Understanding Electrostatics: Conservative Forces and Energy Conservation
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Introduction
In the realm of physics, electrostatics deals with the behavior of electric charges at rest. Through electrostatic principles, we can predict how a test charge will interact with a system of fixed charges. This article delves into the foundational concepts of electrostatics, focusing on conservative forces and the law of conservation of energy.
The Basics of Electrostatics
Electrostatics is concerned with charged particles and the forces acting upon them. When a charge, denoted as q, is placed in an electric field created by other charges (e.g., q1, q2, q3, or a continuous line charge λ), we can analyze its behavior based on established principles. The fundamental objective is understanding how the charge interacts with the provided electric field and predicting its motion using Newton's laws.
 Calculating Electric Field
The electric field E at a point due to multiple point charges can be calculated using the principle of superposition. We compute the contribution of each charge to the electric field and sum them up: E = E1 + E2 + E3 + ... + En
Where E1, E2, ... En represent the electric fields due to each charge.
 E = E1 + E2 + E3 + ... + En
 Force on a Test Charge
The force F experienced by the test charge due to the electric field is given by: F = q * E
Here, F is the force on charge q at a point in the electric field.
 F = q * E
Predicting Motion with Newton's Laws
From Newton's second law F = ma, we can determine the test charge's acceleration:
 a = F/m By knowing the charge's acceleration, we can update its velocity and position over time, leading to its trajectory.
Conservation of Energy in Electrostatics
The conservation of energy principle is foundational in physics, asserting that energy within a closed system remains constant over time. In electrostatics, this principle helps us understand how a charge interacts with an electric field, leading us towards the concept of potential energy.

Kinetic and Potential Energy
The kinetic energy K is given by: K = 1/2 mv^2 Where m is the mass and v is the velocity of the charge. Potential energy U in an electric field is related to work done against the electric force, obtained from:
 W = ∫F dx Thus, the workenergy theorem connects kinetic energy changes with the work done via the electric force.

WorkEnergy Theorem
The workenergy theorem states that the net work done by the forces acting on an object equals the change in its kinetic energy, which is foundational for calculating energy conservation in electrostatics: [ K_2  K_1 = W ]
Conservative Forces
A force is considered conservative if the work done on an object moving between two points is independent of the path taken. This characteristic is crucial for electrostatics, where understanding electric potential aids in the simplification of calculations:

Conservative Forces
If a force can be derived from a potential function U, then it is conservative: F =  ∇U

Identifying Conservative Forces
One can determine if a force field is conservative by checking if the line integral of the force around a closed path is zero: ∮F • dr = 0
Electric Potential and Fields
The concept of electric potential V significantly simplifies calculations in electrostatics. Electric potential at a point in space is defined as the potential energy per unit charge, expressed mathematically as:
 V = U/q
 Thus, the electric field can be derived from the potential by differentiation:
 E =  ∇V
Example: Electric Potential of a Dipole
One practical application is calculating the potential created by a dipole consisting of two equal but opposite charges separated by a distance. The potential at a point in space can be calculated analytically:
 Potential Energy of a Dipole
For a dipole, the potential V is typically expressed as:
 V(r) = (p • r)/(4πε_0 r^3)
where p is the dipole moment.
 V(r) = (p • r)/(4πε_0 r^3)
Conclusion
The interplay between electric charges, forces, and energy conservation forms the basis of electrostatics. Understanding these principles allows predicting the behavior of charged objects in various configurations effectively. Moreover, the identification and characteristics of conservative forces facilitate a streamlined analytical approach to complex electric scenarios, leveraging potential energy computations. In all, the study of electrostatics is foundational for further explorations in physics that may venture into dynamics and field theory.
We can do the following problem. You give me any number of charges which are fixed, anywhere you want.
And instead of these discrete charges, you may have a line charge too, maybe λ coulombs per meter. That guy is here.
which can actually move, allowed to move. The question we ask is, do we know what this charge will do?
And the answer is, yes, we know what it will do, because if it's sitting here and we have already computed the electric field everywhere you know how to compute
That is some electric field sitting here. Then we know, the force on this charge we're interested in, this q times the
electric field at the location of the charge. So at every instant, we know what force is acting on it.
Then from Newton's law, that will give me the rate of change of velocity or the acceleration. You can write it either way.
That's all I need, because using that I can find out the new position, new velocity, right?
The acceleration tells you, if you wait 1 nanosecond, you go to this place, because the initial velocity, I can predict where it will be, and because you know the
acceleration, I know what velocity it will have. Then I keep on updating it and I get my trajectory.
