4.1 Potential from Office of Academic Technologies on Vimeo.

**4.1 Potential**

Now, we’re going to introduce a very important concept that’s associated with the electrical interactions, which is called, electric potential. Before we define the electric potential let us recall universal law of gravitation, and, as well as, Coulomb’s law. As you remember, the universal law of gravitation states that, any two masses in the universe attract one another with a force which is directly proportional to the product of the masses. And, it is reversely proportional to the square of the distance separating them.

Therefore, if you have two masses with N1 and N2, separated by a distance of R from one another, the universal law of gravitation says that these masses attract one another with a force F such that the magnitude of this force, let’s denote that by F sub G, is directly proportional to the product of the masses and inversely proportional to the square of the distance separating them. If we do not the proportionality constant as G, which is also known as universal gravitational constant, then we can express that relationship in the form of an equation such that F sub G is equal to gravitational constant G times the product of the masses divided by the square of the distance separating them.

By the same token, when we recall the Coulomb’s law, it’s simply stated that any two point charges, let’s say light charges Q1 and Q2, repel one another with a force such that the magnitude of this force is directly proportional to the product of the magnitude of the charges, and it is inversely to the square of the distance separating them. And, in this case, introducing the proportionality constant which is, which we know of it as Coulomb’s constant, 1 over 4 pi Epsilon zero. We express this as law in the form of an equation as the force is equal to Coulomb’s constant times the product of the magnitude of the charges divided by the square of the distance separating them. Of course, if the charges were unlike charges, then the direction of these forces, or the nature of the forces, are in an attractive way. They attract one another.

By looking at the mathematical form of these two laws, these two forces, we see that they’re very similar, only the constants are different, and we simply replace the masses with charges, in the case of electrical interactions. We know from gravitation that the gravitational force is a conservative force. It means that, that is work done by gravitational force is independent of the path traveled. Now, by looking at the mathematical similarity between these two laws, these two forces, we can conclude that the Coulomb force is also a conservative force. Coulomb’s force is also a conservative force. In other words, in this case also, work done by this force would be independent of the path traveled.

Well, what is the use of this future? Whenever we are dealing with conservative force fields, such that the work done by these forces are independent of path traveled, then we’re going to be able to find, or we can find that that explicit potential energy function associated with that field. If you recall, again, the gravitational interactions, we know that the gravitational field is pointing in downward direction. From the global point of view, it is pointing towards the center of the Earth. And, it has the value of G, gravitational acceleration. And, so, let us assume that here we have the ground level, and we take an object, with a mass M, and move it from this initial position to a final position somewhere over here. And, let’s say that the difference between these two positions is equal to H.

As we do this gravitational force exerted by the Earth on this object is always acting in downward direction and it has the value of MG, which is the weight of the object. Let’s look at this during this travel along this path at an arbitrary location. So, MG is in downward direction, and we’re moving the object in upward direction, therefore the incremental displacement vector can be defined at DY relative to Y axis. And, the positive Y is in upward direction.

First of all, the conservative future of gravitational forces is telling us that, as we move the object from this initial position to this final position, it doesn’t make any difference whether it moves along a straight line of path, like this, or it moves along a curve, like this, from initial to final point. The work done by the gravitational force will always be the same.

As you recall, work done is force vector dotted with displacement vector. So, in moving the charge, for an incremental distance, or incremental displacement of DY, gravity will do an incremental work of DW, and that will be equal to gravitational force dotted with displacement vector DY. Since that product is equal to magnitude of the first vector which is MG, in our case, times the magnitude of the second vector, which is DY, times the cosine of the angle between these two vectors. And, we can easily see that formation, that angle is 180 degrees, or Pi radians, we will have cosine of 180 over here. Cosine of 180 degrees is Y1, so the incremental work done, by the gravitational force, during this motion, or this placement, will be minus MGDY.

To be able to get the total work done, moving the mass from this initial point to the final point, we will add all these incremental works to one another along that path, from initial to final point, and we will get the total work done. In this case, that’s going to be equal to, since MG is constant, we can take it outside of the integral, an integral of DY is going to give us Y, which will be evaluated at initial and final point. And, furthermore, work done will be equal to, MG times final minus the initial point. And since the distance between the final point and initial point is equal to H, we will end up with this final expression of MGH.

Now, by recalling work-energy theorem, which is, in terms of the potential energy, work done is equal to negative of the change in potential energy where U represents the potential energy. Or, in terms of the kinetic energy, simply change in kinetic energy, is equal to work done. K is the kinetic energy.

We can express, then, the work done which is minus MGH, as the negative of the change in potential, in this case gravitational potential energy. We can cancel these negative signs on both sides of the equation and since potential energy is a relative concept, it is the form of energy associated with the position.

And, we can choose the initial position as the 0 potential energy level. And, final potential energy position as the final potential U. And, in this notation system, therefore, gravitational potential energy of this object becomes MG times H. Which is a familiar expression that you have seen earlier.

Therefore, dealing with a conservative force field eventually leads us to an explicit potential energy function, as in the case of gravitational potential energy. Well, if you go back to electrical interactions now, since we concluded that the Coulomb’s force is also a conservative force, so therefore, the work done by this force will also be independent of the path traveled.

In this case, if you consider an electric, an external electric field, pointing in downward direction. And, if you take a positive charge, and move this charge in opposite direction to the direction of the electric field, as soon as we place this charge inside of this electric field, it is going to feel Coulomb force exerted by this external electric field. And, that force, since the charge is positive, is going to be in the same direction with the direction of the external electric field.

Of course, if the charge were a negative charge, and this force would have been it the opposite direction, to the direction of the external electric field, because it is going to be, in that case, equal to minus QE. Now, again, if you displace this charge, some DY distance, along some DY displacement, starting initially starting from some initial point here, along this path to this final point, a work will be done by the electric field. The incremental work done by the electric field will be equal to Coulomb force dotted with this displacement. Let’s denote the displacement by DL in this case. It doesn’t have to be in the vertical direction. So, any displacement vector DL, along a path L.

And, again, as any case with a gravitational force, here as the charge moves from this initial point to final point. Work done by the Coulomb force will be independent of the path traveled, so it will not make any difference whether it goes along a straight line of path or along a curved path from this initial to final point, we’ll always have this same result.

And, this work done is, therefore, equal to FC dotted with the displacement vector DL, or it’s going to be equal to the magnitude of this force, which is QE, times the magnitude of the displacement vector DL time cosine of the angle between them, and in this case, again, the angle is Pi radians, or 180 degrees. Which is cosine 180 is equal to minus 1.

And, to be able to get the total work done, we add all these incremental works to one another along this path. And, we will do the addition by taking the integral of both sides. Therefore, the total work done will be equal to integral of QEDL and we will take the negative sign out, integrate it from initial to final point. And, this is going to give us minus Q’s integral from initial to final point of EDL.

Well, from work-energy theorem, again, from the work-energy theorem this work done has to be equal to negative of the change in potential energy. In this case we’re talking about electrical potential energy. If we equate these two quantities, U sub of F minus U sub I, will be equal to minus Q integral from initial point to final point of E times DL.

Or if we move, take it one more step, and if the electric field is constant, if E is constant, we can take it outside of the integral and therefore, we end up with minus QE, an integral of DL is going to give us the length of the path that the charge travels from initial to final point.

And, if we call that distance as D, then we can express that distance, the distance traveled as, D. Therefore, the change in potential energy of this charge becomes equal to QED. And, again, we end up with an expression for the electrical potential energy of this charge, or the electrical potential energy that it experiences as it travels from initial to final point.

Well, so, we can express the change in electrical potential energy, U sub F minus U sub I, equal to negative of the work done in moving the charge from initial to final point, from work-energy theorem, or U sub F minus U sub I is equal to minus work done is Coulomb force dotted with incremental displacement vector integrated from initial to final point.

In explicit form, if we express F sub C as QE, then we can rewrite this expression as minus QE dot DL integrated from initial to final point. Here charge is constant; we can take it outside of the integral. And, write down the expression as U sub F minus U sub I is equal to minus Q times integral of E dot DL.

