# Electrostatics

Styrofoam padding material is electrostatically attracted to a cat's fur.
Lightning as a result of electrostatic charge

The static electricity is the branch of physics that deals with static electric charges , charge distributions and electric fields concerned charged body.

The phenomena of electrostatics arise from the forces that electrical charges exert on one another. These forces are described by Coulomb's law . A classic example is that rubbed amber attracts particles (see story ). Even if the forces appear small, the electrical force is e.g. B. extremely strong compared to gravity . The electrical force between an electron and a proton (both together form a hydrogen atom ) is about 40 orders of magnitude greater than their mutual mass attraction .

Electrostatics is a special case of electrodynamics for static electrical charges and stationary, i.e. H. constant electrical fields over time. Electrostatics is analogous to magnetostatics , which deals with stationary magnetic fields , such as those generated by electrical currents that remain constant over time.

## history

It was already known in ancient times that certain materials such as amber, for example, attract small, light particles when rubbed with a cloth or fur (see static electricity ). William Gilbert continued the work of Petrus Peregrinus from the 13th century and found out that other substances can also be electrified by friction and developed the Versorium , an early type of electroscope . He led in his 1600 book, De magnets Magnetisque corporibus, et de Magno Magnets Tellure (German about: About the magnet magnetic body and the large magnet Earth ) to the New Latin borrowed term "electrica" for the appearances, which he in Discovered connection with amber, "electron" comes from the Greek word for amber .

## Overview

The force exerted by a given charge Q on an object is proportional to the charge q on the object. So it can be described by the equation ; E is the field strength of the electric field accompanying the charge Q.${\ displaystyle F = q \ cdot E}$

An external electric field produces different effects in electrical conductors and insulators . The free electrical charges in conductors, e.g. B. the conduction electrons of metals, macroscopically shift in such a way that the electric field disappears in the entire interior of the conductor (see Faraday cage ). This phenomenon is called influenza . On the other hand, the locally bound charges in an insulator, i.e. the electrons and nuclei of the atoms, react by shifting one another, which polarizes the insulator .

The force field F acting from an electrostatic field E on a test charge q is conservative , which means that the potential energy W of the test charge in the electrostatic field depends only on the position x of the test charge, but not on the path along which the test charge takes to x was moved. This also means that the electrostatic field can be represented as a gradient of an electrical potential . The potential energy of a test charge in the potential is . The difference between two electrical potentials corresponds to the electrical voltage . The disappearance of the electric field,, is synonymous with spatially constant electric potential = const. ${\ displaystyle \ phi}$${\ displaystyle W = q \ cdot \ phi}$${\ displaystyle E = 0}$${\ displaystyle \ phi}$

The field and thus also the potential of any charge distribution in a homogeneous insulator can easily be calculated using the regularities derived from Coulomb's law. (The field in a conductor disappears.) Such a calculation is only simple in a few cases with spatial arrangements of conductors, non-conductors and charges.

## The electric field

Illustration of the electric field and the equipotential surfaces around a positive charge in space
Field lines of a positively charged, infinitely extended plane

For the electrostatic special case of stationary magnetic fields ( ) and vanishing electrical currents ( ) follows from the Coulomb's law and the definition of the electric field for a point charge Q at the site excited electric field at the location${\ displaystyle {\ dot {\ vec {B}}} = 0}$${\ displaystyle {\ vec {j}} = 0}$${\ displaystyle {\ vec {E}} = {\ vec {F}} / Q}$${\ displaystyle {\ vec {x '}}}$${\ displaystyle {\ vec {E}}}$${\ displaystyle {\ vec {x}}}$

${\ displaystyle {\ vec {E}} ({\ vec {x}}) = kQ {\ frac {{\ vec {x}} - {\ vec {x}} '} {\ left \ | {\ vec {x}} - {\ vec {x}} '\ right \ | ^ {3}}}}$

