B$ is not arc-connected, and hence $T$ is not locally arc-connected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $17$. Either-Or Topology: $4$
Category:Locally Arc-Connected Spaces
Category:Either-Or Topology
Either-Or Topology is not Separable

== Theorem ==
Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is not a separable space.
== Proof ==
From Limit Points of Either-Or Topology, the only limit point of any set of $S$ is $0$.
So the only set whose closure is $S$ are $S \setminus \set 0$ and $S$ itself.
So these two are the only subsets of $S$ which are everywhere dense in $S$.
Both of these are uncountable.
Hence the result, by definition of separable space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $17$. Either-Or Topology: $3$
Category:Separable Spaces
Category:Either-Or Topology
Either-Or Topology is not T1

== Theorem ==
Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is not a $T_1$ (Fréchet) space.
== Proof ==
Let $x = \dfrac 1 2$.
We have that $V = \set x$ such that $x \in V, 0 \notin V$.
However, by definition of the either-or topology, the only open sets of $T$ containing $0$ also contain $\openint {-1} 1$, and so must also contain $x$.
So we have that $\nexists U, V \in \tau: 0 \in U, x \notin U, x \in V, 0 \notin V$.
Hence the result by definition of $T_1$ (Fréchet) space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $17$. Either-Or Topology: $1$
Category:T1 Spaces
Category:Either-Or Topology
Either-Or Topology is not T3

== Theorem ==
Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is not a $T_3$ space.
== Proof ==
Consider the set $\set 0$ which is closed from Closed Sets of Either-Or Topology.
Let $x = \dfrac 1 2$.
Let $U \in \tau$ such that $\set 0 \subseteq U$.
Then by definition $\openint {-1} 1 \subseteq U$ and so $x \in U$.
So we have found a closed set of $T$ and a point in $S$ which do not satisfy the conditions for $T$ to be a $T_3$ space.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $17$. Either-Or Topology: $1$
Category:T3 Spaces
Category:Either-Or Topology
Electric Charge/Quantum/Examples/60W Bulb at 200V

== Example of Quanta of Electric Charge ==
Consider a $60 \ \mathrm W$ light bulb running at $200 \ \mathrm V$.
Approximately $2 \times 10^{18}$ units of elementary charge flow along the filament of the light bulb every second.
== Proof ==
Let $n$ denote the number of units of elementary charge flow along the filament of the light bulb every second
The power $P$ dissipated in an electrical conductor is equal to the current $I$ flowing through it multiplied by the electric potential difference $V$ applied across it.
By definition, current $I$ equals the rate of flow of electric charge.
We have that a $60$ watt bulb with $200$ volts across it.
== Sources ==
* : Chapter $\text {I}$: Preliminary Survey: $\S 1$. Electrostatics
Category:Electric Charge
Electric Charge in Electric Circuit/L, R, C in Series

== Theorem ==
Consider the electric circuit $K$ consisting of:
:a resistance $R$
:an inductance $L$
:a capacitance $C$
in series with a source of electromotive force $E$ which is a function of time $t$.
:
The electric charge $Q$ in $K$ is given by the linear second order ODE:
:$L \dfrac {\d^2 Q} {\d t^2} + R \dfrac {\d Q} {\d t} + \dfrac 1 C Q = E$
== Proof ==
Let:
:$E_L$ be the drop in electromotive force across $L$
:$E_R$ be the drop in electromotive force across $R$
:$E_C$ be the drop in electromotive force across $C$.
From Kirchhoff's Voltage Law:
:$E - E_L - E_R - E_C = 0$
From Ohm's Law:
:$E_R = R I$
From Drop in EMF caused by Inductance is proportional to Rate of Change of Current:
:$E_L = L \dfrac {\d I} {\d t}$
From Drop in EMF caused by Capacitance is proportional to Accumulated Charge:
:$E_C = \dfrac 1 C Q$
where $Q$ is the electric charge $Q$ that has accumulated on $C$.
Thus:
:$E - L \dfrac {\d I} {\d t} - R I - \dfrac 1 C Q = 0$
which can be rewritten:
:$L \dfrac {\d I} {\d t} + R I + \dfrac 1 C Q = E$
By definition of electric current:
:$I = \dfrac {\d Q} {\d t}$
and so:
:$L \dfrac {\d^2 Q} {\d t^2} + R \dfrac {\d Q} {\d t} + \dfrac 1 C Q = E$
== Sources ==
* : $\S 2.13$: Simple Electric Circuits
Category:Electronics
Electric Field Strength from Assemblage of Point Charges

