In the previous lecture on force and energy, we saw how a force is obtained from
corresponding potential energy by taking its gradient. In this lecture we explode this
little further and then go on to discuss non conservative
forces as a real life example of calculating the force from a potential. Let
us now look at the electric field and the potential or equivalently the potential energy
and the force on a charge an a in an electric dipole.
So, I am going to consider a dipole sitting at the origin with charge with minus q on
negative y x axis and charge plus q on the positive side of the x axis with the distance
between them being 2 a. I want to find the electric field at a general
point r form the origin, I will do, so by taking
superposition of fields due to charge minus q charge plus q adding and then taking the
limit a going to 0. So, that limit is taken in such a manner that p equals 2 q a remains
unchanged that defines for me point dipole setting at the origin. So, I could find the
field at r by superposition of field due to minus
q at r plus field due to plus q at r and get my
.answer in an easier way. This will tell you how talking about potential helps at times
could be that I find the potential energy of a unit charge at point r and take its gradient.
Negative of that will give me force on that unit charge which by definition is the electric
field at r, let us do this as an exercise. .
So, recap I have dipole sitting here I want to find the field at distance r which is x
from the origin and y this way. Finding potential
in this case is easier because I am not doing a
vector sum, therefore potential energy for unit charge q equals 1 at r. The r is given
in 2 dimensional spaces this x and y point is going
to be potential due to the first charge plus the potential due to the second charge. It
is going to be k q over square root of x minus a
square plus y square minus due to the other charge this one, k q over square root of x
plus a square plus y square. Here, I just want to remained to this distance between
the charges is 2 a and k is one over four pi epsilon
0. I am not so much concerned about these quantities more about taking gradient
of this. Since a is very small, so far away from the
dipole I can write 1 over square root of x minus a square plus y square as 1 over square
root of x square plus y square minus 2 ax plus a square. I am going to neglect this
terms because finally I am going to take the limit
going to 0 and keep only q times a term, so I neglect that.
Making the figure again, I am trying to find the potential due to a dipole at a far away
point which is x from the origin this way y this way this is q minus q. I made the
approximation that 1 over the square root of x minus a square plus y square is equal
to one over the square root of x square approximately
plus y square minus 2 a x which I can write as 1 over square root of r square minus
2 ax. This is approximately equal to one over r one plus ax over r square, similarly
I can write 1 over square root of x plus a square plus y square approximately as 1 over
r 1 minus ax over r square. .
.Combining the two, I get the potential for this dipole separated by 2 a minus q plus
q as U at r. Here, r is this point distance y distance
x from the origin as k q over r 1 plus ax over r square I am working under those approximation,
where a is very small minus k q over r 1 minus ax over r square. That gives
me U r to be 2 k q ax over r curbed, now when I take the limit a going to 0 such that
2 q a remains constant equal to the dipole moment that I want to have. So, then in that
limit becomes 2 q a gives me the dipole moment k p x over r curbed, you can check
that the dimensions are alright p is q times the distance this distance divided by this
distance q. So, this is the potential, now I want
to find the electric field due to this dipole at point r.
So, I am finding, so there is a dipole is here of the magnitude p pointing in x direction.
So, let me now make this dipole I am making slightly big is in x direction its magnitude
is p. I am trying to find the electric field at this point r distance away which is x this
way and y this way, I found that U of r or a single
charge. So, this is a potential is equal to k p
x over r cubed, so I can find the x and y components of the force of the electric field
because electric field is same as force on a unit charge is equal to F x is equal to
minus d U d x which is equal to minus d over d x.
I am taking a partial derivative, so when I do that y remains a constant of k p x over
x square plus y square raise to three halves.
When I do that, I find I differentiate this, first I
will get three over 2, this minus sign goes with that k p x over x square plus y square
.raise to 5 halves times 2 x minus k p I differentiate x, so I get divided by x square plus y
square raise to 3 halves, this two cancels. .
So, the expression for the electric field x component that I get is going to be x which
I took to be the partial derivative of U with
respect to x comes out to be 3 k p x square divided by r raise to 5 minus k p over r cubed.
Let me see if it is correct going back to the previous slide this 3 k p x square over
r raise to 5 minus k p over r cubed.