The electric field is a directional vector field . For a positive charge it is directed exactly away from the charge, for a negative charge it is directed towards the charge, which is equivalent to the repulsion of charges of the same name and the attraction of opposite charges. Its strength is proportional to the strength of the charge Q and inversely proportional to the square of the distance from Q . The proportionality factor k (see dielectric constant ) is the Coulomb constant in the SI system of units and in the Gaussian system of units . ${\ displaystyle k = 1 / (4 \ pi \ varepsilon _ {0})}$${\ displaystyle k = 1}$

The measure of the electric field strength in SI units is

${\ displaystyle [E] _ {\ mathrm {SI}} = {\ frac {\ mathrm {V}} {\ mathrm {m}}} = {\ frac {\ mathrm {N}} {\ mathrm {C} }} = {\ frac {\ mathrm {kg} \ cdot \ mathrm {m}} {\ mathrm {s} ^ {3} \ cdot \ mathrm {A}}}}$

The field excited by a set of charges, Q i , is the sum of the partial contributions ( principle of superposition )

${\ displaystyle {\ vec {E}} ({\ vec {x}}) = k \ sum _ {i} {Q_ {i} {\ frac {{\ vec {x}} - {\ vec {x} } _ {i}} {\ left \ | {\ vec {x}} - {\ vec {x}} _ {i} \ right \ | ^ {3}}}}}$

Or in the case of a continuous space charge distribution, ρ , the integral

${\ displaystyle {\ vec {E}} ({\ vec {x}}) = k \ int {\ rho ({\ vec {x}} ') {\ frac {{\ vec {x}} - {\ vec {x}} '} {\ left \ | {\ vec {x}} - {\ vec {x}}' \ right \ | ^ {3}}}} d ^ {3} x '}$

The Gaussian law describes that the flow of the electric field through a closed surface A is proportional to the strength of the area enclosed by the surface charge Q is

${\ displaystyle \ int {\ vec {E}} \ cdot d {\ vec {A}} \ sim Q = \ int \ rho dV}$

The Gaussian integral theorem links the flow and divergence of a vector field :

${\ displaystyle \ int {\ vec {E}} \ cdot d {\ vec {A}} = \ int \ nabla \ cdot {\ vec {E}} dV}$

from which it follows that the divergence of the electric field is proportional to the space charge density:

${\ displaystyle \ nabla \ cdot {\ vec {E}} \ sim \ rho}$

A conservative electric field can by the gradient of a scalar electric potential are described ${\ displaystyle \ phi}$

${\ displaystyle {\ vec {E}} = - \ nabla \ phi}$

From which the Poisson equation follows:

${\ displaystyle \ rho \ sim \ nabla \ cdot {\ vec {E}} = - \ nabla (\ nabla \ phi) = - \ Delta \ phi}$

## Potential and voltage

Since an electric charge experiences a force in the electric field, work is done when it moves through the electric field, or work has to be done to move the charge against the electric field. From the Maxwell equations (more precisely the law of induction ) it follows that electrostatic fields are eddy-free . In the conservative field, the energy required only depends on the start and destination, not the exact route. "Eddy-free" means that the rotation of a vector field is zero (on a simply connected area): ${\ displaystyle B = {\ dot {B}} = 0}$

${\ displaystyle \ operatorname {rot} {\ vec {E}} = 0 \ leftrightarrow \ oint {\ vec {E}} \ cdot d {\ vec {s}} = 0}$

Thus, a potential energy of the charge can be defined. Since the force is proportional to the charge, this also applies to the potential energy. Therefore, the potential energy can be calculated as the product of the charge and a potential which results from the electric field.