== Theorem ==
Let $q_1, q_2, \ldots, q_n$ be point charges.
Let $\mathbf r_1, \mathbf r_2, \ldots, \mathbf r_n$ be the position vectors of $q_1, q_2, \ldots, q_n$ respectively.
Let $\map {\mathbf E} {\mathbf r}$ be the electric field strength at a point $P$ whose position vector is $\mathbf r$.
Then:
:$\ds \map {\mathbf E} {\mathbf r} = \dfrac 1 {4 \pi \epsilon_0} \sum_i \dfrac {q_i} {\size {\mathbf r - \mathbf r_i}^3} \paren {\mathbf r - \mathbf r_i}$
where $\varepsilon_0$ denotes the vacuum permittivity.
== Proof ==
Let $q$ be a test charge in the vicinity of $q_1, q_2, \ldots, q_n$ at $\map P {\mathbf r}$.
For all $i$ in $\set {1, 2, \ldots, n}$, let $\mathbf F_i$ denote the force exerted on $q$ by $q_i$.
Let $\mathbf F$ be the force exerted on $q$ by the combined action of $q_1, q_2, \ldots, q_n$.
We have:
Hence the result.
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.2$ The Electric Field: $(1.4)$
Category:Electric Fields
Category:Point Charges
Electric Field caused by Point Charge

== Theorem ==
Let $q$ be a point charge.
Let $\mathbf r_q$ be the position vector of $q$.
Let $\map {\mathbf E} {\mathbf r}$ be the electric field strength due to $q$ at a point $P$ whose position vector is $\mathbf r$.
Then:
:$\ds \map {\mathbf E} {\mathbf r} = \dfrac 1 {4 \pi \epsilon_0} \dfrac {q \paren {\mathbf r - \mathbf r_q} } {\size {\mathbf r - \mathbf r_q}^3}$
where $\varepsilon_0$ denotes the vacuum permittivity.
== Proof ==
A specific application of Electric Field Strength from Assemblage of Point Charges for $1$ point charge.
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.2$ The Electric Field
Category:Electric Fields
Category:Point Charges
Electric Field induces Surface Charge in Conductor

== Physical Law ==
Let $B$ be a body made out of an electrically conducting substance.
Let $B$ be in the shape of a slab with two parallel surfaces $P_1$ and $P_2$.
Let $B$ be placed in an electric field $\mathbf E$ acting perpendicular to $P_1$ and $P_2$.
Then $B$ will develop a surface charge on both $P_1$ and $P_2$ of a polarity such as to oppose the electric field being applied.
:
== Informal Explanation ==
== Also see ==
* Definition:Induced Charge: the name for the surface charge which develops.
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.3$ Electric Fields in Matter: $1.3.3$ The macroscopic electric field
Category:Electric Field induces Surface Charge in Conductor
Category:Induced Charges
Category:Surface Charges
Category:Electrical Conductors
Category:Macroscopic Electric Fields
Electric Field induces Surface Charge in Conductor/Informal Explanation

== Physical Law ==
== Informal Explanation ==
When the electric field $\mathbf E$ is first applied, there exists a macroscopic electric field $\mathbf E'$ throughout $B$.
Under the influence of $\mathbf E$, the conduction electrons move by Coulomb's Law towards the positive source of $\mathbf E$ and away from the negative source of $\mathbf E$.
Hence the conduction electrons make $P_1$ and $P_2$ charged in the opposite polarity to the electric field at $P_1$ and $P_2$.
From Distribution of Macroscopic Electric Charge within Conductor, it is not possible for there to be different (macroscopic) electric fields throughout $B$.
Hence, away from the surfaces $P_1$ and $P_2$, there is no difference in electric charge throughout the interior of $B$.
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.3$ Electric Fields in Matter: $1.3.3$ The macroscopic electric field
Category:Electric Field induces Surface Charge in Conductor
Electric Flux out of Closed Surface/Examples/Sphere

