we have learnt that the potential satisfies
the poisson's equation 
which is laplacian of the potential is equal
to minus charge density at that point divided
by epsilon zero or the special cases laplace's
equation which tells me del square v is zero
in the places where there is no charge this
has consequences and as we said earlier that
we had also said that ah you know these becomes
the differential equations for the potential
if you want to use these to calculate potential
we should be able to show that these gives
unique answers given a boundary condition
as for the consequences i will just mention
one thing if i take del square u v is equal
to zero this implies there is no minimum or
maximum of potential at a given point right
and therefore at a given point if the potential
is going to a maxim through a maximum at one
point it will go through a minimum on the
other side a one way of good way of visualizing
it look at this part of your hand in this
if you come from the side of your fingers
and thumb you are hitting a minimum but if
you go from the palm to the other side you
hitting a maximum so there is no maximum here
there is known as a saddle point and this
is where del square of this quantity whatever
the height if i calculate is going to be zero
and what it means is there is no equilibrium
point for a charge in a charge free region
let's understand that if there is no minimum
or maximum that means the potential is going
through a maxim on one side a minimum from
the other side if i put a charge here it may
be stable in this this direction from which
it is hitting a minimum on the other side
it will start moving away and this is a consequence
of del square v being zero
mathematical way of seeing this is as follows
suppose there is a point where there is a
maximum then if i take a very small sphere
around it if i take a very small sphere around
it and calculate divergence of grad v over
the surface this is going to be equal to by
divergence theorem ah sorry divergence of
this glad v over this volume this is going
to be by divergence theorem grad of v dot
d s over the surface since this is a maximum
or the other way if it is a minimum then grad
of v is going to have the same sign over the
entire surface right
so if the point is a maximum or minimum of
v then grad v has the same sign over the surface
and this implies integration grad v dot d
s is not equal zero on the other hand look
at this this this is del square v d v and
this is zero so i am hitting a contradiction
and if i am in a contradictory stage that
means my initial assumption that either i
am at maximum or minimum is wrong so they
can be no maximum or minimum physical way
of understanding this is as follows in a charge
free region 
gauss's law tells me that e dot d s is equal
to zero that means in a charge free region
if there's a point where some electric field
lines are coming in some have to move out
all right
and therefore a charge put here and never
being equilibrium because they whatever lines
are coming in they are also going out so a
charge will tell to move in the direction
where the field lines are going out this is
equivalent to saying del square v zero and
all that is equivalent this is know b the
way as earnshaw's theorem that in a charge
free region a charge cannot have an equilibrium
point now let's go to uniqueness of solution
uniqueness of solution of laplace's or poison's
equation for a given boundary condition what
we mean by given boundary condition means
that we have specified that on a given boundary
what is the potential it is specified
so let us take a close boundary it could at
infinity and we have given potential at the
boundary let's call it v naught and let there
be some charge inside this boundary given
by rho r prime so you have given the del square
v is equal to minus rho r a given at that
r over epsilon zero let us assume there are
two solutions v one r and v two are both satisfying
this equation and both satisfying the same
boundary conditions and let a function phi
r v equal to v one minus v two then phi r
satisfies del square phi is equal to zero
and phi is equal to zero on the boundary why
because phi is the difference between the
two potentials where both satisfy this same
boundary condition and both satisfy the same
poisson's equation
let us now look at divergence of phi grad
of phi which you can easily show by taking
components is equal to grad phi square plus
phi grad square phi it is something similar
to just to facilitate it d by d x of a function
f d by d x f is going to be equal to d f by
d x square plus f v two f by d x square it's
similar to that and you can by taking component
you can show that but now i am going to use
this to show the uniqueness let us integrate
both sides this is divergence of phi grad
phi so i am going to integrate this over the
volume this is going to be equal to grad phi
mode square over the volume plus phi del square
phi over the volume however we have seen the
del square phi zero so this terms drops out
on the left hand side i am going to get by
divergence theorem integration phi grad phi
dot d s over the surface is equal to grad
phi square d v but we have all ready seen
that phi zero on the surface and therefore
this term also dropped out as a consequence
what i see is zero equals grad phi square
d v however grad phi square is always positive
is square so grad phi square d v is equal
to zero immediately tells you that grad phi
is zero gradient or the or the or the derivative
of phi zero so phi is is equal to constant
at most v one and v two can differ by constant
but since they are the same on the boundary
this constant better be zero
so this implies v one equals v two what that
tells me is that and given a charge distribution
the potential the answer given by solving
poisson's equations is unique they cannot
be two different potential satisfying the
same poisson's equation and same boundary
conditions and rho equals zero is just special
case and therefore laplace equation also given
boundary conditions gives me the same answer
so del square v equals rho over epsilon zero
has unique 
solution for a given boundary condition all
right so let's see how we arrived at this
we arrives at this by looking at a divergence
of the difference where phi is v one minus
v two and this we used to get that grad phi
must be zero ah related physical phenomena
is that if i have a close surface with no
charge no charge inside and v same throughout
the boundary then field inside is equal to
zero
this is pity much like if i have a metallic
sphere we consider earlier but now let me
take a metallic any shape ah ah any shape
of a metal and i give it some potential that
this entire surface becomes an equipotential
because metals they cannot be any difference
between two parts otherwise charge is flow
from one side to the other and the field inside
the metal is zero so this is very similar
to that that if i create a equipotential a
close volume who's surface all is at the same
potential then the field inside will be zero
let us give that before using divergence theorem
now del square v inside is going to be zero
because there is no charge again let's take
divergence of v grad of v which will be grad
of v square plus v laplace in a v
but since there's no charge inside this fellow
is zero let us now take volume integrals divergence
of v grad v is equal to grad of v mode square
d over the volume the left hand side i can
write as integral v grad v dotted with surface
is equal to integral grad v square over the
volume please do not confuse between this
v and this v now if you look at v grad v over
the surface we are already saying that the
surface has the same potential therefore this
v does not have any over the surface i can
write this as i can take this v out then it
becomes integral grad of v dot d s
what is grad of v this is v minus e dot d
s since there is no charge inside by gauss's
law this term is zero and therefore left inside
vanishes and i interpreting integral grad
of v mode square d v is equal to and zero
this means grad v zero must be zero inside
so if i take something with constant potential
surface close surface constant potential over
the surface then field inside will zero as
an example you can do this experiment at home
you take a metallic tiffin box and put you
mobile inside if you call that mobile the
call will not go through because the field
inside is zero everything shielded the the
waves i cannot penetrate that metallic close
surface so this is also used to make something
called a faraday's cage 
you take a all most ah closed metallic surface
all this made out of mesh and put your instruments
inside and there not destroyed by any electrical
signals from outside because that electrical
and no matter what you do outside the field
inside is going to be zero because for a metallic
mesh or metallic surface the potential all
throughout is constant