The potential difference between two points is called electrical voltage . The product of the charge of a particle and the voltage between two points gives the energy that is needed to move the particle from one point to the other. The unit of electrical potential and electrical voltage is volts . According to the definition of potential and voltage, volt = Joule / Coulomb applies . ${\ displaystyle U = \ Delta \ phi}$

The potential is calculated as follows:

${\ displaystyle \ phi = - \ int {\ vec {E}} \ cdot d {\ vec {s}}}$

The integration limits result from the choice of the zero level . Often this is set arbitrarily at an infinite distance. A point charge that is at the location causes the potential at the location : ${\ displaystyle Q}$${\ displaystyle {\ vec {x}} \, '}$${\ displaystyle {\ vec {x}}}$

${\ displaystyle \ phi ({\ vec {x}}) = kQ {\ frac {1} {\ left \ | {\ vec {x}} - {\ vec {x}} \, '\ right \ |} }}$

In the case of a continuous space charge distribution, the electrical potential is given by the following integral :

${\ displaystyle \ phi ({\ vec {x}}) = k \ int {\ frac {\ rho ({\ vec {x}} \, ')} {\ left \ | {\ vec {x}} - {\ vec {x}} \, '\ right \ |}} d ^ {3} x'}$

If it is not possible to find an analytical solution of the integral, one can expand into a power series, see multipole expansion or Legendre polynomial . ${\ displaystyle 1 / || {\ vec {x}} - {\ vec {x}} \, '||}$

The concept of tension reaches its limits when dynamic processes occur. An induction voltage can still be defined for variable magnetic fields, but this can no longer be defined via a potential difference. Also, the energy required to move the charge from one point to another is only equal to the potential difference between the points as long as the acceleration is negligibly small, since electrodynamics suggests that accelerated charges emit electromagnetic waves that must also be taken into account in the energy balance .

## Electric field energy

In a plate capacitor there is an approximately homogeneous field. Is the charge of the one plate and the other plate corresponding to , and the size is of a plate surface , then the results field magnitude to ${\ displaystyle Q}$${\ displaystyle -Q}$${\ displaystyle A}$${\ displaystyle {\ vec {E}}}$

${\ displaystyle E = {\ frac {Q} {\ varepsilon _ {0} A}}}$, where is the electric field constant .${\ displaystyle \ varepsilon _ {0}}$

If the distance between the plates is constant , and if an infinitesimally small charge is brought from one plate to the other, then against the electric field the infinitesimal work with the amount ${\ displaystyle d}$${\ displaystyle \ mathrm {d} Q}$${\ displaystyle \ mathrm {d} W}$

${\ displaystyle \ mathrm {d} W = \ mathrm {d} F \ cdot d = E \ cdot \ mathrm {d} Q \ cdot d}$

be performed. The energy conservation lead because it must work to increase the energy of the capacitor. However, this can only be found in the electrical field. The field strength increases by the amount due to the charge transfer

${\ displaystyle \ mathrm {d} E = {\ frac {\ mathrm {d} Q} {\ varepsilon _ {0} A}}}$.

Dissolving and inserting into the work results ${\ displaystyle \ mathrm {d} Q}$

${\ displaystyle \ mathrm {d} W = \ varepsilon _ {0} \ cdot A \ cdot d \ cdot E \ cdot \ mathrm {d} E}$.

Now, however, is precisely the volume of the plate capacitor in which the complete E-field is located (in the ideal plate capacitor it can be shown that the E-field disappears outside the plate capacitor, i.e. is there ). Integrating and dividing by gives the energy density${\ displaystyle V = A \ cdot d}$${\ displaystyle {\ vec {E}} = 0}$${\ displaystyle V}$

${\ displaystyle \ varrho _ {el} = {\ frac {W} {V}} = {\ frac {1} {2}} \ cdot {\ frac {C \ cdot U ^ {2}} {A \ cdot d}} = {\ frac {1} {2}} \, \ varepsilon _ {0} E ^ {2} = {\ frac {1} {2}} DE}$,

where is the dielectric displacement . ${\ displaystyle D}$

## Occurrence, generation, applications of static charges

Occurrence in nature and in everyday life:

Generation of high voltages by transporting static charges (in research, teaching, industry):

Applications: