Displacement Vectors - Electrical Engineering Video Lecture - Electrical Engineering (EE) | Best Video for Electrical Engineering (EE)

ah ha ha they just feed all of us really
teach
you
the last time we had almost reached to
the definition of the displacement
vector and today we will finally reach
there let me review what we done
we had a material some non conducting
material a uniform electric field is
applied and because this material is
made of atoms the negative charge in the
cloud of the electrons moves in the
direction of the electric field and so
there is a layer of negative charge at
the top a layer of positive charge at
the bottom and we have already worked
out that if you have equal amounts of
negative and positive charge a uniform
electric field appears that is in the
opposite direction this is what I will
call an induced electric field now how
much was this induced electric field
where we work that out we say that
supposing the slab of electrons got
shifted upwards due to this electric
field by how much will it get shifted it
gets shifted upwards by an amount D
where D is here the nucleus you have the
net electron cloud and they are
displaced with respect to each other
that distance is d so you take this
distance D
multiplied by the charge of the electron
with a minus sign and then you multiply
it by the number of electrons per unit
volume that is a number density that
gives you the amount of charge per unit
area that is present why because minus E
is the charge and n into D is the volume
so this is the charge density so there
is a charge density equal to minus n EE
D and as in charge density equal to plus
n e D at the bottom and it is this
charge density that causes an electric
field we can apply Gauss's law supposing
we had only this charge no applied
electric field then Gauss's law tells us
there is no electric field outside there
is an electric field inside an electric
field inside is directly related to the
amount of charge enclosed so this is
equal to e induced
he IND so en is negative in sign that is
it is pointing upwards that is not very
useful because we do not know D we know
the charge of the electron
we know the number density of electrons
in the material but we do not know D
however from a model of what happens to
an electron cloud we know that the
stronger the applied electric field the
more this cloud shifts so we can plot
saying applied electric field E external
and we can plot the amount of D and we
find that is approximately linear it is
very linear for small values of applied
electric field but for larger values it
deviates from being Li so in this region
where it is in fact truly linear
we can treat this D and therefore this
induced electric field is proportional
to the applied electric field now there
are materials where you can apply an
electric field this way and the induced
electric field will be in some other
direction these are called
anisotropic materials
we won't touch anisotropic materials
right now so we will assume that if you
apply an electric field the electron
cloud shifts in the direction opposite
direction to the electric field and an
isotropic material would be one where I
have an electric field this way my
nucleus is here my electron cloud
shifted in this direction even though
the electric field is this way the
electron cloud deanship directly
backwards it shifted sideways this can
happen because of a crystal structure of
the material and these are more
complicated materials but very very
important from an engineering point of
view for now we will decide to study
only systems where the induced electric
field is exactly opposite in direction
to the applied electric field so we can
write e induced is equal to minus e n
times D and D is a vector now is equal
to some coefficient I called it Alpha
last time times the external applied
electric field this alpha is clearly
negative because the induced electric
field tries to cancel the external
electric field
let us get back to the original problem
I have a material and I apply an
electric field due to this applied
electric field some kind of charge has
developed actually charge is developed
everywhere but the only place were
charged does not cancel out is the top
and the bottom now supposing we knew how
much this charge was if we knew exactly
what this charge was we can actually
solve for the electric field we can say
that the electric field E is equal to 1
over 4 PI epsilon naught times the total
charge integral of Rho total divided by
R minus R prime squared times R minus
1/2 times unit vector DV
what I mean by this is if I actually
knew how much bound charge was developed
by the applied electric field I can go
back to Coulomb's law and Coulomb's law
is always correct if I knew all the
charge the only problem is I do not know
all the charge all right now there is
one other thing if I take this II
induced it is proportional to the
electric field so I can take the
divergence of both now this expression
tells me divergence of E is equal to the
total charge density divided by epsilon
naught no matter whether the material or
not if I could somehow calculate the
total charge not just the charge I put
there but including the charge that is
induced then this is true let me take
divergence of this so if I take
divergence of this equation I will get
divergence of e induced is equal to
divergence of alpha e
which can be broken though it is not
very useful to do it alpha times
divergence of E + P dot grad alpha the
only reason for doing it is we can see
that divergence of E is already related
to something the road total over epsilon
not so you could write it as alpha Rho
total over epsilon naught plus e dot
gradient alpha
now what is more useful is to look at