== Example of Use of Electric Flux out of Closed Surface ==
Let $B$ be a spherical body in space with radius $R$.
Let $Q$ be the total electric charge within $B$
Then the total electric flux out of $B$ is given by:
:$F = \dfrac Q {\varepsilon_0}$
where $\varepsilon_0$ denotes the vacuum permittivity.
== Proof ==
Let the center of $B$ be located at the origin of a spherical polar coordinate system.
Let $\mathbf r$ denote the position vector of an arbitrary point $P$ on the surface of $B$.
Let the spherical polar coordinates of $P$ be denoted $\polar {r, \theta, \phi}$.
The electric charge density and electric field caused by $B$ are expressible in terms of $\polar {r, \theta, \phi}$ as:
:$\map \rho {r, \theta, \phi}$
and:
:$\map {\mathbf E} {r, \theta, \phi}$
Let us define $\d \mathbf S$ as being the infinitesimal area element of the surface of $B$ demarcated by the arcs on $B$ subtending:
:the polar angles $\theta$ and $\d \theta$ and 
and:
:the azimuthal angles $\phi$ and $\d \phi$.
where $\d \theta$ and $\d \phi$ are likewise infinitesimal.
From Area Element in Spherical Polar Coordinate System:
:$\d S = R^2 \sin \theta \rd \theta \rd \phi$
Integrating over all space, we have:
from which the result follows.
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.4$ Gauss's Law: $1.4.2$ The flux of the electric field out of a closed surface: $(1.15)$
Category:Electric Flux out of Closed Surface
Electric Flux out of Closed Surface surrounding Assemblage of Point Charges

== Theorem ==
Let $Q = \set {q_1, q_2, \ldots}$ be a set of point charges.
Let $S$ be a closed surface surrounding $Q$.
The total electric flux through $S$ is given by:
:$\ds F = \dfrac 1 {\varepsilon_0} \sum_Q q_i$
== Proof ==
From Electric Flux out of Closed Surface surrounding Point Charge:
:$F_i = \dfrac {q_i} {\varepsilon_0}$
where $F_i$ is the part of $F$ brought about by $q_i$.
The result follows from Electric Field satisfies Principle of Superposition.
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.4$ Gauss's Law: $1.4.2$ The flux of the electric field out of a closed surface: $(1.14 \, \text a)$
Category:Electric Flux out of Closed Surface
Category:Electric Flux
Category:Point Charges
Electric Flux out of Closed Surface surrounding Body with Continuous Charge Distribution

== Theorem ==
Let $B$ be a body of matter which has a continuous macroscopic charge density $\map \rho {\mathbf r}$.
Let $S$ be a closed surface surrounding $Q$.
The total electric flux through $S$ generated by the electric charge on $B$ is given by:
:$\ds F = \dfrac 1 {\varepsilon_0} \int_V \map \rho {\mathbf r} \rd \tau$
where:
:$V$ is the total volume enclosed by $S$
:$\d \tau$ is an infinitesimal volume element
:$\mathbf r$ is the position vector of $\d \tau$
:$\map \rho {\mathbf r}$ is the macroscopic charge density of the macroscopic electric field at $\mathbf r$
:$\varepsilon_0$ denotes the vacuum permittivity.
== Proof ==
Let $\d \tau$ be an arbitrary infinitesimal volume element.
$\d \tau$ can be considered as an infinitesimal point charge $\d q$.
From Electric Flux out of Closed Surface surrounding Point Charge:
:$\d F_\tau = \dfrac {\d q} {\varepsilon_0}$
where $\d F_\tau$ is the infinitesimal part of $F$ brought about by $\d q$. 
By definition of macroscopic charge density:
:$\map \rho {\mathbf r} = \dfrac {\d q} {\d \tau}$
Hence:
:$\d F_\tau = \dfrac {\map \rho {\mathbf r} \rd \tau} {\varepsilon_0}$
Integrating over all space:
:$\ds F = \dfrac 1 {\varepsilon_0} \int_V \map \rho {\mathbf r} \rd \tau$
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.4$ Gauss's Law: $1.4.2$ The flux of the electric field out of a closed surface: $(1.14 \, \text b)$
Category:Electric Flux out of Closed Surface
Category:Electric Flux
Category:Point Charges
Electric Flux out of Closed Surface surrounding Point Charge

