Electric Flux and Gauss’s Law | Electronics Basics #6

Summary

This video provides an in-depth explanation of electric flux and Gauss's law. It begins by defining electric flux as the flow of an electric field through a surface, and uses examples to show how the angle between the electric field and the surface affects flux. The video also explains how Gauss's law relates the electric flux through a closed surface to the enclosed charge. Several symmetries, such as spherical, cylindrical, and planar, are explored to demonstrate how the law can be applied to various geometries. The video concludes with the electric field between parallel plates.

Takeaways

• 🔋 Electric flux measures the amount of electric field that penetrates a surface, which can be open or closed.
• 📏 The electric flux through a small area element is calculated using the dot product of the electric field and the area vector, considering the angle between them.
• 🌀 The total electric flux is found by integrating the differential flux over the entire surface, resulting in a scalar quantity that can be positive or negative.
• 🔄 The unit of electric flux is newton meters squared per coulomb (N·m²/C), indicating the amount of field per unit charge.
• 📚 Gauss's Law states that the total electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of free space (ε₀).
• 🌐 For a symmetric charge distribution, such as a point charge within a spherical surface, the electric field is radially outward and uniform, simplifying flux calculations.
• 📉 The electric field's magnitude at a point is independent of the distance from the charge when using a symmetrical Gaussian surface, like a sphere around a point charge.
• 🔄 Understanding different types of symmetry (spherical, cylindrical, and planar) is crucial for applying Gauss's Law to calculate electric fields in various scenarios.
• 📊 The electric field's behavior changes based on the symmetry and charge distribution, as seen with hollow spheres, infinite lines of charge, and infinite planes.
• 🔧 Gauss's Law is a fundamental principle in electromagnetism, providing a way to calculate electric fields due to various charge configurations without direct integration.

Q & A

What is electric flux?

Electric flux is the rate of flow of an electric field through a given surface. It represents the amount of electric field passing through that surface, which can be either open or closed.

How is electric flux through an open surface calculated?

Electric flux through an open surface is calculated using the dot product of the electric field (E) and the differential area (dA), multiplied by the cosine of the angle (θ) between them. Mathematically, the total flux is the integral of E · dA over the surface.

What determines if the electric flux is positive or negative?

If the electric flux goes from inside to outside a surface, it is positive. If it moves from outside to inside, the flux is negative.

What happens to electric flux when the surface is perpendicular to the electric field?

When the surface is perpendicular to the electric field (angle θ = 0), the electric flux is maximized, and the flux equals E times the area (dA).

How is the electric flux affected when the angle between the electric field and the surface is 90 degrees?

When the angle is 90 degrees, the cosine of 90 degrees is zero, resulting in zero electric flux, meaning no electric field passes through the surface.

What is Gauss's Law?

Gauss's Law states that the total electric flux through a closed surface is proportional to the total charge enclosed within the surface, divided by the permittivity of free space (ε₀).

How does Gauss's Law apply to a point charge inside a sphere?

For a point charge at the center of a sphere, the electric field is radially outward, and the total flux through the sphere's surface equals the charge (q) divided by ε₀, regardless of the sphere's radius.

What happens to the electric field inside a hollow sphere with no charge inside?

If there is no charge inside the hollow sphere, the electric field inside the sphere is zero, and there is no net flux.

How is the electric field determined for cylindrical symmetry?

For cylindrical symmetry, like an infinite line of charge, the electric field at a distance (r) from the line is calculated using Gauss's Law, resulting in E = λ / (2πrε₀), where λ is the linear charge density.

What is the electric field between two parallel plates with opposite charges?

Between two parallel plates with opposite charges, the electric field is uniform and equal to the surface charge density (σ) divided by ε₀. The field points away from the positive plate and towards the negative plate.

Outlines

00:00

⚡ Understanding Electric Flux and Gauss's Law

In this section, the concept of electric flux is introduced as the rate of flow of the electric field through a given surface, which can be either open or closed. The electric flux is defined as a dot product of the electric field vector and the area vector, influenced by the angle between them. Various scenarios illustrate the calculation of flux: when the area is perpendicular, at an angle, or parallel to the electric field. Additionally, the calculation for flux through a closed surface is discussed, including the conventions for normal vectors and how the total flux can be positive, negative, or zero depending on the electric field distribution.

05:01

🔍 Applying Gauss's Law with Charges and Symmetry

This paragraph extends the understanding of electric flux by applying Gauss's Law, which states that the electric flux through a closed surface is proportional to the enclosed charge divided by the permittivity of free space. Examples demonstrate how flux is independent of the surface size when centered around a point charge. The discussion covers how symmetry plays a crucial role in applying Gauss's Law and the conditions under which it simplifies calculations of electric fields for spherical, cylindrical, and planar symmetry. The importance of symmetry in determining the electric field distribution within and around various surfaces is highlighted.

