in the previous lecture we defined electrostatic
potential and what we saw was that e dot d
l was equal to minus v two we write integral
from one to two v two plus v one its differential
form was that electric field it was equal
to minus the gradient of electric potential
let us see how do we use these equations to
get electrostatic potential for a unit charge
you already know the expression that if i
have a unit charge at the origin then at a
distance r from it potential is given as one
over four pi epsilon zero q over r let us
get this using these definitions
so if i take integral d l going from a point
r let us say r vector to infinity this would
be equal to minus v at infinity plus v at
r so now let me make a picture this is my
point charge and i am going from a distance
vector r all the way upto infinity and i'll
go along this radial line if you are not comfortable
with radial coordinates let us say i move
along the x axis i'll talk about the coordinates
in ah next couple of lectures because ah just
to review spherical and cylindrical coordinates
so i am moving from a point x to infinity
along the x axis and then if once i calculate
at x since everything is spherically symmetric
everywhere around at the same distance the
potential would be the same
so i am calculating integral e dot d x going
from distance x to infinity and this will
be given as minus v infinity plus v at x this
i can write as going from x to infinity is
one over four pi epsilon zero q over x square
x unit vector dot d x x unit vector which
is equal to x to infinity one over four pi
epsilon zero q over x square d x and this
this you see right over is one over four pi
epsilon zero q over x which for a distance
r i can write as q over r that is by moving
radially i can always take this radius to
be along the x axis so what we learn is that
v at infinity plus v at point r is equal to
one over four pi epsilon zero q over r
as i said earlier it is defined only upto
a constant so i choose a reference point so
that v at infinity we take to be zero if i
take v at infinity to be zero i define v r
as one over four pi epsilon zero q over r
and what it means is if i take a charge bring
it to point r from infinity i'll have to do
this much work sometimes while doing that
integral people get confuse so let me just
do that also if i have a charge q at the origin
and i am moving bringing a charge from infinity
again i'll have now i am moving from x equal
to infinity to a point x e dot d l which will
be equal to v at x with a minus sign plus
v at infinity because point one is infinity
i'll still take d l to be d x along x because
that i am moving in from infinity to x is
actually covered by the limits therefore i
don't have to put a minus sign d l equals
minus d x x here you still take it to be d
x x i am integrating over x and since x is
integrated from infinity to x that movement
in is already taken care of and this is nothing
but again q over four pi epsilon zero d x
over x square infinity to x which is equal
to minus v x taking v infinity to be zero
and this gives you the same answer as this
so this is the potential due to a point charge
q at a distance r from it
so we have calculated potential due to a point
charge q at a distance r from it what about
a charge distribution suppose i am given a
charge distribution so that the density is
rho r prime as we saw in the case of electric
field electric field is superimposed right
it ah follows the principle of superposition
so if i calculate the work done in that electric
field that work done can also superimposed
so the potential at point r is going to be
potential due to a charge out here added to
potential due to charge nearby potential due
to charge nearby and therefore this can be
written as one over four pi epsilon zero integration
rho of r prime over r minus r prime d r prime
o this is d i keep writing it like this this
is actually the volume integral d v prime
and what this formula is obtained under the
condition of that v at infinity we are taking
to be zero so this is in free space so in
free space if i have a charge distribution
which has this charge distribution rho r prime
then the potential at r is going to be given
by this formula v r equals one over four pi
epsilon zero rho r prime over r minus r prime
d v prime notice that this is simpler than
the corresponding formula for the electric
field where we had a vector out here adding
vectors or calculating components is much
more calculating three components is more
difficult than calculating only one quantity
and then if i take the gradient of this electric
field would come out from it
so we prefer to work with potential rather
than electric field and keep in mind that
this formula is in free space and therefore
now consider again a situation i took earlier
if i had say a metallic sphere and i grounded
it by grounding we mean that we made potential
equal to zero then if i put a charge somewhere
near the surface or even at the centre and
i wrote v r is equal to you q over four pi
epsilon zero one over r this will not be correct
because this does not give you zero at the
radius at r equals r you can say well at the
way i can define it is that v r is equal to
q over four pi epsilon zero one over r minus
one over r that would be fine
but what if i take a different geometry suppose
i take a box 
that means make all the surfaces be equal
to zero now put a charge inside what will
be the potential that'll be difficult question
to answer and therefore what we need to do
is develop to calculate the potential we need
to develop a differential equation for v r
and solve it with appropriate boundary conditions
and that should solve the answer suppose i
knew the that should give me the answer suppose
i knew the differential equation then i would
solve the differential equation with these
boundary conditions and that'll give me the
answer off course i have to prove that that
answer is going to be unique so let us see
what is that differential equation for v
we have e r equals minus grad of v we also
have divergence of e is equal to charge density
at that point had the epsilon zero we also
have curl of e is equal to zero but these
two equations are one and the same thing because
from this follows that i we can define a potential
and curl of a gradient is zero so the only
equation we are left with this is this and
now we put the formula for v here it comes
out to be rho r over epsilon zero and that
gives me minus i am going to write this del
square v is equal to rho r over epsilon zero
let see what this quantity del square is
so del dot del v with the minus sign is the
minus sign is there we don't worry about it
is x d by d x plus y d by d y plus z d by
d z acting on x d v by d x plus y d v by d
v plus z d v by d z which is nothing but minus
now it comes out to be a scalar function second
derivative of v with respect to x because
x dot x gives me one x dot y gives me zero
so there are no cross terms plus d two v by
d y square plus d two v by d z square this
is known as the laplacian of v so this del
square term is the laplacian and therefore
the equation differential equation for v we
get is del square of v r is equal to rho r
over epsilon zero with a minus sign in front
this is known as the poisson equation
so if i am given some charge density inside
a box or some other close volume some charge
density i solve this equation with the boundary
condition and that gives me the answer for
v on the other hand it could so happen i may
have a situation for example if i take this
box put this surface at some potential v and
i put all the other surfaces at zero potential
i'll still have electric field inside because
electric field will be coming out of this
line what is the potential inside then i'll
be solving there is no charge density inside
delta square v equal to zero with appropriate
boundary conditions and this is known as laplace
equation so the electrostatic potential either
satisfies the poisson's equation if there
is charge density or laplace's equation or
you can just write one equation if rho r equals
zero it goes over to the laplace's equation
and then solve with the appropriate boundary
conditions and you get your answer
so in the future now on one would like to
solve for v r because this is easier equation
to solve and then get the electric field e
r as minus grad of v r off course we know
once we solved out for v r what the interpretation
is difference between two v r's is actually
the work done in a person taking a unit charge
from that point to the next point
