Electric Charges and Fields

Introduction

• Particles are the fundamental building blocks of the universe
• Particles possess an electric charge that can be either positive or negative
• "Charges" are defined as particles that carry electrical energy in the form of positive or negative charge
• This property enables attraction or repulsion between two charged particles or electrically charged particles
• Objects composed of charges are electrically charged objects
• Depending on the charge, objects placed close enough to each other experience attractive and repulsive forces without physically touching
• These forces (called "electric forces") are mediated by electric fields surrounding a charged object and extend into space, visualized using electric field lines
• A key concept is that charge is neither created nor destroyed, but transferred or transformed

Electric Field

• An electric field is an invisible force that an electric charge generates around itself
• If a charged particle is placed at a point P in space where an electric field exists, the particle will experience a force

Electric Field Formula

• Formally defined as E = F/q₀, where:
  • E is the electric field
  • F is the force acting on q₀
  • q₀ is the charge in question
• The electric field indicates the amount of force experienced by each unit of charge
• The direction of the field is the same as that of the force, but the direction changes depending on the charge of q₀
• The electric field is ALWAYS outgoing from the charge
• A uniform electric field is one that has the same value in magnitude, direction, and sense at every point

Electric Fields and Charges

• Electric fields involve:
• A charge or system of charges generating an electric field in the surrounding space
• A second charge q placed in that space experiences a force given by F = qE
• Note that this is not the electric field of the charge itself, but an existing field
• The intensity of the electric field varies with the distance from the charge
• If r represents the distance, the magnitude of the electric field is calculated as E = (1 / 4πε₀) * (q / r²)
• This calculation applies to a point charge and represents the electric field of q

Electric Field Lines

• Electric fields can be visualized through electric field lines
• Formally, an electric field line is defined as the line connecting a point P and the charge (or the center of the object) that generates the electric field
• The electric field is therefore radial

Examples of Electric Field Lines

• Emanate outwards for a positive charge
• Point inwards for a negative charge
• The density of the field lines indicates the intensity of the field
• Combining two like charges (+, +) yields a different pattern compared to two opposite charges (+, -)

Electric Field of a Dipole

• An electric dipole is the configuration of two charges of equal quantity but opposite sign, placed at a distance d from each other

Components

• The electric field lines are similar to those previously seen
  • The components of E+ and E- along the x-axis cancel each other out
• The component of E+ along the y-axis is E+y = 2 E+ cos θ
• The component of E- along the y-axis is E-y = 2 E- cos θ
• The magnitudes of E+ and E- are equal, thus E = E- = (1 / 4πε₀) * (q / (r² + a²)) E = 2E cosθ = (1 / 4πε₀) * (2qa / (r² + a²)3/2)

Dipole in an Electric Field

• Consider a rigid electric dipole of length d within an external electric field E
• At either end of the bar, because these charges are connected to an immobile object, the charges will not fly away from eachother due to the external electric field.
• The external electric field will either attract the charge or repel it
• These two forces generate:
  • A dipole moment p from the center of mass of the dipole, creating an angle θ with the electric field vector
  • A torque τ which causes the dipole to move
• The net torque can be calculated using the torque formula, considering the center of mass as two distances: distance x from one charge and d - x from the other.

Torque Formula

• In general t = F x sin θ + F(d – x) sin θ = F d sin θ
• Given that we have the electric field E and the dipole moment p and since F = qE and d=p/q, we can rewrite this as the following: t = p E sin θ
• It is the vector product of the two vectors. In the case of an outgoing torque, the sign is negative and clockwise τ = -p E sin θ.

Potential Energy of an Electric Dipole

• A dipole placed in an electric field can have potential energy
• The intensity of the same potential energy is heavily dependant on the rotation of the dipole:
• The minimum state (the force of the system is 0N) is when The dipole moment is aligned with the electric field. Note that the system is not moving and can stay there indefinitely
• The greatest state is when all of the cases are not lined up
• Since they all act like a pendulum, the change of the potential energy can be easily observed and predicted.

Energy Potential

• In physics, only the differences in potential energy matter, one can chose libarally to set the 0 point for the potential energy, it's only relevant to whether or not we are measuring it when p forms a 90 angle with it's vector with the vector of the Electric Field.
• One can find out the potential value of U with this equation, for an arbitrary theta value ∆U= -L U = − L =- - ∫90° θ t dθ = ∫90° θ pE sin θ dθ. U = -p E cos θ Given the above equation, one can note:
• The dipole is the is minimum energy when the electric field is aligned and concurrent
• The dipole is at max potential energy when the electricfield is sign inverted

Microwave Ovens

• The energy generated by the dipoles in the electric field causes the following
• A dipole placed in an external electric field generates torque
• This torque then displaces the dipole moment of the dipole
• This movement then generates energy
• Transformed to heat Water is the dipole that enables this in microwave ovens:
• Continouos alternation of the electric causes molecules to move
• Heat is generated
• Breaking the bonds between water molecules
• Cooled

Electric Flux

• Flux is generally defined as the scalar product between the vector v and a surface S
• Vector v is used to signify speed.
• Can be described a graphic that represents a flow is shown to the right Ф =VnS =VnS *cosθ

Electric Flux - Video

• Similarly, the electric flux is defined in the same way, only instead of the velocity there is the electric or static field itself such that::
• Ф =Static FieldnS,
• The amount force is constant, or same angle, the electric is constant at all points
• In cases where angles an the static field change then Integrals need to used
• Imagine dividing, infinitely subdividing the area into infetisml areas, with dS
• It now possible to calculate for each square because they all have an S and corresponding E.
• The sum must be added, and that is exactly what an intergral does.