10:02

🌀 Exploring Symmetry: Spherical, Cylindrical, and Planar Cases

The final section dives deeper into the three primary types of symmetry—spherical, cylindrical, and planar—and how they simplify electric field calculations using Gauss's Law. It explores specific examples, including a uniformly charged hollow sphere, an infinite line of charge, and an infinite plane with uniform charge density. For each scenario, the appropriate Gaussian surface is chosen to leverage symmetry and simplify the integral calculations. The behavior of the electric field in more complex setups, such as between two parallel plates with opposite charges, is analyzed using the superposition principle. The discussion emphasizes how field lines and magnitudes are influenced by these symmetrical configurations.

Keywords

💡Electric Flux

Electric flux is the rate of flow of the electric field through a surface. It represents how much electric field passes through an area and can be calculated for both open and closed surfaces. In the video, the flux is illustrated using a rectangular surface in an electric field, and the amount of flux depends on the angle between the electric field and the surface.

💡Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed within the surface. It states that the total electric flux through a closed surface is proportional to the net charge inside the surface, divided by the permittivity of free space. The video uses a sphere with a charge inside to demonstrate how Gauss’s Law is applied.

💡Differential Area (dA)

The differential area (dA) is a small element of an area used in calculating the electric flux through a surface. In the video, the area is broken into small elements to understand how the electric flux behaves over small surfaces, and the vector direction of dA is always perpendicular to the surface element.

💡Dot Product

The dot product is a mathematical operation used to calculate the flux as a product of the electric field and the area vector. In this case, the dot product of the electric field (E) and the area element (dA) gives the flux through a small surface. The video emphasizes the role of the angle between these vectors in determining the total flux.

💡Closed Surface

A closed surface is one that completely encloses a volume, such as a sphere. In the context of Gauss’s Law, the video explains how electric flux through a closed surface is related to the total charge inside. The video uses a closed sphere to show how the electric flux can be positive, negative, or zero depending on the charge distribution.

💡Permittivity of Free Space (ε₀)

The permittivity of free space (ε₀) is a constant that appears in Gauss’s Law and other electromagnetism equations. It represents how much resistance is encountered when forming an electric field in a vacuum. The video explains how it is used in calculating the total electric flux and shows its value in the Gauss’s Law equation.

💡Symmetry

Symmetry plays a crucial role in simplifying the calculations for electric fields using Gauss’s Law. In the video, three types of symmetry—spherical, cylindrical, and planar—are discussed. These symmetries help determine how the electric field behaves around charged objects, making it easier to apply Gauss's Law to different shapes.

💡Electric Field (E)

The electric field (E) represents the force exerted by a charge and is used in the calculation of electric flux. The video demonstrates how the electric field interacts with different surfaces, influencing the flux. The electric field’s strength and direction are central to understanding how charges affect their surroundings, as shown with spherical and cylindrical surfaces.

💡Uniform Charge Distribution

A uniform charge distribution means that the charge is spread evenly over a surface or volume. In the video, this concept is used in the examples of a uniformly charged sphere and an infinite line of charge. This uniformity allows for easier application of Gauss's Law, as it simplifies the electric field calculations.

💡Superposition Principle

The superposition principle states that the total electric field created by multiple charges is the vector sum of the electric fields produced by each charge individually. The video explains how this principle applies to the electric fields between two parallel plates with opposite charges, resulting in the cancellation of fields outside the plates.

Highlights

Introduction to electric flux and Gauss's law as key concepts in understanding electric fields.

Electric flux is defined as the rate of flow of an electric field through a surface, which can be open or closed.

Illustration of electric flux through an open surface using a rectangle in a uniform electric field.

The flux calculation involves dividing the area into small elements and using a dot product of electric field and area vector.

Positive flux occurs when the electric field flows from inside to outside, and negative flux when it flows from outside to inside.

The unit of electric flux is Newton meters squared per Coulomb.

In scenarios where the electric field is parallel to the surface, the electric flux becomes zero.

For closed surfaces, the normal vectors point outward, and the total flux through the surface is determined by the net electric field.

Gauss's law states that the total flux through a closed surface is equal to the sum of all enclosed charges divided by permittivity of free space.

In spherical symmetry, the total electric flux is equal to charge divided by permittivity, independent of the surface size.

The electric field outside a charged hollow sphere decreases with distance, following an inverse square law.

In cylindrical symmetry, the electric field around an infinite line of charge is proportional to the linear charge density.

For planar symmetry, the electric field near an infinite plate of charge is uniform and independent of distance.

In a system of two parallel charged plates, the electric field between them is constant and zero outside due to field cancellation.

Superposition principle helps in understanding the total electric field created by multiple charge distributions, such as two parallel plates.

