Field long string uniformly charged wire - Electric Charges & Fields Video Lecture - Class 12 | Best Video for Class 12

hello friends this video on electric
charges and fields part 31 is brought to
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exam please make sure that you have
watched all the videos to the part 30
before going ahead with part 31 so the
first application which we will look at
is the field due to infinite long
straight uniformly charged wire if you
remember we have already talked about
the electric field because of a
uniformly charged conductor right so
here we will look we will try to
calculate the electric field with the
help of Gauss law now before applying
Gauss law how do we apply goshawk we
choose a surface we have to choose a
closed surface in such a way that the
closed surface encloses this uniformly
charged wire to some extent now with
that we will apply Gauss law so in this
case let us suppose I choose the Gauss
sheíll surface as a cylinder a cylinder
which is enclosing the long straight
uniformly charged wire let us suppose
this is my cylinder which encloses the
charged wire now let us calculate the
electric flux through this surface
because what is I mean electric flux has
to be through some surface right so let
us calculate the electric flux through
this surface so that now the total
electric flux will be equal to the sum
of the electric flux through the three
surfaces here you have three surfaces
this circular surface is surface 1 this
is surface 2 that is the body of the
cylinder and again this lower circular
surface is the surface 3 so the flux
total flux through this cylinder will be
equal to Phi 1 that is flux through this
surface plus flux through this surface
plus flux through this surface right so
we will calculate each of them one by
one now what about the electric field
what is the direction of electric field
C this is the
positive charge right so the electric
field is in this direction because the
electric field will originate from the
positive charge so this is the direction
of the electric field and what is the
direction of the normal vector or the
area vector area vector is always
perpendicular to the area so when I
consider that this one the area vector
is along this direction when I consider
surface three the area vector is outward
normal that is this when I consider
surface to the area vector is along this
direction here and again this direction
here right so if we look at it for
surface one that is for v 1 the area
vector e is perpendicular to n cap right
and I told you before that if electric
field is normal to the area vector then
the flux is equal to zero so here since
E is perpendicular to area vector
therefore flux is equal to zero
similarly for surface three also the
electric field vector is normal to the
area vector therefore v 3 is also equal
to zero therefore the total electric
flux through this surface is equal to
the electric flux through the curved
surface of the cylinder so let us
calculate the electric field through the
curved surface so this will be equal to
integration of e dot d s so this will be
equal to e dot n tap D s so this will be
equal to integration of e dot n cap over
closed surface into D s now the angle in
this case that is theta is equal to zero
because n cap and E are in the same
direction therefore we can write it as e
integration of D s so this D s on
integrating becomes the total surface of
the cylinder what
the curved surface of the cylinder
curved surface of cylinder is given by 2
pi R L where R is direct this radius
this distance is R and L is the length
of the cylinder so the length of the
cylinder is n therefore Phi is equal to
e into 2 PI R n now according to God's
law Gauss law states that file should be
equal to the total charge enclosed that
is Q divided by epsilon naught so using
this we can say e into 2 pi RL is equal
to Q by epsilon not right so we can say
electric field E is equal to Q divided
by 2 pi epsilon not R into L now since
just now we have told that this wire is
uniformly charged wire that is let us
suppose that the linear charge density
is lambda so this lambda will be equal
to Q divided by L that is charge per
unit length therefore using this we can
write E is equal to lambda divided by 2
pie epsilon not R so this is the
expression for electric field at any
point for an infinitely long uniformly
charged wire so why do we see we see
that electric field is inversely
proportional to R that is it is
inversely proportional to the distance
so as the distance of the point
increases the electric field at that
point decreases right so here you saw
that we did not need thee there is no
need to do lot of calculations we just
choose a suitable Gaussian surface and
with the help of Gauss law we were very
easily able to calculate the electric
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