Electric field axial point due to diapole - Electric Charges & Fields, Physics, Class 12 Video Lecture

hello friends this video on electric
charges and fields part 25 is brought to
you by exam feel calm no more fear from
exam please make sure that you have
watched all the videos still part 24
before going ahead with part 25 so let
us start with the axial point so we will
evaluate here the electric field at any
axial point due to a dipole now let us
suppose we have to calculate it at any
axial point let us say this is our
actual point this point P so this point
P is located at some distance X from the
dipole so as I mentioned from where will
be measured this distance X obviously
from the center of the dipole that means
this distance is X right so let us say
this is minus Q and this is plus Q and
by convention we generally consider the
distance between the two dipole as 2a so
me consider it as 2a
now we have to calculate the electric
field at this point again we will follow
the same rules which we used to follow
previously by calculating electric
fields so what would be the electric
field net electric field at Point P so
the net electric field would be because
of this charge as well as because of
this charge so let us mark this position
as a and this position as B so we have
to calculate the electric field at P due
to a and we will also calculate the
electric field at P due to D I mean we
are following the same steps are the
same sequence as we did during
calculating electric fields at other
points so electric field at P due to a
is equal to K Q divided by r square
voters are R is the distance between P
and a so looking at this diagram you can
see distance between B and a is X plus
ay
this distance is a so it is X plus a
whole square right God would be the
direction of this electric field this
discharge is a negative charge and at
Point P we again assume our test charge
that is imaginary charge which is
positive so this positive charge will
try to attract it towards itself
therefore the electric field at this
point will be along this direction so a
will try to attract P towards itself so
the direction of e.p.a will be towards a
so this will be along P a now similarly
let us calculate the electric field at
Point P due to B so this will be K Q
divided by what is the distance between
point B and point P it is nothing but X
minus a whole square right so this will
be K Q divided by X minus a whole square
and what would be the direction of this
this is positive this is also positive
so B will try to push P away from itself
so the electric field that is e PB will
be along this direction now let us call
this point at some point in so this will
be along P n so now we need to calculate
the net electric field at Point P now
looking at the values of EPA and EPB
what do you think where what where what
will be the direction of the net
electric field one electric field is
going towards right and the other one is
going toward slip obviously the one
which has a larger value will dominate
so which has a larger value the one if
you look at both the values they have
the same numerator so the one with a
lesser numerator will have a greater
value so which is a lesser numerator
obviously EPB so very clearly you can
see that EPB is less than ETA that is
electric field at Point P due to B is
lesser than the electric
field at Point P due to a I'm sorry it
does greater than okay so that means EPB
will dominate so the direction of EPB
would be the direction of the net
electric field at point B right so what
would be the net electric field let us
denote the net electric field as EP so
the direction of EP will be along in and
what will be the value it would be e PB
minus e a so along p.m. so let us put
the values now so it will decay Q into 1
by X minus a whole square minus 1 by X
plus a whole square right now if we
solve this we get KQ in common and then
X plus a whole square minus X minus a
whole square divided by X square minus a
square whole square so now I am using
simple mathematics to solve the equation
so this is KQ in to break it down in the
formula of a plus B whole square so this
is X square plus a square plus 2 ax
minus X square minus a square plus 2 a X
divided by X square minus a square whole
square so this X square and a square
will get cancelled with the respective
terms so we get K Q into two a X divided
by X square minus a square whole square
now if you look at it Q multiplied by 2
a that is product of either charge into
the perpendicular distance between them
what is that that has nothing but dipole
moment so we can replace it with dipole
moment P so this is KP
X divided by X square minus a square
whole square right so this comes out to
be I'm sorry I think I missed out a
factor 2 2 ax + 2 x was for a X right so
there was not a factor of 2 so we get to
K P X divided by X square minus a square
whole square therefore we can say that
the net axial electric the net electric
field at an axial point is given by this
expression and what would be the
direction of this electric field it
would be along P n that is the same
direction at which the electric field
due to B exists right so if you look at
it the approach of reaching to the
expression for electric field still
remains the same there is nothing new
the only new thing that was introduced
was the dipole and how to measure the
distance from a dipole other than that
the each and every step remains the same
so now in a very similar way we will
also calculate the electric field at an
equatorial point due to a typo so there
also we will take a similar approach so
before proceeding with equatorial point
let us have a few points of discussion
on what we did just now that is on the
axial point now looking at the
expression which we arrived at just now
we will discuss two points so what did
what was the expression that we got ye
axial is equal to 2 K P X divided by X
square minus a square whole square right
now what happens if X is very very
greater than a that means if the point
where we want to calculate the electric
field is at a very far off distance from
the dipole so what would be the value of
the electric field at time in that
if you look at this in the denominator
since X is very very greater than a so
you can just ignore the term a right
because a very small term getting
subtracted from a very large term it
doesn't make much difference to the
large term so we can write this e axial
as two K P X divided by X square whole
square so this becomes equal to 2 K P
divided by X student so this becomes the
value of electric field in that case so
what what are the results that we arrive
at here the electric field is inversely
proportional to X s cube that means as
the point where we want to find the
electric field is moved away from the
dipole the electric field at that point
decreases that is very obvious as well
right
if as you go away from the dipole the
electric field will keep decreasing and
ultimately if you go to a very far off
point the range of the electric field
will get over so you will not experience
any electric field at all right so as
you go away from the dipole the electric
field will decrease and we also observe
that E is directly proportional to P
that is as the dipole moment increases
the value of electric field at any point
also increases so the dipole moment also
contributes to the value of the electric
field and what about the direction now
as I already mentioned the direction of
dipole moment is always from minus Q 2
plus Q and so is the case for the
electric field in this case but here I
am talking about only this case when I
am talking about an axial point thank
you please visit exam dot-com to watch
free educational videos try 3 online
tests get the best quality study
materials study from the best tutors and
mentors and much more thank you once
again