Gauss’ Law - Video

• This must be imaged as a cage of the CHIUSA Surface The source of the fluxis NOT inside the cages The surface of the Flux cage is anything with a non-zero surface.

• Because:

• The amount of flux that leaves also corresponds to the amount the enters, so the net total is 0

• It' possible that there dominance of flow that exists internally must have source, so it exists within the cage. Because the flux net difference is Not 0.

Electric Charge

• Contextualize the above cage with the electric charge, by stating Gausses theorem: The flux of the surface is equal to total charge enclosed by the “cage” divided by Eosilon 0 Therefore one can easily apply what value can be obtained by measuring charge flow directly.

• Allows to better measure and directly look at the surface can measure the charge within.

Electric Field Flux

• The flow of the of the field is when applied to surface A more mathematically inclined one can immediately measure at that surface

• An easy way to start is to make each value a flat plane and field

• Surface are can then immersed into a uniform “E” static magnetic field, such that the E field penetrates

• Delimit a point that the point the E exits and call it change of area small or AA

• One can apply correspond the x and y coordinates the points

Surface

• The component value intersecting face has a magnitude Ex COS(0)

• The expression can then be easily solved. Assuming the vector is applied perpendicular to the areas.

• All small amounts must then added up to receive the true, or full effect. Its is an integral. Ф = ∫ staticmagnetic *Area

• If facing in a positive direction then, its flux is positive

• If facing into a negative direction, it is a negative flux

• If parallel it’s a zero flux

• Because non flat plane and static magnetic field non-uniform then its must be applied within general scenario.

Principals

• In order to correctly derive this one must consider area, in order to measure.

• Angle is measure relative to AA, negative, positive or 0

• Apply the above method.

• Then one must reduced area to very precision and apply intergral.

• Thus it is possible to easily measure area

Gauss’ law

• Relates to field to a point in space

• Relates to with change enclosed in the by point

• Expressed, the “StaticMagnetic Force * epsilon* = to change enclosed”

This law is also equal to what previously stated

Causses' law and Coulumabs law

• The most common method is to use the Static electric field due to charge

• Since its field is symmetrical around a shape then is possible to enclose with Guass

• Consider a positiving charge, the coating, “covering” at the centre is R

• Therefore module and set all point point outwards evenly Using vectorial notation, then the static field has value of 0 (Theta) degrees.

Constant

• It follows static magnetic * AA = Charge total , this is EQ 23. 8

• Each Static E, remains the same independent of the area elements of the field, hence we intergrate the magnetic field, by brining it outside

This has a formula of : StaticmagneticForce change small = Charge total

• And if we use all elements: Integral value represents surface, with length 4 pi Radius squared. EQ 23.0 This allows us to determine using EQ 23.10 via the coulombs equations

Isolated Conductive

• Gausses law demonstrates an important relative related conductive Because one only has to supply to isolasted isolator

• No surface is capable of moving the conductive

• Since each of the sign are are capable of being displaced they push till equiliburm is attained

• Gauss therm allows us to understand for there is no net static on the other side without the inclustion, there’s many vector E, its static charge in null is the case or point inside

• Because Static Charge will always create to flow out of the sphere.

• Therefore Staticmagnetic Field by the method of Gauss result’s “no change”

• Because the surface that is created Gauss has to also become surface

• Because charge net enclosed is to 0 , there the amount charge enclosed from outside of EQ , means its by law should be on the conductor

EQ

• Given that internal field , the cavity is generated, static change is supposed to be be set to an internal environment than gauss should be set to 0, or it not within Gauss laws.

Non Surface

• In norm conditions, the surfaces does not have uniform or singular value.

• The challenge and measuring static field and applying change using Gaussian laws its immediately measurable

• For this demostrated that static is in eletric magnetic equiliburim.

• There this exists 2 cases

• One portion of the area is low enough to be nearly planar

CYLINDRICAL

• The surface of the guass is perpendicular, “or parralled “ to the shape with one part being outside the conductive.

• Consideration to the direction to the incoming or Static with AA is the most efficient. Because this causes the surface with charge moving side to side this implies its perpendicular to magnetic

Total EQ

Because gauss with be able to determine Static E. AA the formula = 0 for it is.

• At a point The net effect of Static flowing, or crossing over with formula of QA . Guass now now turns

Formula E* area= surface area . E= value / Epsilon

Guass Laws: symmetrical cylindrical

• Understand the of the plastic, and try to find the modules

• If Symmetrical cylindrical then Gauss law can be used.

• Use the with a cylindrical length of the h, based on the fact that is radially outwards on the plane and parallel field. , we are able to use this formula

• Because The field is uniform thus there out the integral and it a surface we can use for a cylindrical 2 * pi Radius Height Because Gausses in second formula has charge , we use this value. Resulting that it has length/ surface and one must it by length epsilon *Q = Surface change

• Che and we do this in the following manor . epsilon( Q) surface change à H results with a change and gives us a value of epsilon * pir adus E with length surface change

Gausses law

Lets imagin, apply some value with charge at fixed radius, we can apply that gauss at “r”

Now what will it be to have cylidrically on its plane at location r

Because value is zero.

Area will have is the the field static

The other cases we cant get values

Static field+ A + A surface = change total

E= charge value / 2 surface epsilon

Guass symetrical spherically

• Surface symmetrical

With the field Static at r