Gauss law proof - Electric Charges & Fields Video Lecture - Class 12

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Gauss law is a very important law in
electrostatics just like Coulomb's law
like house Coulomb's law play a very
important role when we talk about
electric field when we talk about
electric force similarly Gauss law means
a crucial role by talking about electric
flux so Gauss law basically makes our
calculations to evaluate electric field
easier so let us first see what is Goss
law
Gauss law states that the total electric
flux over a closed surface is equal to
one by epsilon naught times the total
charge enclosed by the surface it says
that if you have a closed surface let it
be of any shape doesn't matter so any
closed surface if you want to calculate
the total electric flux over that
surface that is equal to 1 by epsilon
naught times the total charge enclosed
by the surface now when is a total
charge enclosed by the surface it can it
includes any charge located anywhere
inside the surface for example I have a
positive charge Q inside this spherical
inside this sphere so this sphere being
a closed surface according to God's law
the total electric the total electric
flux over the sphere will be equal to
integration of e dot d s over the closed
surface right so according to God's law
this electric flux is equal to Q divided
by epsilon not because Q is the charge
enclosed by the sphere now if instead of
Q there with u 1 Q 2 Q 3 so this would
have become equal to Q 1 plus Q 2 plus Q
3 divided by epsilon not so it doesn't
matter where are these individual
charges located inside the surface any
charge located inside the surface
contributes to this
so let us prove Gauss law let us suppose
we have a sphere which is a closed
surface so we will calculate the flux
through this sphere let us say there is
a positive charge plus Q inside the
sphere so we consider now before
considering anything letters first
denote the direction of electric field
what would be the direction of electric
field we studied in the slide referring
to electric field that the electric
field will point radially outwards for a
positive charge because electric field
lines always originate at the positive
charge so this would be the direction of
electric field now let us suppose I
consider a small element D s we consider
a small element D s let us say this is d
s right so what would be the normal to
this area element the normal to this
area element will also be along this
direction right so let us calculate the
flux file which is denoted by e dot d s
right so from definition of electric
field we know that E is equal to KQ by r
square K is 1 by 4 PI epsilon naught Q
divided by R square if this distance is
R that is distance of the element D s is
R from the source charge so this is Q by
R square into R cap R cap denotes the
direction of the electric field now we
substitute this value of electric field
in the expression for v so therefore phi
is equal to 1 by 4 pie epsilon not Q by
R square R cap dot d s right now what r
dot d s is rd is cos theta so what is
the theta between the 2 R cap and D s
are along the same direction so theta is
equal to zero Degree here therefore we
can write it as 1 by 4 PI epsilon not Q
divided
by r-square integration of D s over the
entire surface s so what is D yes this
is the this is a sphere so what does the
surface area of the sphere surface area
of a sphere is nothing but 4 PI R square
all right so we can write it as 4 PI R
square so this 4 pi and 4 pi will get
cancelled R square R squared will get
cancelled so we are left with Q divided
by epsilon naught therefore we found
that Phi is equal to Q divided by
epsilon naught so what is Q Q is nothing
but the total charge enclosed by this
sphere therefore we saw that the Gauss
law proves out to be correct so for any
closed surface no matter whatever the
shape may be the total flux over the
surface is equal to 1 by epsilon naught
times the total charge enclosed by the
surface it is not necessary that the
total charge has to has to be a single
entity it can be multiple charges but
they all should be located inside the
surface now let us try to look at the
situation the other way around
let us suppose we have a closed surface
such that there is no charge inside the
surface so according to Gauss law the
total flux through this surface should
be equal to 0 because there are no
charge enclosed so how do we prove this
let us suppose this is a closed surface
now since there are no charge include no
charges are it enclosed by this closed
surface so let us suppose it is the
direction of electric field is as shown
in the figure so here at all the points
on the surface of the sphere the
direction of electric field will remain
the same now let us suppose if I
consider a particular element say D s at
this point so if I consider an element D
s at this point the vector normal vector
will be along this direction now if I
consider a D
is along at this point the normal vector
will be along this direction so what
will happen is that all the elements by
symmetry the contribution of electric
flux by this element will get canceled
by the other element so similar thing
will take place for all of them and as a
result the total flux are the net flux
through the closed surface will be zero
so therefore we can see that we can
logically also prove the truth of
Gauss's law right now we will see how
can we derive Coulomb's law from Gauss
law see all the law all his laws in
electrostatics are some way or the other
related to each other so we will see how
do we derive Coulomb's law from
Gauss law so we will start our
derivation from Gauss law according to
Gauss law if we have a closed surface
and the charge enclosed by the closed
surface is plus Q in that case Gauss law
states that the total flux through the
surface is equal to integration of e dot
d s over the closed surface is equal to
Q divided by epsilon naught now we
already know that P dot d s is nothing
but e d s cos theta is equal to Q
divided by epsilon naught now in this
case at every point the elect direction
of electric field and the direction of
the area vector will be the same that is
theta will be equal to 0 degree at all
the points on this on the surface of the
sphere therefore cost 0 is equal to 1 so
this we can write it as e integration of
D s is equal to Q by epsilon not so what
is e now this D s can be written as 4 PI
R square because 4 PI R square is the
surface area of the sphere so this is 4
PI R square is equal to Q by epsilon
naught or we can write E is equal to Q
divided by 4 PI epsilon naught R square
now what is
1 by 4 pie epsilon-not that is nothing
but a DQ / R square right now let us
suppose that Q naught is a test charge
at a distance R from Q so Q is the force
charge here because Q is the one which
is generating the electric field now let
us suppose that I have a test charge
that is an imaginary small charge at a
distance R from Q so let us suppose I
have Q naught which is a test charge at
distance R from Q so what will be the
force experienced or what will be the
force experienced by this Q naught or
what would be the force of interaction
that will be equal to Q naught into the
electric field e so this Q naught into
electric field E will be substituted by
this value so this is equal to K Q Q
naught divided by R square so that is
Coulomb's law so we reach at this
expression that F is equal to K into Q
naught and Q divided by R square so this
is the statement of Coulomb's law so
here we saw that we started with Gauss's
law and we ended up at Coulomb's law
that means all the laws are consistent
with each other and they are all true
for all these situations and
electrostatics so we discussed Gauss
from now we will see how has Gauss law
helpful to us I mean why do we need to
study Gauss law because we saw that we
can find out the electric field of any
object at any point so how is Gauss law
helpful so what do we conclude about
Gauss law this law is true for any
closed surface irrespective of their
shape the second thing is this law
facilitates easy a calculation of
electrostatic field when the system has
some symmetry so that is what I will
tell you why why I will be studying
ah slaw we don't have that much of free
time to study any law given by any
scientist right so obviously this law
must be having some advantages because
of which we are studying about it
so I told you long back when I was
talking about electric field that how do
we calculate electric field at different
points for different kinds of conductors
right now I will tell you the same thing
but with the help of Gauss law and you
will see a marked difference in that
calculation there we had to do a long
calculation first we have to consider
this triangle then that triangle then
sine theta cos theta and things like
that but now with the help of Gauss law
we will just have to choose a suitable
Gaussian surface we will just choose a
suitable surface and we will apply Gauss
law and we will reach the result in in
the spur of a moment so now we will look
at certain applications thank you please
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