Transcripts

hello dan here from

howtomechatronics.com

in this video we will learn about

electric flux and gauss's law

in order to understand gauss's law first

we need to understand the term electric

flux

electric flux is the rate of flow of

electric field through a given surface

it is the amount of electric field

penetrating a surface and that surface

can be open or closed

first we will take a look at an example

of electric flux through an open surface

these red lines represent a uniform

electric field

we will bring in that field a rectangle

which is an open area and we will divide

the area into very small elements each

with size d a

the a is called a differential of area

now we are going to make the area d a a

vector with a magnitude d a

the vector direction is always

perpendicular to the small element d a

the electric flux that passes through

this small area d phi also called a

differential of flux is defined as a dot

product of the magnitude of the electric

field e and the magnitude of the vector

area d a times the angle between these

two vectors theta

the total flux is going to be the

integral of d phi or the integral over

the entire area of e dot d a

it is a scalar quantity and the end

result can be positive for negative

if the flux is going from the inside to

the outside we can call that a positive

flux and if it's going from the outside

to the inside that's a negative flux

the unit of electric flux is newton

meters squared per coulomb

to get a better understanding of what

electric flux is i will bring into this

electric field three rectangles in fact

these rectangles represent one rectangle

with different orientations now let's

explain the flux through each one of

those open areas

in the first case the area is

perpendicular to the electric field and

the angle between their vectors theta is

zero cosine of zero is one so the

electric flux is going to be e times d a

here we have the maximum flux

in the second case the angle between e

and d a theta is 60 degrees and cosine

of 60 degrees is 0.5 so the electric

flux will be half of e times d a

in the third case the area is parallel

to the electric field which means that

their vectors are perpendicular to each

other and the angle theta between them

is 90 degrees cosine of 90 degrees is

zero so the electric flux here will be

zero

this means that nothing goes through

this rectangle so the flux is zero

now let's take a look at a surface that

is completely closed

how do we define flux

here we can put some normals the a's in

different directions

by convention the normal to the closed

surface always points from the inside to

the outside

now we can calculate the total flux

going through this closed surface

the total flux is equal to the integral

of d phi over the entire surface which

we write as the integral over the closed

surface of e dot d a

the total flux can be positive negative

or equal to zero

if the same amount of flux is entering

and leaving the surface we have zero

total flux if more flux is leaving than

entering the surface then the total flux

is positive opposite if more flux is

entering than leaving the surface we

have negative total flux

let's take a look at another example and

see how the electric flux is related to

gauss's law

we have a point charge plus q in the

center of a sphere with radius r

now we will take a small segment d a

which vector is perpendicular to the

surface and is radially outward

the electric field generated by q at

this point is also radially outward

this means that d a and e anywhere on

the surface of this sphere are parallel

to each other and the angle between them

theta is zero and cosine of zero is one

the differential of flux through the

small surface area d phi is equal to e d

a

the total flux phi is going to be the

integral of d phi which is the integral

over the closed surface eda

the magnitude of the electric field

everywhere is the same because the

distance from the charge is the same at

each point so we can pull that out of

the integral and we are left with e

times a

the total area of the sphere is 4 pi r

squared and the total flux through this

closed surface is simply e times 4 pi r

squared

from the previous videos we know that e

is equal to k times q divided by r

squared which is equal to q divided by 4

pi epsilon not r squared

here we can cancel out 4 pi r squared

and we can notice that the total flux is

equal to q divided by epsilon naught

where epsilon naught is the permittivity

of free space

the flux doesn't depend on the distance

r we would get the same result no matter

the size of the closed surface around

the point charge

what if we bring more charges inside the

closed surface

the equation should also hold for any

system of charges inside

this leads us to the gauss's law which

says that the electric flux going

through a closed surface is the sum of

all charges q inside the closed surface

divided by permittivity of free space

epsilon naught

if the flux is zero that means there is

no net charge inside the shape

there could be positive and negative

charges inside the shape but the net is

zero

no matter how weird the shape gauss's

law always holds as long as there is a

perfect symmetry in the charge

distribution inside the surface

so in order to calculate the electric

field you need a symmetry and there are

three types of symmetry spherical

cylindrical and planar symmetry

we will start with the spherical

symmetry this is a thin hollow sphere

with radius r and we will bring a

positive charge q into the thin shell

which is uniformly distributed

now we need to find the electric field

inside the sphere at a distance r1 from

the center and outside the sphere at the

distance r2 from the center

to do that we need to determine our

gaussian surface

in this case we will choose concentric

spheres as gaussian surfaces one smaller

with radius r1 and other larger with

radius r2

now we need to use two symmetry

arguments that will help us calculate

the electric field

the first symmetry argument shows that