this picture picture we've been drawing
for a while now for example to look at
this and say that the only place charge
develops is where this factor-alpha
changes abruptly because we look out
here this is air alpha equals 0 there is
no induced charge this is my glass alpha
equals something this is air again alpha
equals zero so if you look wherever
alpha is constant nothing is happening
but where alpha is zero changing from
constant to zero that is where there is
charge developing so we could think of
this as representing a charge itself
wherever e induced changes abruptly that
is a place where there must be charge
and this picture agrees there because I
apply Gauss's law he induced went from
zero to positive this charge and closed
he induced remained constant there is no
charge enclosed he induced went from
positive to zero this is negative
charging closed so it looks like
divergence of e induced represents some
sort of charge in fact it represents the
induced charge so I will write this
these two equations divergence of E is
equal to total charge density over
epsilon not divergence of e induced is
equal to Rho bound power epsilon not the
B means this is the induced charge so it
is it is bound to the individual atoms
of the material it is not something V
placed there it is something that
appears only in the press
is of an applied field the problem in
engineering the problem in all of
electromagnetics is we do not know it
this row bound is simply not known to us
so what we are going to do is instead of
working with row total which is a sum of
the charges we have placed which I am
calling a row free and the induced
charges row bound or working with E
induced which deals with row bound I
would like to work with some kind of
electric field that only relates to row
free that is only row I know about the
remaining rows are self consistently to
be determined so I am going to take the
difference because I can see I can
cancel out row bound so I will take
divergence of e minus e induced is equal
to row free divided by epsilon not
what is more we have already written
down that we induced is equal to some
alpha times e now alpha is not the
symbol that is used but I am just show
keeping it there for placeholder so I
can write this as divergence off and I
am going to take this epsilon naught to
the other side it is harmless it is a
constant times 1 plus alpha times the
electric field is equal to root 3 now
this probably looks like magic but what
it is is actually a kind of a fancy sort
of cheating and I will go over it again
so that you feel more comfortable with
what we have done but we have what we
have achieved as a result of all this is
we have identified a new kind of field
it is related to the electric field but
it has some material properties in it
this alpha represents the relation
between the applied field and the
induced field in other words it is
talking about the slope of this line
now
this is a it is a electric field like
quantity because it depends on electric
field but it is multiplied through by
some material properties and it's
divergence is equal to Rho F only this
quantity is what we call the
displacement vector the name is purely
historical but its usefulness is
extremely high because if you write this
equation down you know the right-hand
side completely you place the charge
there so you know exactly what charge of
placed and therefore you know mostly
what this vector field is going to look
like the problem of course is then how
from this displacement vector do I
obtain the electric field now this all
comes down to finding this alpha and if
you look back to see what is this alpha
II induced is nothing but minus sorry he
induced is nothing but minus e and D
okay this is the Q bound and from
Gauss's law
we found that E is equal to minus E and
D and E and n a material he is a
constant n is just the number of atoms
per unit volume d has to do with the
materials tendency to deform in the
presence of an electric field so all of
this is
can you put down as fixed material
property of the medium
so what does that mean it means that the
relation between D and E is something we
can put down in a handbook and just use
it isn't something to be solved for if I
make a plastic then a manufacturer or
the plastic four will subject that
plastic to tests he will find out if I
apply one volt
how much is the induced charge if I put
2 volts how much is the induced charge
and he will give me a plot of e induced
versus E and if the material is a linear
material that is he induced is
proportional to e he will give me this
whole number and this number is called
the dielectric constant now let us take
go back go back and review if we apply
an electric field to a material induced
charges appear these induced charges for
a field for a slab like this form at the
top and the bottom of the material in a
middle they tend to cancel out if you
ask how much is formed it has to do with
the thickness of this layer of charge
the thickness has to do with the amount
of electron cloud shifted and the amount
the electron cloud shifted
experimentally has been seen to be
linear there is no proof that it always
has to be linear that it does not always
have to be linear there are non linear
portions to this curve but where most
engineering applications happen it is a
linear curve so I can write the induced
electric field as something times the
applied electric field
start with this this is the first
equation now the second thing is if I
knew every charge present in the entire
universe including the charges I
deliberately placed and the charges that
are induced on surfaces of materials in
the bulk of materials when you all those
charges then Coulomb's law is still true
because Coulomb's law says when you all
