Electric Flux, Gauss's Law & Electric Fields Video Lecture | Electromagnetic Fields Theory (EMFT) - Electrical Engineering (EE)

in this video we're going to focus on
electric flux so let's talk about what
it is and how to calculate it imagine if
we have a surface with an electric field
emanating out of the surface and that
surface has an area which we'll call a
the product of this perpendicular
electric field it's perpendicular to the
surface the product of the electric
field times the area is equal to the
electric flux Phi is the angle between a
normal line and electric field because
the normal line and electric field are
parallel to each other
the angle Phi is equal to zero and
cosine zero is one now what if the
electric field is not perpendicular to
the surface what happens in that case so
let's draw the normal line and here's a
line parallel to the surface and let's
say the electric field is at an angle so
this is the angle Phi this is the angle
theta so Phi is between the normal line
and electric field vector in this case
the electric flux is going to be equal
to e a cosine Phi the electric flux has
its maximum value when this angle is
zero now what happens if the electric
field vector is parallel to the surface
or perpendicular to the normal line
so let's say the electric field vector
is here in this case the angle between a
normal line and the electric field is 90
the electric flux is equal to EA cosine
Phi cosine 90 is equal to zero so the
electric flux is zero in this case let's
try this problem what is the electric
flux through a sphere of radius four
meters that contains a 50 micro Coulomb
charge and negative 50 micro Coulomb
charge at its center so let's draw a
picture so here we have a sphere and
there's a positive charge inside the
sphere now the electric field will
always be perpendicular to the surface
so the flux is simply the electric field
times the area in the case of a sphere
it's always perpendicular to the surface
so it's going to be e times a where a is
the surface area of a sphere so let's
say the sphere has a radius R the
surface area is going to be 4 pi times R
squared now the electric field produce
by a sphere you can treat as if it's a
point charge particularly if you want to
calculate the electric field outside of
the sphere and so that's equal to K
times Q which is the charge in the
sphere divided by R squared
so notice that we can cancel r-squared
NK is 1 over 4pi times epsilon sub not
so we can also cancel the 4pi so
therefore the electric flux and the
sphere is equal to the charge enclosed
by the sphere divided by epsilon sub
naught this equation is basically
associated with Gauss's law which states
that the electric flux is proportional
to the charge enclosed by the object so
it's just going to be 50 times 10 to the
minus 6 divided by 0.85 times 10 to the
minus 12
so you should get 5.65 times 10 to the
6th power
Newton's time square meters per Coulomb
so that's the electric flux now here's
the question for you is the answer
positive or negative the answer is
positive because if you place a positive
charge inside the sphere the electric
flux is an outward flux for Part B the
magnitude is the same but if you put a
negative charge inside the sphere then
the electric field enters the sphere so
we have a negative in word electric flux
so the answer for Part B is the same
value but negative five point six five
instead of positive five point six five
now let's say if we have a horizontal
disc let me draw it better and it's the
shape of a circle and this this has a
radius of let's say three meters here's
a normal line and let's say the electric
field is in this direction it passes
through the disc all throughout the
surface of the disc but in that
direction
the angle between the electric field
vector and the plane of the disc let's
say it's 30 degrees calculate the flux
the electric flux passing through the
surface feel free to pause the video and
let's say in the electric field
let's give it a value of 100 Newton's
per Coulomb so the electric flux is
equal to the electric field times the
area of the disc times cosine of Phi now
the angle that we have is Theta and not
5/5 this angle is the complementary
angle so Phi is going to be 90 minus 30
so it's 60 degrees the area of the disc
is PI R squared so this is going to be e
times PI R squared times cosine Phi the
electric field is 100 R squared is 9 3
squared is 9 and then cosine 60 cosine
60 is one-half and 100 times 9 is 900
half of 900 450 so the electric flux is
450 pi so this is the answer
now let's try another problem
so let's say if we have a cube that
looks like this
now we're going to place a positive
charge at the center of the cube what is
the total electric flux that emanates
outside of the cube how can we calculate
the total electric flux now for a
problem like this you don't want to find
the electric field that passes through
each surface that's going to be
difficult because in some areas the
electric field will be at an angle and
so that's going to be very difficult the
best way to answer this problem is to
use Gauss's law where the electric flux
is simply equal to the charge that is
enclosed by the surface divided by
epsilon sub naught this equation makes
these problems a lot easier to solve so
let's say the charge is 30 market looms
so all you got to do is take the 30
markleham charge which is 30 times 10 to
the minus 6 and divided by epsilon sub
naught so that's going to give you the
total electric flux that passes through
the cube and this is an outward electric
flux so the answer is positive so you
should get 3.39
times 10 to the 6 so that's the electric
flux now what if we wanted to calculate
the electric flux through one of the six
faces and the cube if you need to do
that simply multiply your answer by one
over six or divided by six so the
electric flux that passes through each
surface or each face of the cube is five
point six five times ten to the five
Newton's times square meters per Coulomb
now let's draw another cube
so this time there's not going to be any
charge inside the cube now the length of
each side of the cube is three meters
and let's say that there's an electric
field that passes through the sphere
from the bottom face to the top face and
let's say the electric field is also
perpendicular to the plane of the top
part of the cube in the bottom part of
the cube which means that it's parallel
to the normal line so what is the
electric flux through this object how
would you calculate it so the electric
flux on the right side and on the left
side is zero because it's not
perpendicular to the plane of those
sides the electric flux in the front and
in the back is also zero now what about
the top and the bottom face what is the
electric flux in those regions so let's
focus on the top part the electric flux
that passes through it is going to be
the product of e times the area of that
surface so it's going to be positive
EA the reason why it's positive is
because the electric field is leaving
the closed object it's escaping from it
so it's an outward flux so it's positive
but what about the bottom surface notice
that the electric field vector enters
the object so it's an inward flux so the
electric flux is e times a but it's
going to be negative EA so to find a
total electric flux it's the sum of
these two values positive EA and
negative EA they cancel and so the net
electric flux
is therefore equal to zero because the
flux that enters is equal to the flux
that leaves and it makes sense because
based on Gauss's law the total electric
flux that passes through a closed object
is equal to the charge enclosed divided
by epsilon zero now there is no charge
enclosed in this object so Q is zero and
if Q is zero the flux has to be zero