Similarly, I can find E y which is minus partial of U with respect to y and this comes out
to be minus partial of U with respect of y k p x over x square plus y square raise to
3 halves and that gives you three k p x y over
r raise to 5. These are the two components with electric field you can check for yourself
that this is consistent with the general expression of the electric field which is
given as 1 over r cubed rather k over r cubed p
dot unit vector in r direction times 3 r minus p.
Getting this directly by superimposing the fields due to 2 charges would be slightly
more difficult. You see that taking gradient of
the potential which is slightly easier to find is an
effective way of finding the field. Having talked about all these quantities, you may
be now wondering what happens to the minimum
maximum of the potential in three dimensional cases.
In one d cases, I knew that the force was 0 when d U d x was 0 if d 2 U dx square was
greater than 0, then I had a minimum of the potential. Similarly, d 2 U dx square less
than 0 gave me maximum of the potential and we have talked in the
past this gave me an equilibrium point
or I should say a stable equilibrium point. If I moved a body from this
point it will come back and this gave me an unstable equilibrium point, the body had
force 0, but if I moved it away from there, we will just take off it will not come back
what about in 3 d cases? .
.In 3 d cases, I am talking about a three dimensional force which is the gradient of
U off course if gradient of U is 0 it implies that
the force is also going to be 0. As far as the
minimum and maximum and such things are concerned, these are slightly more
complicated in 3 d cases. For example, although a point would be where the gradient is
0, it does not mean whether that it will be a maximum or a minimum for sure in three
dimensions. For example, if I come along the x line suppose at the origin the gradient
is 0 if I come along the x line x axis U could
go through a maximum. On the other hand, you could have a minimum
along the y direction, so to find the minimum maximum in 3 d, case one really has
as to see for example, for a minimum whether in all directions that point the U
is minimum. For a maximum, whether coming from all directions that point is a maximum
there could be more complicated cases, I just discussed coming in one direction would be
a maximum and the other could be minimum the which is known as a settle point.
You will learn about these things more in your electrodynamics course I thought I would
just give you a gist of it. .
Now, let us talk about an issue that we have been avoiding. So, far conservative forces
and non conservative forces how to deal with non conservative forces this you already
dealt with. To work with non conservative forces, we said earlier I should not be able
to define a potential energy since the work depends
on the path one simplest example in one dimension is frictional force.
.More you move more work you do you come back to the same point the work done is
not 0 and therefore you cannot define a potential energy for a frictional force in one
dimension how about in three dimensional case. Let us take to 2 examples and learn
through examples, let us take 2 forces F equals x I plus y j and try to construct potential
energies for these and this equals minus y I plus x j.
So, I am taking 2 forces let me just draw a line here F equals to x I plus y j and F
equals minus y I plus x j in this case you see that
the force lines are going radially outward. If
we do not see directly write F as x is r cosine theta y is r sin theta, so I can write this
as cosine theta I plus sin theta j and by now
having learnt polar coordinate this is nothing but r in the radial direction. On the other
hand, in this case the force line is going to be in
theta direction at all points, so they are going to look like this.
Again, if you do not go see them clearly write this in polar coordinates and you will see
that this is nothing but r minus sin theta I plus cosine of theta j which is nothing
but r in theta direction. So, further out you go the
force magnitude increases with respect to r and
this direction is in theta direction, let us try to find the potential energy for the
Again, let me draw a line F equals to x I plus y j which is minus partial of U with
respect to x I minus partial of U with respect to
y j and this implies partial of U with respect to x
is equal to minus x or U could be equal to minus x square plus some function of y.
Therefore, d U d y is equal to d f d y is equal to minus y and this after integration
implies that U is equal to minus x square minus y
square plus some constant of integration C. Let
us try to do the something for F equals minus y I plus x j, so partial of U with respect
to x is equal to minus y.
I can therefore write U as minus y x plus some function of y and partial of U with respect
to y therefore, is going to be equal to minus x plus some d f d y, which is a strictly a
function of y, but, d U d y cannot be minus x because d U d y is given to be plus x.
Therefore, U cannot be defined, this is a conservative force this is a non conservative
force, let me give the same take the arguments further.