the charges just write down Rho over R
squared with appropriate direction and
integrate over all space which and from
Coulomb's law we can always derive the
divergence theorem divergence e equals
the same charge density divided by
epsilon naught so now we have this
equation where this total charge is Rho
free plus Rho Brown that is Rho the
charge we placed plus the charge that
was induced now the induced electric
field is clearly related to the induced
charge because that is where it came
from so divergence of E induced is
nothing but the bound charge divided by
epsilon not but we do not know what it
is but I take these two equations and I
say I can eliminate Rho bound you
subtract the two equations I have got
rid of Rho bound that is what I do I
take divergence of e minus e induced and
it is only related to Rho free and when
I write out induced in terms of the
applied electric field I get this
equation divergence of some other vector
is only equal to Rho free
so crucial equation and it it is a bit
of a tricky equation the sense that you
can just write it down and use it
without realizing what it represents let
us take an immediate example and see
just how this applies it's a trivial
example but it is worth understanding I
have a charge Q and I place it inside a
sphere this is at the center here it is
air here if glass and here it is air the
inner radius of the glass is a outer
radius is B I want to know what is the
electric field and the potential as a
function of R alright
now there are two ways of going about it
one way of going about it is to say well
up to this point I am completely in air
so I ignore the electric field from zero
to a
nothing but Coulomb's law
now this electric field when it enters
this meat material is going to cause
induced charges and I can drive in write
down a relation between e induced and
the applied electric field and I can try
to work out what is going to happen
inside the material however there is an
easier way of going about it the easier
way is to simply say divergence D is
equal to Rho free so if I take if I
apply Gauss's law not to the electric
field but to D I can take any radius
sphere
the integral of divergence D is nothing
but surface integral over that we
surface of that volume d dot d s but
this is equal to volume integral of Rho
free now I do not know what is happening
inside this material but whatever is
happening inside this material is
induced response that is the charges are
all induced the only free charge I know
about is this Q so it is equal to Q say
spherically symmetric problem so surface
integral d dot d s is nothing but 4 PI R
square D of R D in the r direction which
is a function of R it is equal to Q
so I know Dee Dee has only an R
coordinate and the d r is Q over 4 PI R
square I haven't solved anything else
but regardless of what the dielectric
constant of glass is regardless of
whether I have one glass or many glasses
as long as the problem is symmetric in
all directions spiracle asymmetric I
have already solved the problem I have
found out D sub R how does it help me
well I will look up my handbook for the
glass and I will find out that this
handbook tells me epsilon glass is equal
to two point two five or it could be two
point five or it could be three it's
some number it is written by the
manufacturer of the glass so in that
case I know that D sub R is equal to
epsilon glass P sub R so let us just
look at the plot and make it clear this
is the radial direction this is a this
is B
if I plot D D is nothing but Q over 4 PI
epsilon sorry Q over 4 PI R square so it
is going to be 1 over R square
this is d what is e well instead of
talking about what is e I am going to
talk about epsilon naught e epsilon
naught E is going to be identical to D
in these two regions but in this region
where there is glass I know that epsilon
naught E R is equal to D R divided by
two point two five so this part of the
curve alone is going to be a smaller
value so this is epsilon naught here
what has happened is surface charge is
developed here and surface charge is
developed here so that the electric
field has been reduced the applied
electric field has been reduced by some
induced electric field so the net
electric field is smaller inside the
material then what you would expect it
to be once I have got this electric
field then getting the potential is easy
potential Phi is equal to integral from
minus infinity or I should say from R to
infinity of ER d R which is going to be
equal to Q over 4 PI epsilon naught R
for R greater than B because this is
nothing but Coulomb's law up to here for
R less than B is going to be Q over 4 PI
epsilon naught B that is the value at
this point plus the integral of this
curve because this is what the electric
field is so there will be Q over 4 PI
epsilon naught epsilon naught times 1
over R minus 1 over B this is for a less
than R less than B similarly when you go
inside or equals a
the point of this example was to show
you the usefulness of displacement
vector it is not just a arbitrary field
that we invented we created the
displacement vector because when we
worked out what divergence e is we find
divergence e contains an unknown charge
so we constructed a field which depends
only on known charge now this field
therefore is complicated it is related
not to Coulomb's law but to something
else
luckily for linear materials the
relation between D and E is very
straightforward D and E are related to
local material properties and that is
very important you can see here in this
picture the material properties are
changing I have air glass air
however D in here D is equal to epsilon
naught e out here D is equal to epsilon
naught into 2.25 into E and D is equal
to epsilon naught