the magnitude of the electric field is

the same at any point since the charge

here is uniformly distributed

the second symmetry argument shows that

if there is an electric field it must

point either radially outwards or

radially inverse in this example we have

a positive charge which means that the

field is pointing outwards

from the previous equations we know that

the surface area of a sphere which is 4

pi r squared times the magnitude of the

electric field e is equal to the charge

inside the sphere q divided by the

permittivity of free space epsilon not

however we don't have a charge inside

the smaller sphere so the electric field

is zero

if a closed surface has no net charge

enclosed by it then the net flux to it

will be zero

now let's see what happens with the

larger sphere

the symmetry arguments hold for this

sphere as well

but if we take a look at the equation we

will notice that q is not zero because

there is a charge inside that sphere

so the magnitude of the electric field

will be equal to the charge and closed

divided by 4 pi epsilon not r2 squared

if we draw a graph with the distance on

the x-axis and the magnitude of the

electric field on the y-axis we can

notice the following

up to the point r which is the radius of

our initial sphere we have no electric

field but then it reaches its maximum

value and decreases as the distance

increases

second type of symmetry is cylindrical

symmetry

let's say we have an infinite line of

positive charge with uniform linear

charge density lambda and we want to

figure out what the electric field is at

some point above the line at distance r

here we will choose a cylinder as a

gaussian surface with a center along the

line of charge

we don't have an electric field through

the end cups the electric field will be

pointing out through the walls of the

cylinder

also we have symmetry here which allows

us to use the gauss's law in order to

calculate the electric field

we can calculate the flux using the same

equation that we used previously but now

we need to find the surface area around

the cylinder including the wall without

the end caps for that purpose we need to

cut the cylinder along its length and we

will find out that the area is equal to

2 pi rl

so 2 pi rl times e is equal to the

charge enclosed divided by epsilon

naught the charge density lambda is the

total charge q per length l so the

charge enclosed is equal to lambda l

so 2 pi r l e is equal to lambda l

divided by epsilon naught the electric

field is equal to lambda l divided by 2

pi r l epsilon naught

l cancels out so the electric field is

equal to lambda divided by 2 pi r

epsilon naught the last type of symmetry

is planar symmetry

in this example we have a flat infinite

large horizontal plate

we'll bring a charge onto this plate

with a uniform charge density sigma

sigma is actually an amount of charge

per area and is expressed in coulombs

per squared meter

now we want to calculate the electric

field in the surrounding area of this

plate let's say at a distance d

in this case we are going to choose a

cylinder again as a gaussian surface

the cylinder intersects the plate and in

that intersection we have a charge

enclosed

in order to be able to calculate the

electric field we need to meet three

conditions

first the cylinder end cups with an area

a must be parallel to the plate

second the walls of the cylinders must

be perpendicular to the plate

third the distance from the plate to the

end cup's d must be the same above and

below the plate

now that we met the symmetry

requirements we can calculate the

electric field using the gauss's law

we are not going to have any horizontal

components of the electric field only

vertical coming out of the two end cups

sigma is equal to the charge divided by

the surface and from this equation we

can see that the charge q is equal to

sigma times the area

the flux from the wall of the cylinder

is equal to zero so the total flux

consists of two components the flux

through the top cup plus the flux

through the bottom cup of the cylinder

this is equal to q enclosed divided by

epsilon naught or sigma a divided by

epsilon not

but also the flux through the top and

the flux through the bottom can be

expressed as ea so the total flux is

equal to 2 ea

finally the electric field is equal to

sigma divided by 2 epsilon not

if the plate is positively charged the

electric field would be pointing outward

and if it is negatively charged the

electric field will be pointing inwards

if we draw a graph with the distance d

on the x-axis and the electric field on

the y-axis we can notice that the

electric field has a constant value of

sigma over 2 epsilon not and it doesn't

depend on the distance from the plane

now let's take a look at another more

complex situation of two infinitely

large parallel plates the first plate

has a surface charge density plus sigma

and the plate below has a surface charge

density minus sigma the distance between

them is d so what is the electric field

anywhere in the space

the positively charged plate has an

electric field pointing away from the

plate equal to sigma divided by two

epsilon naught it doesn't depend on the

distance from the plate so it continues

below

the negatively charged plate has an

electric field pointing towards the

plate also equal to sigma divided by 2

epsilon naught in order to calculate the

total electric field we are going to use

the superposition principle by adding

vectors

the vectors that are in the opposite

direction cancel out so the electric

field there is zero the vectors between

the plates are in the same direction so

the electric field is sigma divided by

epsilon naught

here's how the electric field lines

would look like they will be pointing

away from the positively charged plate

and towards the negatively charged plate

and the electric field outside will be

zero

okay so that would be all for gauss's

law i hope this video was helpful and

you learned something new don't forget

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