Electric Flux and Gauss’s Law | Electronics Basics #6
Summary
TLDRThis 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
⚡ 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.
🔍 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.
🌀 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.
Mindmap
Keywords
💡Electric Flux
💡Gauss's Law
💡Differential Area (dA)
💡Dot Product
💡Closed Surface
💡Permittivity of Free Space (ε₀)
💡Symmetry
💡Electric Field (E)
💡Uniform Charge Distribution
💡Superposition Principle
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
to subscribe and for more tutorials
visit my website howtomechatronics.com
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