So, in the case of F equals x I plus y j I found that U is equal to minus x square minus
y square plus C. I can take as a reference point
U at 0, 0 to be 0, then the C becomes 0 and the constant and the surfaces are like this
U becomes more and more negative as you go further and further out. Therefore, the force
is in this direction is v you can already it is
radially out, on the other hand when I take F to be minus y I plus x j I cannot really
find a function that will give me gradient of U to
be equal to minus F. Another way of looking at it is you see from d U dx equals minus
y I find U should be of the form of minus x y.
On the other hand, if I look at d U d y to be equal to x I find U to be of the form of
x y both cannot be true, so this is a non conservative
force I cannot really get any potential for this. Now, the question we asked do I
always have to try to find the potential energy and if I cannot find it, then I say it is
non conservative or is there any easier way to look
at it. If I all the time I have to integrate the force it becomes a difficult task at it,
so happens that they happens to be a differential
way of doing the same thing and now I like to talk about that it also gives me the definition
of a quantity called curve.
So, go back to the definition restricting myself to 2 dimensions and generalization
into three dimension is quite easy I go from point
one to point 2 and if I find that the work done is independent of which two paths I choose
is going to be a conservative force. If it is not independent or it depends on the path,
which I choose, then it is going to be non conservative force also put another way. If
I go from point one, go around this path and come back to the same point if the work done
is 0, then it is a conservative force if it is
non 0, then it is not a conservative force. So, let us try to do that let me go from point
x y to point 2 which is x plus delta x y plus delta y and without any loss of generality,
let me take this path to b 1 a 1 be let me take
this path to be 2 a 2 b. So, work along path 1 a since I am moving along the x axis, this
is path one a is going to be F x at x y times
delta x plus some correction because F x itself might change I am I am going to go all the
way to second order. So, some correction let us call it delta F as F x changes along the
x axis delta x square and some constant, this is
going to cancel. So, I am just putting some constant k k could to be one half or
something. As I move along path one b that is from here
to here only thing that is changes y, so this is going to be F of y at x plus delta x plus
y has not change plus some correction, let us
call it k one k 2 as I move along y F y may change delta y square. I am going to do this,
there should be a delta y here because I am let has a distance, so let me expand this.
So, I get delta w along one a is equal to F x x x y delta x plus some constant k 1 partial
of x delta x square is going to be F x at oh
sorry F y at x plus delta x y delta y plus some k 2
d F y d y delta y square. If I expand this further I get F y at x and y delta y plus
delta F partial of y component of F with respect to
x changes. Since I moved by delta x delta y
by Taylor series expansion plus a second order term k 2 y square, so I am keeping all the
terms to second order and there could be some higher term. Let me remind you again I
am calculating this work in going from point 1 to point 2 along path 1 a and 1 b. I could
do similar excise while going from path 2 a in 2 b liking go this you can take similar
steps and I leave this as an exercise.
You would find that the work done when I go from along this path comes out be along 2
a comes out be F y at x y delta y plus you can write k 2 delta F y del y del y square
and along 2 b this is 2 a this is 2 b. It comes
out be F x at x y delta x plus partial of F x with
respect of y delta x delta y plus k 1 partial of x with respect to x delta x square plus
higher order terms. So, let me flash the previous expressions again this is delta W 1 a one
b expanded it looks like this. This flash back and you see when I did the work along
path one I got these contributions and I worked
along path 2 I got these contributions comparing you will see that this term appears
in this case. Similarly, this term and this term, so I will
remove this I will focus only on this this term
and this term appears in this case also. This term appears and this term appears only
terms that I am left with are this and this, so if I were to take the difference moving
along path one a and one and path 2.
I will find that work done along path 1 minus work done along path 2, which is delta W
1 a plus delta W 1 b minus delta W 2 a minus delta W 2 b and let me flash those things
back again. This was delta W 1 a this delta W one b these terms canceled, similarly in
this case this is delta W 2 a delta W 2 b the only terms which were left were these
this also canceled. Therefore, I get this to be
equal to partial of F y with respect to x minus
partial of F x with respect to y delta x delta y. If this were to be 0 that is if force is
conservative, then this term must be 0 this implies partial of F y with respect to x minus
partial of F x with respect to y must be 0 otherwise not.
.So, I get delta W 1 minus delta W 2 is equal to partial of Fy with respect to x minus
partial of Fx with respect to y delta x delta y. This delta W 1 minus delta W 2 is nothing
but this is delta W 1 is the work done in moving from this point to this point and minus
delta W 1 is the work done and moving back along path 2. So, this is a work done in
moving the entire closed path and if this is non zero, then the force is non conservative.
I can give similar arguments for other dimensions
movement along the plane of y z or movement along the plane of x z.
I find in general that if delta F y over delta x minus delta F x over delta y or delta F
z over delta y partial y minus partial y over
partial z. Let me write it here partial F x with
respect to z minus partial of F z with respect to x if they are any one of these is non zero
you have a non conservative force. If these are all 0, you have a conservative force is
there a compact way of writing this the answer is yes.
These are all components of a vector form by this which is known as curl of F and you
can see this I j k this is by definition F x F y F z determinant, this gives me all
three components that I wrote in the previous slide.
This is a z component this is the x component this is the y component of that curl that
I wrote here.
If any of these components is non zero, force is non conservative, if they are all 0 the
force is conservative. So, I found given a force I do not really have to calculate the
potential energy I can straight away take its curl if it is non zero non conservative
force if it is 0 conservative force. This also tells
you that by the way that curl of a gradient must
be 0 because if I define a gradient; that means, I am taking about a conservative force.
.What does this curl mean, can I get more feeling for it, the answer is yes, as a name
suggest curl is something that describes a curly sort of thing.
Let us again go back to our two examples F equals x I plus y j and F equals minus y I
plus x j and take the curl, curl is going to be I j k partial with respect to x partial
with respect to y partial with respect to z of
x y and z and you can see this 0.This is 0 here, on
the other hand if I look at the curl of the other example j k partial of x partial of
y partial of z this is minus y x 0. You will see that
it gives k component or z component which is
equal to 2, so this is non zero, let me again make how these force will looking.
So, F equals x I plus y j the force was looking something like this and in this case F
equals minus y I plus x j the force field was looking something like
this in this case curl of F is equal to 0. In this curl of F is not
equal to 0, so this quantity curl seems to give
you a non zero value when things seem to be moving around in circles or they are curling
they turning around. On the other hand, if they do not they go in straight line curl
is 0 you can see in this case very easily if I
take a part and then go around in a circle work
done is going to be non zero. So, it is a non conservative force, this is a conservative
force to get more feeling for this let us look at some more examples.
.Let us take a force field F I just take make pictures which is something like this as you
go out it increases as you go on the negative side it again increases. Let us take another
one as you go up it increases as you go down it increases when the same direction. You
can see if I take a particle around here like this I do positive work when I go this way,
no work same positive work and come back here
I have gathered some work. Work done in going around in a closed path is non 0 that
means, curl of F is not equal to 0. Similarly, here if I go around the origin
in symmetric way you will see I pick up some positive work 0 work negative work equal to
this positive work and 0 again. So, around the origin the work done seems to be 0, but
if I take a closed path here, I do some work here 0 work less work because of force is
become less and come back here. So, I picked up some work in going around a circle, so
on this point curl of F is not equal to 0 around
the origin near the origin curl of F is equal to 0. Again, you see just looking at the
pictures the curl seems to give you a feeling for this moving feeling for a rotation on
this point this seems to be the lines are becoming
smaller and smaller. So, this sort of curling around they are curling
around here, they are not curling around they are changing the same way on both sides
and this is really the meaning of curl just to quantify this. Let us say this could be
something like F is equal to as you increase y F
magnitude increases in x direction. So, this could be a function that is described by this
sort of geometric shape, in this case as you no matter which way you go on the y
direction F could be y square i, so let us see if really our feeling of about curl is
correct or not.
If I make it again in this case I had force field like this and I wrote F equals to y
I, then I also had F equals y square I and the force
filed looks something like. If I calculate the
curl is I j k partial with respect to x partial with respect to y partial with respect to
z this is y this is 0, 0 and you will see that it
has a k component which is non 0, this gives you
minus 1 k, this is 1. So, this is non 0 all the time, on the other hand in this case if
I take y square, so I j k partial with respect to x
partial y partial z y square 0 0 it gives me a curl
which is equal to 2 y k. I am calculating curl of F I am calculating
curl of F and this is nonzero for nonzero y,
but, 0 at y equals 0. So, our feeling that at the origin curl is 0, so this gives you
curl of F is equal to 0 for y equal to 0, but non 0
at any other y for y not equal to 0 curl of F is not
equal to 0.
Let me make it clean once more F equals y I curl of F as we can calculated
previously is not equal to 0 F equal to y square I curl
of F is equal to 0 for y equal to 0 not 0 for y not
equal to 0. So, you can see that the curl seems to give you a feeling here again the
some sort of rotation involved here, there is no
rotation involved it gives you a feeling for a
rotational kind of thing, these quantities curl and so on. What introduced initially
in the studies of fluids that is where the origin
lies and you can get a feeling again a physical sort of feeling if you think in those terms
think of any given force field. .
.For example, the one that we took this or this or this or the radial one and imagine
these lines to be representing velocity of a fluid
flowing and think take a small match stick put
it somewhere. So, I take a match stick put it somewhere here if you feel that the match
stick rotate then curl is not 0 if you feel match stick would not rotate curl is 0. So,
for example, in this case no matter where you
put the match stick it will tend to rotate if this
is the velocity of the fluid, curl is non 0 anywhere in this case. No matter where I
put the velocity is becoming larger and larger as
you go outside the match stick would tend to
rotate curl is non 0. In this case, if I put the match stick at
the centre it is being pushed by the same force on
the same velocity on both the sides it will not rotate curl is 0, but if I put it somewhere
here, it will tend to rotate like this curl is non 0 in this case. No matter where I put
it will not rotate curl is non 0, so this sort of
physical way of looking at the quantity curl having
learnt about curl and getting some feel for it. Let us talk about little more, so that
you get a better feeling for it, I will first discuss
a fantastic theorem about curl that really arises
from the way we derived it. .
So, this is known as Stokes theorem and what it states is that if I take a closed path
and given a force or a vector function F which
I multiply over dl where dl is a line element path and integrate over this closed path.
.For example, I could be calculating the force in sorry this should be the other way force
in going around this close path from this point and coming back to the same point. So,
I am going this way and coming back to this
point and I want to calculate the work done of course is the force is conservative that
means, a curl is 0 the work done is 0. If it is not
conservative curl is non 0 and this integral is given as the curl of the force field is
perpendicular component to this the this surface on which the curve is being made times
the area d a. Let me elaborate a little bit I am taking
a close path, so I am calculating F dot dl along
this path this is related to the curl. So, I can think of a small area da on the surface
find the curl there multiply with a component of
curl perpendicular to x surface with that delta area small area and added up. It is
a very useful theorem and you can see right away how curl becomes very important quantity
for us. The proof is also very nice and I would like to go where it, so that you sort
of get a nice feeling about curl. .
For proving, I will go back to the way we obtain the quantity curl where we obtained
was it was by taking a small rectangular path
going around it starting from x y going to x plus
delta x y going up to x plus delta x y plus delta y coming back to x y plus delta y. Then,
getting back to where we restarted from, in that case we found when I go counter
clockwise then the work done F dot dl along this path close path was equal to delta F
y delta x minus delta F a partial F x partial
y times delta x delta y.
.This is nothing but the perpendicular component of F on the x y surface times the small
area delta a sort of Stokess theorem and a miniature and infinitesimal version. Now,
would I can do is take a big path and divide it into this small infinitesimal areas and
for each area I can go around like this for this
area I will go like this. For area I go this I will
go like this and in doing, so finally, when I add
F dot dl around each area you will see that opposite going paths cancel each other.
Finally, the only place where they would not cancel is this outside line, so finally, if
I add all these F dot dls, I am going to get F dot
dl around this path, let me call this C one to distinguish it from this one.
For each path, I know it is delta F y over delta x minus l partial F x over partial y
or perpendicular to of the curl times delta.
So, this is going to be a sum over curl of F
perpendicular times delta x or delta x delta y or delta a, which is nothing but I can write
this in integral forms it is sum over small infinitesimal areas F delta x. You have to
be careful in these because we are talking about
directional quantities in that you follow the
convention when I go counter clockwise and then the thumb. If I curl my fingers around
the counter clockwise direction, the thumb gives me the direction along which I should
take the component of the curl of F. .
So, let me rewrite this what I did was I took a close path divided this into small grids
or infinitesimal areas I took this F dot dl around
each path and added it all up when I added it all up all these counter going paths canceled
the contribution of F dot dl. Finally, I was
.left with the only this, but the area integral that is, so for each small one I had F dot
dl is small one let me say it C i when I summed
it over it gave me F dot dl around this path C
1. This is C i and when I summed over perpendicular component of F with the area it
gave me integral curl of F perpendicular delta a.
These two are equal and that is a Stokes, here again I remind you be careful about the
direction we are talking about directional quantities. I am going to follow right hand
rule when a fingers curl right hand figures curl
thumb gives me the direction along which I should take the component of the curl.
As I told you earlier, this Stokess theorem F dot dl is curl of F times delta a integral
over that surface really comes from the way we
defined initially the curl quantity. So, it is a
very useful quantity when I want very useful theorem when I want to see curl in different
situations or curl in different coordinate systems. As an example, let me take now what
will be the expression for curl in my polar coordinates, remember in the beginning of
the previous lecture. When I was talking about
gradients I had warned you that when I write gradient as I d by d x, let me just confined
myself to two dimensions. It does not mean, it does not imply that in
polar coordinates it is going to be r d partial with respect to r plus theta partial with
respect to theta. In fact, this is not even dimensionally correct, same thing is true
for all other quantities and as an example I will
.just derive curl for you using this definition or this theorem stokes theorem which really
initially introduced in infinitesimal way the quantity curl.
So, if I am talking about polar coordinates I am going to choose my path in polar
coordinates as this following right hand rule I go counter clockwise. So, for the curl the
component will be coming out of this screen, this point is r theta this point is r plus
delta r theta this point is r plus delta r theta
plus delta theta and this point is r theta plus delta
theta. I will go quickly over this if I want to calculate F dot dl over this path is going
to be Fr at r delta r that will be the contribution
from this plus F theta at r. However, as I go out theta component changes
it changes by partial of F theta by partial of r delta r and the distance I cover is this
one and this distance is r plus delta r times delta theta, now I am coming back. So, this
becomes minus F r, however, this is Fr at theta plus delta theta, so this is going to
be minus Fr at r theta. I should also have theta
here minus how the radial component of F changes with respect to theta delta theta and
the distance I cover. Let me make it plus put the brackets here delta r and finally,
when I come along this path is going to be minus
F theta at r theta delta theta times r. You can check for yourself these are the only
linear terms the other terms, if I really look
at the other variations of these quantities they will give the second order terms. So,
those higher order terms third order terms in delta
theta and delta r, so I am not worrying about those. Now, you can see this will cancel with
this and similarly, F theta r delta theta will
.cancel with this one part of this the other part I will get F theta at r theta delta r
delta theta. So, let me just write it here then
I will take it over to the next screen. I am going to get this equal to F theta delta
r delta theta the other component F theta r
delta theta cancels with this. Then, I am going to get plus partial of F theta with
respect to r delta r, r delta theta minus this term
will be left which is partial of F r with respect to
theta delta theta delta r. .
So, let me rewrite it what I have determined is F dl and this comes out to be I went along
this path, let me copy this comes out to be F theta delta r delta theta plus del partial
theta partial r r delta r delta theta minus partial
Fr with respect to theta delta r delta theta. This
is F dot dl by definition or by Stokes theorem this must be equal to curl of F coming out
of the screen this is x this is y. So, it will be z component times the area which is
going to are delta theta times delta r and that gives
me the z component of curl of F and z direction in terms of polar coordinates.
This comes out to be one over r F theta plus partial F theta over partial r minus 1 over
r partial F theta with respect to a partial
Fr with respect to theta, which in short I can write
as one over r partial r r F theta minus partial Fr over partial theta. So, you see I have
been able to derive using the basic definition
the curl in different coordinate system and use
the basic theorem of this.
.This is just to give you a feeling for these quantities you will obviously, be using the
advanced version of these or using it more and more in your coming courses in
electrodynamics advance mechanics and so on. So, you should practice a lot of problems
on this play around try to get may be the other components of curl in cylindrical
coordinates or in spherical coordinates. To conclude, I will give you potential energy
in three dimensional how potential energy and
force are related. .
How we can think of a conservative and non conservative force and how we can test
them using differential form of differential way of taking curl and this sort of some of
our introduction and some details of work energy potential energy and so on.