== Theorem ==
Let $q$ be a point charge.
Let $S$ be a closed surface surrounding $q$.
The total electric flux through $S$ is given by:
:$F = \dfrac q {\varepsilon_0}$
== Proof ==
=== Lemma ===
From this lemma, we can transform the surface integral for the flux through the entire surface $S$ into a double integral between angles:
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.4$ Gauss's Law: $1.4.2$ The flux of the electric field out of a closed surface: $(1.12)$
Category:Electric Flux out of Closed Surface
Category:Electric Flux
Category:Point Charges
Electric Flux out of Closed Surface surrounding Point Charge/Lemma

== Theorem ==
Let $q$ be a point charge located at the origin of a spherical polar coordinate system.
Let $S$ be a closed surface surrounding $q$.
Let $\delta \mathbf S$ be an area element of $S$.
Let $\delta \Omega$ be the solid angle subtended by the projection of $\delta \mathbf S$ from $q$.
The total electric flux through $\delta \Omega$ is given by:
:$F = \dfrac q {4 \pi \varepsilon_0} \delta \Omega$
== Proof ==
Consider the area element $\delta \mathbf S$ at the position $P$ with position vector $\mathbf r$ whose spherical coordinates are $\polar {r, \theta, \phi}$.
Let $\delta S_p$ be the projected area of $\delta \mathbf S$ to the plane perpendicular to $\mathbf r$.
:
Let us define $\delta \mathbf S$ as being the subset of the surface of $S$ demarcated by the arcs on $S$ subtending:
:the polar angles $\theta$ and $\delta \theta$ and 
and:
:the azimuthal angles $\phi$ and $\delta \phi$.
From the above diagram:
:$\delta S_p = r^2 \sin \theta \rdelta \theta \rdelta \phi$
Let $\mathbf E$ be the electric field generated by $q$ at $P$.
According to Coulomb's Law of Electrostatics:
:the magnitude of $\mathbf E$ is:
::$E = \dfrac q {4 \pi \varepsilon_0 r^2}$
:the direction of $\mathbf E$ is along $\mathbf r$ away from the origin.
Hence the electric flux in the direction away from the origin through $\delta \mathbf S$ is:
where $\Omega$ is that solid angle so subtended.
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.4$ Gauss's Law: $1.4.2$ The flux of the electric field out of a closed surface: $(1.11)$
Category:Electric Flux out of Closed Surface
Category:Electric Flux
Category:Point Charges
Electric Force is Gradient of Electric Potential Field

== Theorem ==
Let $R$ be a region of space in which there exists an electric potential field $F$.
The electrostatic force experienced within $R$ is the negative of the gradient of $F$:
:$\mathbf V = -\grad F$
== Proof ==
The electrostatic force at a point of $R$ is in the direction of the greatest rate of decrease of electric potential.
That is, it is normal to the equipotential surfaces.
It also has a magnitude equal to the rate of decrease.
Hence the result.
== Sources ==
* : Chapter $\text {IV}$: The Operator $\nabla$ and its Uses: $2$. The Gradient of a Scalar Field
* 
* 
Category:Gradient Operator
Category:Electric Potential
Category:Electric Fields
Electric Potential over Conducting Surface is Constant

== Theorem ==
Let $S$ be a conducting surface.
The electric potential $V$ over $S$ is constant.
This can be expressed using the Laplacian:
:$\nabla^2 V = 0$
and is thus seen to satisfy Laplace's equation.
== Proof ==
== Sources ==
* : Chapter $\text I$: Introduction: $\S 1$. The origin of special functions
Category:Electrostatics
Electromotive Force in Closed Path in Electric Field is Zero

== Theorem ==
Let $\mathbf E$ be an electric field acting over a region of space $R$.
Let $\Gamma$ be a closed contour in $R$.
Then the electromotive force in $\Gamma$ is zero.
== Proof ==
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $3$. Line and Surface Integrals
Category:Electrostatics