Student: I think it should be dt squared. Prof: Yes, thank you. So everyone understands this is how you proceed.
This charge will of course push those guys, but we're not worried, because we don't let them move. They are just fixed.
And if there are much such charges, we can find what they all do in this environment, which is the background electric field.
That's all we can do right now, this is electrostatics. In the real world, everybody is moving and all kinds of stuff happens, but you cannot use
electrostatics for that. This asymmetry between the background and the test charge that we are studying is in the static situation.
Guys that move and guys that don't. Now this is something you could have done on the first day, because I think on the very first day I told you about
Coulomb's law and the principle of superposition. If you put them together, you know the force on a charge q.
Then you use F = ma, you predict its trajectory. But in principle, that's all you need to do electrostatic problems, but in practice you want to
clean up the machinery a little bit, make it more efficient. For example, we found Gauss's law.
Gauss's law follows from Coulomb's law, but we found it's quite useful. For example, if we give you a ball of
charge, spherical charge, going back to Coulomb's law and first principles is very difficult. But if you use Gauss's law and the symmetries,
you're able to find the electric field around the sphere, or an infinitely long cylinder, or an infinite plane or simple geometries like that.
So it's not enough to say "I've got the basic equations." You always want to see if there are ways to speed up the
computation to make it more efficient. So let me remind you what we did in the first part of the course.
There, if you have the following question  here's a spring and here's a mass, the spring likes to have that length and I pulled it to way over there and I released it,
and right now it happens to be here, going at some speed. I want to know what velocity it will have when it is here,
I want to know v_2. That's a reasonable thing you can ask. Unlike the electric force, the force on this mass is due
to the spring. Apart from that, it's the same situation. So one way to do that is to say, I know F = ma,
so the instant you start the problem here, I know the acceleration because the force of the spring is k times x_1.
Once I know the force, I know the acceleration. I can find the new velocity, new position, keep on doing it, or I can solve the equation,
d^(2)x/dt^(2) is −kx/m, and I can eventually end up here. Then I can find the velocity here.
Anybody in the last row want to tell me what the shortcut is? Yes? Student: You can use conservation of energy.
Prof: Conservation of energy. So conservation of energy is a consequence of Newton's laws and certain properties of the force.
I'm going to remind you how it's done. One attitude I could take is, you've already done this before, but usually it's better to make sure that you're all on
the same page when we do this. So let me remind you how it works there. You go to Newton's laws and we write m dv/dt =
Then I multiply both sides by v. On the left hand side, if you know basic calculus, this is really d/dt of
mv^(2)/2, by the chain rule of calculus. The d/dt derivative is the v derivative, followed by the derivative of v with respect to
This becomes F dx/dt. This means in every tiny instant of time, the force times the change in the position is equal to the
change in this quantity, which we know is called kinetic energy. So if you multiply both sides by dt,
then you find this relation, and if you add up all the changes, we get this relation, mso let's see what we are integrating from a starting
And this is called the work energy theorem. The work energy theorem tells you something you may expect, namely, when a force acts on a body, it's going to accelerate
You're going to ask, "How much velocity change do I get for a certain action of the force?" and the statement is, the force pushes the body from
x_1 to x_2, then this is the change in this quantity called kinetic energy. This is true no matter what the force is.
It's true for friction, it's true for spring, it's true for everything, because Newton's law is always valid.
But if you want to extract from this a law of conservation of energy, you have to do a certain manipulation and that's what I'm going to describe.
So let's go back here and try to extract from it a law of conservation of energy. So I'm going to write here k_2 
k_1 is integral F(x) dx, from x_1 to x_2. Now it's a fundamental result in calculus that integral of any
where G is any function whose derivative is the F I'm integrating. For example, if F = x,
You can add x^(2)/2 a constant, but the constant is not important, because they'll cancel between this and this.
That tells you the combination K  G does not change as the particle moves, but this doesn't look very nice.
We want to get the old law of conservation of energy in the standard form. You define the function U(x),
So the relation between U, which is called the potential energy, and the force, is the following.
The force is  the derivative of the potential energy. The  came because G is the integral of F, but U is G, so that this formula,
Again, the  sign is present because F is defined to be  the derivative of U. So you can go back and forth from potential to force and
If you knew the potential, you can differentiate it to get the force, and more importantly, you have this great conservation law,
which is very helpful in the spring problem, because if you know the potential energy, once and for all, it's kx^(2)/2,
you balance the kinetic and the potential, you can find the velocity at one other point. Now why doesn't this work for friction?
Prof: It's not conservative. It's just a definition. But what's wrong with my argument?
Where in this argument does it fail? It looks like any functionyes? Student: Because the direction of the force depends
on the direction that the particle's moving. Prof: Very good. The point is that frictional force is not a function just of
exert a force kx if you are displaced to the right by an x, it exerts a force kxwhether or not you're moving to the right,
away from the equilibrium or towards it. Thus given x, it gives you a given force, but friction is not like that.
Friction does not have a definite direction. The only thing friction wants to do is to slow you down. If you want to go right, it wants to act left.
If you want to go left, it wants to act right. I will resist all political analogies. I'll just point out to you that friction does not have a
definite sign, therefore it's not a function of x, but as she said, it's a function of x and of x dot,
which is the velocity. So it doesn't have a unique answer. That does not mean you cannot find the work due to friction.
If you know it's moving to the right, then the force is to the left. You can integrate that force, as long as it is moving to the
right. But if it's going right, left, right, left and oscillating, then you cannot integrate it
once and for all. That way doesn't work for friction. Now here's an even more powerful example of the law of
conservation of energy. You have this famous roller coaster and you are here. I say "you are here," because you'll never get me on
top of one of these guys. So you are here and you got some speed, and I want to know your speed when you are here.
Again, it follows from Newton's laws. You can calculate from Newton's laws, because at every point, wherever you may be,
there's the force of gravity acting this way and there is the perpendicular force from the track acting this way, and the perpendicular force is such that it cancels this part
of gravity, and this part of gravity will accelerate it and you can find the velocity here. Keep doing this, you can find the velocity at
the end. But that's not how we do this. We use the fact that the gravity force allows us to
define an energy, and I'll come to that in a moment. Then we say kinetic potential is kinetic potential in the
beginning and at the end, and we can simply find the velocity here if you knew the potential energy here. All right, so the law of conservation of energy is also a
very profound result, which has survived all the revolutions of quantum mechanics. After quantum mechanics, we don't like to think in terms
of force, and we don't like to think in terms of trajectories. You don't even know for sure where the particle is. But amazingly, the total energy is always
conserved in the problems we study. So the law of conservation of energy is very profound and it's also very useful for computations.
Now the next question I have is, if I go to two dimensionsand it's the same as in three dimensions. I'm just going to go to two dimensionswhat does it take to
get the law of conservation of energy? So I'm going to mimic this derivation in 2D and see if I can pull out a law of conservation of energy.
So here's what I will do. So in two dimensions, let me start this time with the kinetic energy.
Then let me ask, what's the rate of change of this kinetic energy? You've got to take derivatives.
Now you may be a little queasy about taking derivatives from this dot product, but you can check by expanding the component, just like ordinary product,
and you can take derivative of this guy times that guy, derivative of that guy times this guy, and they will both be the same, and you will find it's equal to
In other words, the change in kinetic energy in one dimension is just force times distance for tiny distances.
In higher dimensions, like in 2 dimensions, if the particle is moving in the 2 dimensional plane along some path and it moves a distance dr,
and at that point, the force is in that direction, you should not simply take the product of the length of the force times the length of the distance,
but that times the cosine of the angle. Why should it take that? Because that's what gives you the rate of change of kinetic
energy. Just go to the kinetic energy and it suggests to you that you take this product and define it to be the work,
because if you define it that way, just like in 1D, the work done by the force will be the change in kinetic energy. You guys with me now?
In spite of this little mess up here, this ma is my force. The dts got canceled. The dK is added up to K_2 
K_1 and this is the integral. All right, so now it looks pretty good, except for the complication.
This is say where you started, and maybe you ended up here at r_2. You've just got to integrate this F from
r_1 to r_2. Let's imagine there is no friction. Let's imagine F depends only on the position.
Prof: Your path. See, in 1 dimension, if I want to go from here to there, I have no choice, okay?
Just move along that line. In 2 dimensions, if you want to go from one point to another, I can think of many ways.
You can do that, you can take the straight line path, or you can take some other path. That's an ambiguity we have in higher dimensions.
So even though the work energy theorem is valid, and no matter which path you take, the work done by the force will be the change in kinetic energy,
Because if it had this form, we are done. You agree? If it had this form, then again you write
If it was true, you are in business. But it's a very unreasonable thing to ask of this force, that its integral from some point to another point,
does not depend on what path you integrate it on. In other words, you can take that path, break it up into segments dr, and on each segment,
the force is some variable. You take F⋅dr of that little segment, add it up over all the segments and make the segments
vanishingly small. That's the meaning of this integral. And you're supposed to get the same answer, no matter what you
do, including going like this, or anything else. How can they all give the same answer? In fact, I'll give you a simple example where it's simply not
true, so the answer depends on the path. In fact, just about any force you pick randomly will not have the property, the answer does not depend on
the path. Answers typically will depend on what path you take. So let's take a simple force, F =
So here's a simple one  force depends on position. It points in the x direction only and it grows with y.
And I'm going to do two calculations. I'm going to find the integral from the origin to the point (1,1) first by going this way and then by going that way.
They are two possible paths. And you will see, when I do the calculation, the answer is different.
Therefore you cannot write the answer as a difference of something here  something here, because the answer is not a function only of the starting and ending points,
So let's take this segment here. What is the work done? The work is the force which is i times y times
the distance. If you move horizontally, perhaps you guys can see dr for a tiny segment simply i times dx.
it looks like this. It gets bigger and bigger and bigger as you go further off. And if you go to negative values, it looks like this.
nonzero, but notice, the force is like this and the dr is in the y direction and they are perpendicular, so the dot product is 0.
0. If you take this path, the work done is 0. Now let's take the other path, go like this and then like
You're moving straight up. dr and F are perpendicular. But on the last segment, from here to here,
goes from 0 to 1, and y is frozen at the value 1, because you're on this line. Frozen at the value 1.
Just put 1 for it. And i⋅i is 1, integral of dx is 1. Therefore the work done by the force going up and then to the
right is 1. Going to the right and then up is 0. You can take other paths and keep getting other answers.
So this is a force for which it's not true, and I bet you that if you just randomly selected forces in the 2 dimensional plane, they will never have the
property that the integral is path independent. So it looks like we are onto a hopeless task, that the answer cannot depend on the path.
So maybe there is no problem in which you can derive a law of conservation of energy. Of course, you guys know, you can derive a law of
conservation of energy for a family of forces. In fact, the answer is, there's an infinite of numbers of forces you can devise for which this is true,
namely, the answer does not depend on the path. Before writing those forces, let me explain to you one other way to write the condition.
The condition, namely of a conservative force, is that the integral from here to here is the same as the integral from there to there, from 1 to 2 along path A
and 1 to 2 along path B. They must be equal for any starting point and any ending point and any two paths joining them.
Shifted it to the left hand side. But then we use the fact that when you go from 1 to 2, with a  sign, this second guy is same as from
2 to 1 along B of F⋅dr. In other words, the line integral going this way is  the line integral going backwards,
Therefore it tells you then that the line integral from here to there and back is 0. We write that by saying the line integral of
F⋅dr in a closed path is 0 for all paths, for all starting points, all ending points, all paths.
And that is the force that we want. That's the conservative force. I've just rewritten.
I've not gotten any closer to finding conservative force, but I'm saying there are two ways to write it. Answer doesn't depend on the path, or the answer on any
closed loop, the line integral on any closed loop, is 0. That means sometimes it's got to be positive,
sometimes it's got to be negative. So question is, where am I going to get a force with these amazing properties?
And I'm just going to give you the answer. The theoremif you want, you can call it a theorem, you can call it whatever you like.
It's a factnumber 1, here is a recipe for finding conservative forces. Take any function U(x,y).
You can write your own ticket. Write whatever function you want, time hyperbolic sinh(y) times cosh(x),
force whose x component is  the x derivative of the function I picked and the y component is  the y derivative of the function I picked.
So in my example, F_x will be x derivative partial means keep y as a constant, take the derivative with
3x^(2)y^(2) and 2x^(3)y. Let me see if I got that right, because you guys have been catching me too many times.
Okay, I think that's all right. This force is guaranteed to be conservative. I will show you why this is true.
Why is this magic working? But let me write it in another way that's more compact. So the force I want to get, written as a vector,
That's the force I claim is conservative. We're going to write that as  grad U. This is called the grad and it means gradient,
and this symbol is shorthand for this. There's nothing more to it. Gradient is a machine.
you give it x squared, it gives you 2x. Gradient is a different kind of machine. You feed into it a function U of x and
y, and it gives you a vector in all of the xy plane obtained by taking the two derivatives, one along x and one
the force we have = i times 3x^(2)y^(2), j times 2x^(3)y. Amazingly, it is claimed that this force, when you integrate
it from one point to another, answer will not depend on the path; it will depend only on the end points.
So we have to understand what makes this work. Before I show that, I will make another statement. I will show you why this works, but the second statement,
I will not prove, which is that all conservative forces are of the form  grad U. In other words, not only is this an example of
conservative force, that's it. There are no more examples. Every conservative force you cook up will always beyes?
Student: Does the j component ___________? Prof: j component should be what? Oh, that's negative.
So how are you guys doing with this? I'm always worried. You are inscrutable, and I don't know what's going
on in there, whether silence means "I'm with you" or silence means, "I'm so far behind, I don't even know where to
If you don't speak up, you don't get service. You have to ask. Don't assume people next to you know anything.
All right. So it's like one of these psychotherapy sessions. Turn to the person next to you and insult him or her,
because that's what you want to do. Don't worry about them, okay. When you go to physics seminar, you never assume the speaker
has a problem. I mean, you assume that you don't have a problem. If you don't follow it, it's a problem with the
speaker. That's what makes seminars exciting. It's the closest thing we have to gladiator fights in the old
days. You bring a speaker from somewhere, maybe Harvard, then you put them on the stage and you just roast them for the
Somebody else did it hundreds of years ago. You should relish that challenge. If this was a small seminar, you will have time to do it.
I don't have that much time, but I also don't have that little time that if you don't follow me, you cannot intervene. Okay, you have a question?
fashion is conservative, okay? So let's calculate integral F⋅dr for this force from some startingI'm just going to call
It is F_xdx F_ydy. That's the meaning of the dot product. I hope you know what I mean by this.
I'm going from here to here, that vector dr is some amount of dx and some amount of dy. It's got jdy and idx.
And F is some other thing, pointing in that direction, so F⋅dr is this. Just look at this part.
This is what you're integrating. That becomes, except for this  sign, dU/dx times dx dU/dy
What do you think it stands for? In 1 dimension, if I took df/dx times dx, what am I calculating?
Come on guys, you know what the derivative means. When you multiply the derivative by some dx,
Prof: You're finding the change in the function f, right? The change in the function f is df/dx
that's all it is. That's the meaning of the rate of change. If you multiply the rate of change by the change you get a
change in the function. If you're in 2 dimensions, if you move from one point to another point, you've got a function U
that depends on x and on y. Therefore it changes, because you change x and it changes because you change y.
It's changing because of two reasons. This is the change due to change in x; that's the change due to change in y.
For very small, infinitesimal dx and dy, that is the change. That is just the change in the function U.
It follows then that if you add up all these F⋅dr's from 1 to 2, this integral will give me U_1  U_2.
Therefore integral F⋅dr from 1 to 2 is U_1  U_2. So the magic is the following  you pick a function U,
you can think of the function U as measured perpendicular to the blackboard. It's like a height or something coming out of the blackboard.
You cooked up a force which is related to the rate of change of the function U, so that if you add all the F⋅dr`s, you're getting the change in
the function between two points. And that change in the function is independent of how you got there.
Think of U as a height of a mountain, sitting on top of the xy plane. Then F is really proportional to the rate of
change of the height of the mountain. And F⋅dr is the change when you move a distance dr, a vector distance dr.
You add all the changes, what are you adding? You're adding the height, your elevation. And when you're done, the total change is the height
difference between the final point and the starting point. It doesn't depend on how you climb that mountain. You can take a long path, you can take a short path.
As long as you're keeping track of how many feet have I climbed, you're going to get only one answer. So not every F has this property, but an F
derived from U by that trick of taking the gradient has this property. Therefore for that F, you might say this is for
In other words, the U that you began with is the potential energy for that problem. So what you're manufacturing is pairs of potentials and forces.
For every U that you dream up, there's a force, which is obtained by gradient,  gradient of U. And in that problem, where that force is acting on
particles, the energy that is conserved will be the kinetic, the function U at the starting point, will be equal to kinetic plus
potential at the end point. And the interesting thing is, there are no other conservative forces, except forces you can get by taking a U and
taking its gradient. So all conservative forces fall under this category, and they all have a conserved quantity where the potential
energy is the U from which the F is derived. So we can say it as follows  F =  gradient of U.
And U at 2  U at 1 is  integral F⋅dr from 1 to 2. This is the 2 dimensional generalization of something I
F⋅dr, rather than the F_x dx. So you can get conservation of energy in higher dimensions for
those forces that come from a potential. U is called a potential. U is called a potential energy.
So now we can ask the following question  suppose I give you a force. I know if it's conservative or not.
How are you going to find out? Now there are two options. One is, you are so smart, you can look at the function
and say, "Hey, this is the gradient of some other function U." If you do that, then you are done,
because if it's the gradient of some function, we know it's conservative. But what if you cannot see that right away?
With simple polynomials, you can easily guess. Like the example I gave you, if F_x is 3x^(2)y^(2), F_y is
2x^(3)y, maybe you can guess that U is x^(3)y^(2). But for more complicated functions, you won't be able to
Then consider dF_y / dx, which is d^(2)U/dxdy. And it's the property of partial derivatives that the
cross derivative, second cross derivative, is independent of whether you take dydx or dxdy. So let's go to our test case, U = x^(3)y^(2).
this y derivative. That's written as d^(2)U/dydx. y derivative of the x derivative gives me
Prof: Yes, that's correct. So in 2 dimensions, this is the only test you have to apply.
The test you have to apply is, are the cross derivatives of the force equal? Namely, is the y derivative of the force
= dF_y/dx. If that is true, it is conservative. And it comes from the fact that if F is the derivative of
some function U, then this mix derivative being equal is a diagnostic for that. In 3 dimensions, there are 2 more equations.
We're not going to use them, but I'll just mention what they are. You get that by moving every index 1 notch.
There's one more condition. And another condition, z goes back to x, x goes zlet's see, y goes to z,
but I do want you to know in 2 dimensions, you have to remember this condition. So this mathematic extends to all dimensions,
but I'm doing it for the case of 2D. So let's take stock of what has been done. What has been done is to think from scratch on how to get the
law of conservation of energy in high dimensions. You start with the kinetic energy and you ask, why does it change?
And you find it changes due to the force, and you find the change in kinetic energy is the line integral of F⋅dr from start to finish.
But that doesn't give you a work energy it gives you a work energy theorem, but not a law of conservation of energy,
because this integral can depend on the path. Then you say, maybe there are some functions for which the answer depends only on the end points.
Then we get a law of conservation of energy, and we found out that we can manufacture such functions at will, by taking any function of
U, taking its gradient, or  the gradient. Therefore that's a way to generate conservative forces and
they are the only conservative forces there are. There's no other conservative forces which are not gradients of something.
And the relation between the potential and the force, you should know now. It's what you learned in 1D, suitably generalized to higher
dimensions. Change in potential is this and the forces obtained from the potential by taking the gradient,
the change in potential is obtained by integrating the force. If you ever get confused about sign and  sign and so on,
go back to the harmonic oscillator, for which the thing is 1 overis kx^(2)/2. So let me give you one simple example that we all know,
which is the force of gravity near the earth. So here is the earth, so there is some x and y coordinates, and here is z.
So I might as well write down the answer, because you guys all know this. U for gravity is mg times z. And
kmg, because it doesn't depend on x, it doesn't depend on y. It depends only on z and the z derivative is
trivial. Force is mg pointing down. So the force is the gradient of the potential and you can ask,
is the potential difference between two points, U_2  U_1, is it equal to integral of F⋅dr from 1
then the integral of mg times dz from start to finish, with a  sign is just mgz  0. So indeed, it is true.
And if you took two points, not on top of each other, but something like this, you can always do the integral from here to here, where there's no change in
potential, and do the integral there, which will give the same answer. So gravity's a very trivial example, a very easy case,
but you can see that gravitation is a conservative force. And you can see why it's a conservative force,
because if you took a path where you went like this, let's say, and you went like this, where you get the same answer as here and here.
Because in this path and this path, you don't do any work, because F and dr are perpendicular, right?
F is like this, and the displacement is like that, and it's the same thing, whereas here, if you go like this,
are the same height so you get the same answer. Okay, but now, why did we spend quite a bit of time on conservative forces, because that's not what the
primary focus is? It was electricity and electrostatics. So we've got to ask ourselves, does the electrostatic force
meet the test of a conservative force? Well, we know it will, because otherwise I wouldn't have spent this much time building up the stuff.
But let's just verify that. It looks like a very formidable task. Let's see why it's a formidable task to verify it's
conservative. I'm claiming that you've got all kinds of these charges producing a field here and I want to prove that the line
integral of this force on any closed loop is 0. Or I want to prove the integral on one path is the integral on any other path, for any 2 paths between any 2
points. I don't have time to show all that, right? If the theorem is wrong, do you agree it may not be so
If you pick any 2 paths, should be the same end points, so always the integral is not equal, that's it. But the fact that they were equal doesn't mean that somebody
else will not find some other two paths with some other starting points which they are not equal. So how are we going to do this?
How are we going to test for every configuration of electric field, every element of charges is conservative? So what do you think a strategy will be?
and use that test to see Prof: Right, but to find the electric field and see if it works, you have to calculate the electric field as a function of
all charges, because of supersposition. Prof: There we go. If it works for one charge, it will work for any
arrangement of charges. So you may not have appreciated fully the power of superposition, but you can see now.
If you did not realize that, the problem looks insurmountable. Who can handle every arrangement of charges?
But if charge by charge it is true that the electric field due to charge 1 is conservative, electric field due to charge 2 is conservative.
other will give 0 on any closed loop. The trick is going to be just do it for one charge. That's it.
If you can see it for one charge, you get by superposition that it's true for any charge, because the field is additive and the integral is additive, okay.
That's the way integrals work. If each one vanishes on a closed loop, you're done. So I'm just going to take one charge.
Here's one charge and let me draw for convenience all the field lines coming out of it. Then I want to see, if I do a line integral,
So the first integral is from 1 to 2 on this path. Then I'm going to take another one, which is different from it, where you can see the answer is
the same, then go on making more and more changes, and the answer will not change. So I'm going to take another path where I go tangent to the
lines, along a circle. Then I go here, till I get to that radius, then I turn around and do that.
I claim the integral will be the same on this path, this one, that one and that one. So I want you to think about why it's going to be true,
okay? Think of it, why it's okay to do that. You've got a new path but it doesn't make a difference.
So other people know why it's true. I'm going to ask the other people. Okay, why is it true?
Yes? Student: It's true, because when you're moving on the paths this way, they're radial,
so the radius is constant, which would mean that there's no _______. Prof: There's no what?
Student: There's no change in force because the radius is constant. Prof: That's not a good reason.
Student: It's always perpendicular to the force. Prof: The force and the displacement are perpendicular in this section, because force is radial,
displacement by definition is tangential along the circle. So dr is perpendicular to F. Oh, by the way, I should tell you something.
The electric force in any situation on a charge is q times the electric field. You will agree that if the electric field is conservative,
multiplying by a number q is not going to change the fact. So I'm going to prove the electric field is conservative,
then the electric force on any charge is just proportional to it. It will also be conservative.
So really taking force on a unit charge and seeing what happens. So this section is for free.
This sectionwe'll come to thatthis section is also for free, because F looks like that and dr looks like that.
Here you can pair the sections, so that that section is paired with that section, because you're moving the same radial distance with a radial force.
F⋅dr, the dot product, is the same here, section by section. Any work you're doing here, you're doing here,
therefore the answer is the same. Then you can see that you can go on adding all kinds of changes, radial and angle,
radial or angular, and when you're done, all the radial integrals will give you that part, and the angulars won't count.
So as you move around, it's only when you're moving radially you're even aware that you're doing work. Going in the angle doesn't matter.
And the net radial change in going from here to here is that distance, no matter how you move. Now you might say, prove to me that that integral
is the same as that integral, because this was made up of radial and angular sections. This doesn't seem that way, but we can always make a very
fine mesh in which every displacement is approximated by radial angle or radial angular. So you can approximate any contour by sequence of radial
and angular moves. Angular moves are free. Radial moves keep track of the change in the distance times the
force and you can pair them with this reference. Every dr we move there we have a dr here. So that is the pictorial proof of this.
So let's now calculate what is the potential due to that electric field. So we're going to do the calculation.
You see, we know it is conservative. I've shown you with this argument. So if you want to find the potential energy between 1 and
2, you can take any path you like between 1 and 2. So let's pick the points. Here is point 1 and here is point 2.
You can pick any path you like, so here is what I'm going to do. I'm going to go radially out and then I'm going to go along
the arc. So this is 1, this is 2, but I'm going to start at some intermediate point 3, then go right.
So work is done here; no work is done there. So let's just find the work in going from 1 to 3, which is e_r (is a vector in the radial
A tiny section here, dr, has a magnitude equal to the change in the radius, and a direction equal to the radius unit rate
Prof: Don't wait. Don't wait even one second. I told you what happens to me, right?
When I go home, they tie me to a stool and they force me to watch this tape. And it's just awful when you think you dropped a symbol,
and you're waiting for somebody to say, "Hey, what about q?" So don't have anyif you hesitate, it's only to wait,
not to stop me. All right, so look, I'll get the q because I know the formula from the time I was in a sandbox.
I know something is wrong, but I don't want to get to that last stage, then go back and change everything. So I forgot the q sitting at the origin here.
1/r_2 − 1/r_1. Again, sorry, I made a mistake now. The change in potential energy is really q times
e, but q is the test charge, so I'm going to call this V. So it's hard for you guys to do it on the notebook,
because I want to distinguish between potential energy and potential. In other words, the forcenow some other test
charge q is the electric field times q. And the force is going to be  the gradient of the potential energy.
Student: Oh, okay. Prof: So the upper limit, then there's lower limit. So let me correct what I said.
I'm trying to find the difference in what's called the potential. It's not the potential energy.
Potential difference between some point and some other point is the work done on a unit charge. If you've got some other charge, you should multiply it
by the charge. So the energy of some other charge in this field, the potential energy, is q times the
So we're comparing this expression to this expression, we can conclude that V at any point r is q/4Πε _0(1/r)
any constant. Any constant, because you cannot immediately say this guy is equal to this guy and that guy is equal to
that guy, because even if you add a constant to both, it will still be true. And we can pick the constant any way we like.
It is common to pick the constant so that V goes to 0 as r goes to infinity. That's what we will do.
That's the convention. Then we can write here, what is the V(r). This is the V due to a single charge,
and the U(r) = to some q_0. I don't want to call q again, because q is the guy producing the potential.
q_0 is the guy experiencing the force. So q_0 times V is the potential energy of charge q, of charge q_0
sitting at that point. Now let's also verify, if you like, that this V(r) does produce the electric
I want to go back, just to get you familiar with this. Getting the electric field from the potential.
Can you do that in your head? Multiply this by i, and do similarly for y and z and add them up.
times r over r. Put r over r, position vector over r, is the unit vector in the radial direction.
I'm doing this a little fast, because this is something you can go home and check. I don't want to spend too much time doing it.
I claim that if you took this potential and took its x derivative or y derivative or z derivative, you get the x,
y or z components of the electric field. Okay, if you didn't follow this, you should go home and check that this is true.
All right, so this is not that important for me, this checking, because we know it's going to work.
What is important for me is for you to know that we've now found that the potential due to a single charge leads to a conservative force.
The field due to a single charge leads to a conservative one, therefore by superposition, if you add any number of charges,
if you find the potential due to all of them, then you take its gradient that will give you a conservative electric field.
So what have you got so far? Our gain is the following  we have a law of conservation of energy, which is the following 
½mv^(2) q (this q is the q of the body you're interested in) times V at the point r_1 = ½ mv_2 ^(2)
This equals q_i which is charge number I divided by the length of the vector from r_i to the r where I'm finding the
is sitting there and so on. And the potential at the point that I'm interested in is obtained by taking each q divided by the distance from
that point, and this is the usual 1/4Πε _0. That is the complete story, and the charges are not
discrete but continuous you can write an integral. So every problem, you first find the potential by adding the potential from all the charges.
Then its gradient will give it the electric field. But we have now a law of conservation of energy with this as the conserved energy.
The second advantage is that it is easier to work with V than with E. So why do you think it's easier to find V and then find
Prof: I'm sorry, I didn't hear the last part. Student: It's always easier to take derivatives than to integrate Prof: Very good.
you simply add the numbers coming from each one. There are no arrows, there are no vectors. Notice the formula for V has no vectors in it.
Each one contributes a number. Each charge contributes a number to the point they're interested in.
Add them all up, find the number here, there, everywhere else. Then take derivatives of your answer to get the field.
That's going to be a lot easier than adding the vectors. So I'm going to illustrate that with one simple problem and that's the last thing for today.
q at a. And we want to find the field there. So let's call that distance r_ and let's
call that distance r_. These are not vectors; these are just lengths. If you want to find the electric field directly,
you know what you have to do. You have to find the vector here and you've got to find the vector there due to the second guy, add the two vectors and
find the result. Of course, you can do it, but it's going to be very tedious, because you have to find the
direction of the vector here and the direction of the vector here, in terms of all these coordinates.
What we will do instead is to find the potential everywhere, and take the x and y derivatives of the potential to get the E_x and
Just wanted to simplify the expression a little bit. So this is true for any separation, but I want the limit in which r_ and
r_ are much bigger than the separation between them. So let's define an r which is the position vector of
the point you're interested in from the center of the dipole. So we've got to find r_ and r_.
r^( ) is the length of the vector r_ . That's equal to the square root of the length squared. The length squared of r_ squared = root of