so point by point the relationship
between D and E is only through a
different constant so the advantage of
this is I have got a new field which is
essentially the electric field if I knew
that field I know the electric field but
this new field satisfies a very simple
equation it is satisfies an equation
whose right-hand side I know completely
and that is an enormous advantage and
you can see it in this example because I
know Q I do not know the induced charge
but even so I found the answer
all right
now I want to go on to one more problem
and that problem is supposing I had my
slab and I had the electric field coming
at it we worked out if there is a
induced charge and we know that there is
a reduced electric field now if we apply
Gauss's law we have an incoming electric
field and an electric field inside the
material so I will call this e to have
called this e1 so e to minus e1 it is
normal so I do not have to keep the
vector notation is equal to Rho bound
divided by epsilon not
is that clear that is the amount of
induced charge is the amount by which
the electric field reduced but I know
that this e 1 is e 2 is related to the
displacement vector because if I look at
the displacement vector nothing changed
displacement vector didn't even notice
this material because there is no free
charge as only bound charge so we know
that if we scaled up this second
electric field by the dielectric
constant epsilon of the material and we
scaled up T 1 by the dielectric constant
of air okay so these are there is
nothing but D 2 is nothing but D 1 these
two cannot really be any different
because the same Gauss's law I can apply
to divergence D I can apply divergence D
is equal to Rho 3 and I know that Rho
free is zero
I haven't placed any charge there
whatever charge has developed has
developed because it was induced so I
know that these two must be equal
because if they are not equal
I would not get 0 when I worked out the
total amount of enclosed charge
let me repeat because I think is a
subtle point the amount of electric
field entering an amount of electric
field leaving a Gaussian cylinder an
imaginary cylinder is not equal which
means there is charge enclosed how much
charge is that that is nothing but the
induced charge but if I apply the same
thinking to D then I have that D 2 minus
D 1 is equal to Rho free divided by
epsilon not but Rho free is zero so it
is equal to zero so we get this special
condition if you want to be more
accurate and we do what we should say is
that supposing the electric field is
coming at some angle
we do not exactly know what angle the
electric field will be inside we know
there will be less we can draw a
cylinders and when we draw the cylinder
if this direction is led eetu Z minus e1
Z would be Rho bound over epsilon not
it's only the normal component that
matters if you look at the sideways
component there is flux entering but the
same flux will leave so only the top and
bottom contribute similarly when we do
the argument here we get D to Z minus D
1 Z is equal to 0 for D 2 Z is equal to
D 1 z the way this is put is normal
component of D
is continuous
across interfaces
I say it's continuous across interfaces
but it's really continuous everywhere
okay it is continuous here also it is
continuous there also but that is
assumed to be true but in a place where
things are changing abruptly you can see
that the electric field is not
continuous e to is not equal to e one
why because there is this row bound if e
two are equal to e 1 e 2 Z minus e 1 0
would be 0 but instead eetu Z minus e 1
z is equal to the induced charge divided
by epsilon not however even in such
places the displacement vector is
continuous for the displacement vector
to be not continuous you should have
placed charge on this surface so let us
just emphasize that and go on to next
point supposing I had a slab with an
epsilon say glass epsilon equals to 0.25
and supposing I deliberately put charge
how much are let's say you know ten to
the minus six columns per meter square
some charge I have put so I know this
charge and therefore this charge is row
free because it is it is charge I can
account for it is present even if there
is no applied electric field now I apply
an electric field this applied electric
field will create an induced charge so
there will be a small amount of negative
charge and small amount of positive
charge
so on the top after these this negative
charge and this positive charge combined
these two will cancel so what you will
be left with a little bit of negative
charge
when you try to work out the pro this
problem you will have divergence D is
equal to row three which means D to Z
minus D 1 Z is equal to row free but row
free is no longer 0 row free is 10 to
the minus 6 Coulomb per meter square
so you will find that now your
displacement vector is not continuous it
is not continuous because we
deliberately put charge there if we
hadn't put charge there then we would
have had a continuous displacement
vector there are quite a few problems
we're working with displacement vector
will actually simplify your analysis and
that is why it is an extremely important
formula
now let me remind you of the other
condition that we always have been using
which is that integral e dot dl from any
point any point is independent of path
and i mentioned in a class earlier and
if I have a point two and 0.1 and I take
two different paths going from one to
two is like minus of going from two to
one so this is nothing but integral 1 to
2 T dot DL minus integral 2 to 1 e dot
DL is equal to 0 ok sorry plus integral
2 to 1 is equal to e dot is equal to 0
or
if I take any loop whatsoever and
integrate II around that loop I get zero
this statement which we made in the very
beginning to obtain potential and this
statement are identical as saying the
same thing what I am saying is I will go
this way and then I will come back this
way since both gave me the same answer
adding up the two answers this is the
negative so this is nothing but integral
1 to 2 e dot DL minus integral 1 to 2 e
dot DL equals 0 but this is a very
interesting equation because supposing I
have my slab and let us say my electric
fields have got a sloping electric field
and I want to know what this electric
field is going to do is going to do
something
what I will do is I will take a loop
like this and I will try to integrate e
dot DL on this loop this way the
vertical pieces I will keep very very
thin so I have only large horizontal
pieces and if I take this large
horizontal piece what I get and call
this side one this side too
I get e one this direction is Z this
direction is X so e 1 x times DL minus e
two x times DL is equal to zero that is
the vertical parts do not contribute
much because every thin II one the X
component multiplied by the length of
this - E - X component multiplied by the
length is equal to zero that is what
this equation says or it says D 1 X is
equal to e 2 X
so I have two equations and let me put
them prominently I have that D normal is
continuous and E tangential is
continuous
these are the equations we have which
describe everything we know from the
start of this course from the start
we obtained that if you have charges you
can derive from the electric field
potential and the relation is electric
field is minus the gradient of potential
from whatever we have done now we know
that there is a new field very useful
field called displacement vector which
is the electric field times the local
dielectric constant epsilon is
dielectric constant
furthermore looking at the displacement
vector and applying Gauss's law we know
that the normal component at an
interface of D is continuous and looking
at the fact that the electric field can
be derived from potential through e
equals integral minus integral 1 to 2 of
5 DL sorry Phi is equal to e dot DL
using this I have also obtained electric
field tangential component is continuous
let me finish the class by trying and
using these equations supposing I have a
slab with epsilon and I have an electric
field that is arriving at a slope this
angle is Theta
so the normal electric field e normal is
equal to the magnitude of e say the
magnitude is e naught cos theta and E
tangential is equal to e naught sine
theta so this is e normal this is e
tangentially
now I know that if this is e naught and
this is e 1 I know that V naught cos
theta times epsilon naught that is the
normal component of D is equal to
epsilon times e1 times cos theta whereas
I also know that II not times sine theta
is equal to e1 times sine theta sine
theta 1
so I have two unknowns V naught e 1
sorry II not is known a 1 and cos theta
1 I have two equations if I divide these
two equations I will get epsilon naught
and theta is equal to epsilon naught
inverse epsilon inverse and theta 1
all I have done is I have divided this
equation by this equation so sine theta
or cos theta is tan theta sine theta 1
over cos theta 1 is tan theta 1 and
epsilon naught an epsilon going to the
denominator so I put epsilon naught to
the power of minus 1 and epsilon to the
power of minus 1 this is not quite the
same thing but it is like Snell's law
essentially saying the same thing now we
can also work out what happens to e 1
itself for that all we need to do is we
need to eliminate the theta 1 so
supposing we divide the first equation
by epsilon square it square the second
equation and add them up
what do you get the right hand side will
be e 1 squared cos square theta 1 plus e
1 squared sine square theta 1 cos square
theta 1 plus sine square theta 1 is 1 so
it is equal to e 1 square but what
happens on the left hand side I divided
this equation by epsilon so I get
epsilon naught over epsilon times e
naught whole square cos square theta
plus
he not square sine square theta so it is
a mess here but the mess does not matter
I know what theta is I know what Epps
not is so left hand side is a known
thing is equal to e1 square so
essentially if you look at what this is
saying it is saying that the portion
that is tangential portion that is along
X does not change it contributes fully
to e1 with a portion that is pointing
into the slab that part gets reduced in
the proportion epsilon naught over
epsilon so if you finally solve this you
get e 1 is equal to e naught square root
of epsilon naught over epsilon 1 squared
cos square theta plus sine squared
so if I had normal incidence which is
theta equals zero
sine theta goes off cos theta is one
that will say P 1 equals epsilon naught
over epsilon 1 e naught so the electric
field inside the material reduces in
proportion to the dielectric constant
which is exactly what we had come up
with earlier because we found that
induced was proportional to the applied
electric field and D and the quantity we
came out with the displacement vector
was a minus e induced so obviously if D
is continuous then you must have a
reduction in e wherever the dielectric
constant is large whereas if theta is
large then what will start happening is
this term will dominate this term and
you will have that II 1 is equal to e
naught
this example should tell you the sort of
an introduction to the idea of what we
call matching conditions and the
equations here these two are matching
conditions for electrostatics
namely the normal component of D is
continuous tangential component of E is
continuous this more or less establishes
all the building blocks we need to solve
problems in electrostatics
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