Now, electrical potential energy, naturally, depends not only on the electric field, here, but also on the charge Q. On the other hand, if we look at the ratio of potential energy, per unit charge. In other words, if we divide both sides by Q, then we will eliminate the Q from the right-hand side, and that ratio will have a unique value at any point in the field, which will be independent of the magnitude of the test charge.

Let’s write down this point over here by saying that electric potential energy depends, not only on the electric field E, but also charge Q. On the other hand, potential energy, per unit charge, that is U over Q, would have a unique value at any point in the field which will be independent of the magnitude of the test charge. This quantity is called electric potential at the point in question. Therefore, electric potential, or simply potential, is electrical potential energy per unit charge. We’re going to denote electric potential by capital V, therefore, V is going to be equal to electrical potential energy U per unit charge. And, if we look at the units of this quantity in SI units system, it will be equal to joule per coulomb. And, we have a special name for this ratio, it is called Volt.

Here, now we can introduce commonly used, another energy unit, which is called electron volt. It is aggregated with EV, and one 1 EV is the energy equal to the work required to move a single elementary charge of E, which the magnitude of electron charge through a potential difference of 1 volt.

All right, now, if we go back to the change in potential energy expression, which is from work-energy theorem, is equal to the negative of the work done in moving the charge from some initial point to a final point. And as we just mentioned, whenever we deal with the potential energy, then we have to know the electric field as well as the charge.

But, if we divide both sides by charge Q, then energy per unit charge will have a unique value at every point in that medium. And, we define this energy, electrical potential energy per unit charge as the electric potential, therefore, this quantity will be equal to the final potential, and this will be the initial potential of the charge.

So, the difference will give us the potential difference and it will be equal to minus integral of E dot DL, which is a path integral from initial point to final point. Therefore, if we know the electric field in the region of interest, and if we move the charge in this field from some initial point to a final point, then the potential difference that the charge will experience can be obtained from this relationship.

All right, potential energy is a form of energy associated with the position. And, the position is a relative concept, in other words, we define the position of an object with respect to a [core in a] system relative to an origin. It depends on which point that we choose the origin.

As a result of that, potential energy itself is a relative concept. Whenever we are dealing with the difference, potential energy difference, the origin is irrelevant. Whatever origin that we choose, whenever we calculate the change in potential energy, irrelevant of the origin, we will always end up with the same result.

But what if we ask, the potential energy at a specific point in space. In the case of gravitation, it is easy, because in general we choose the 0 potential energy level as the surface of the Earth. And, we look at the height of the object with respect to that level and, then, calculate the potential gravitational potential energy of that specific mass.

But how are we going to define our origin, our zero potential energy level for the electric charges? In order to do that we will look at the Coulomb’s law and we see that the electric field associated with the charge becomes 0 if you go infinite distance away from that charge, in other words, from the source. And that is going to be, therefore, our starting point. In other words, we will say that the 0 potential, associated with the charge, will be located at infinity.

Let’s say a point test charge moves from one point to another in an electric field. The difference in the electric potential energy of that test charge, if we look at those points, is the negative of the work done by the electric field on that charge during its motion, which is given by this expression over here.

When we talk about the potential energy, then we say that the potential energy U, of a test charge, let’s say Q0, at any point, is equal to the negative of the work done on the test charge by the electric field as the charge moves in from infinity to the point in question. So, if we go back to our expression, such that the change in potential energy is equal to negative of the work done in moving the charge from initial to final point. And the change in potential, which is U sub F over Q minus U sub I over Q, the final potential energy per unit charge, and let’s say per unit test charge in this case, minus the initial potential energy per unit test charge, that will give us the potential difference, or change in potential, which is going to be equal to negative of the work down in moving the charge from initial to final point per unit test charge.

Well, if our initial point is located at infinity, then we can call V sub I as when the charge is at infinity, and assume that the potential is zero at that point, and the corresponding work done, in this case, is going to be moving the charge from infinity to the point of interest. And V final is equal to V. And, in terms of this new notations, then our expression will take the form of that V, the final potential V, V sub I, is when the charge is at infinity, which is 0, equal to negative of the work done in moving the charge from infinity to the point of interest per unit charge. So, V sub infinity represents work done by the electric field in moving the charge from infinity to the point of interest.