4.3 Calculating potential from electric field from Office of Academic Technologies on Vimeo.

**4.3 Calculating Potential from electric field**

Calculating the potential from the field. As we have seen earlier, if we have an external electric field inside of the region that we’re interested, something like this, and if we’re moving a charge from some initial point in this region along a path to a final point, at a specific point along this path, our test charge q0 naturally will be under the influence of Coulomb force generated by the field. At this point for example, the field is going to be tangent to the field line passing through that point and the force that it will exert on this charge, which is Coulomb force, is going to be equal to q0 times the electric field.

If we represent the displacement vector along this path with dl, incremental displacement vector, then the work done is going to be equal to integral from initial to final point of f dot dl. So work done, in moving the charge from initial to final point, will be equal to, replacing f with q0 e, we will have q0 times integral from initial to final point of e dot dl. Of course, we can take q0 outside of the integral since it is a constant. Dividing both sides by this charge work done, in moving the charge from initial to final point divided by q0, is going to be equal to integral of e dot dl integrated from initial to final point.

Well this quantity over here is going to give us the potential difference since work done per unit charge is by definition the electric potential. Therefore here we will have the change in potential or potential difference is going to be equal to minus integral from initial to final point of e dot dl.

Since work done is equal to negative of the change in potential energy and on the left-hand side therefore we have minus u sub f minus u sub i divided by q0 from work energy theorem. U sub f over q0 is v final and u sub i over q0 is v initial, then we simply just move the negative sign to the other side of the integral. Therefore using this expression, we can determine the potential difference that the charge will experience in this electric field by calculating the path integral of e dot dl from initial to final point.

Well if the initial potential is equal to the potential at infinity, which is equal to zero, and final potential is equal to v, then the potential will be equal to minus integral from infinity to the point of interest r in space of e dot dl or we can represent that dl in radial incremental vector dr.

Using these concepts, let’s do an example. Let us assume that we have an electric field pointing in downward direction in our region of interest and a charge displaces from some initial point to a final point such that the length of this distance is equal to d. At an arbitrary location along this path r positive q is going to be under the influence of Coulomb force generated by this electric field, which will be equal to q times e.

Now here the change in potential that it experiences will be equal to minus integral of initial to final point of e dot dl. Dl is an incremental vector along this path. That will be equal to minus e magnitude, dl magnitude times cosine of the angle between these two vectors. Here again dl and electric field are in the same direction so the angle between them will be zero degree. Therefore we will have cosine of zero in the integrant of this integral.

Cosine of zero is just 1 and v sub f minus v sub i is going to be equal to minus, since electric field is constant, we can take it outside of the integral, e times integral of dl from i to f and that is going to give us minus e times l evaluated at this initial and final point, which is going to be equal to minus e times final point minus the initial point and that distance is given as d. This will be equal to minus ed volts in SI unit system.

Now let’s take the same example that we have, let’s call this one as our first case. This is the second case. We have the same electric field pointing in downward direction and our charge is going to displace again from initial to final point, which are d distance away from one another. In this case, it is going to make the displacement such that first it will go to this intermediate point of let’s say c, and then from c to the final point f. It will follow a trajectory of this type instead of going directly from i to f.

Here if we look at the forces acting on the charge whenever it is traveling from i to c part, there, the electric field is in downward direction and the incremental displacement vector here, dl, is pointing to the right, and the angle between them is 90 degrees. From c to f, dl is going to be pointing in this direction and again the electric field is in downward direction when the charge is just right at this point.

The potential difference that it experiences through this path, again the potential at point f and the potential at point i, initial point, v sub f minus v sub i, is going to be equal to minus, first the charge displaces from initial point i to point c of e dot dl and then we have plus it goes from c to f, so we have again a negative sign over here. C to f of e dot dl.

If we move on, v sub f minus v sub i will be equal to the angle between displacement vector dl and electric field for the first path is 90 degrees, therefore we will have dl magnitude times cosine of 90 integrated from i to c. Then we have minus, from the second part, integral from c to f of e magnitude and dl magnitude. To make it easier, let’s say that this path is also equal to d. If that is the case, then this angle over here is going to be 45 degrees. Therefore this angle will also be 45 degrees.

We will have cosine of 45 degrees and the change in potential, or the potential difference, will be equal to, electric field is constant, we can take it outside of the integral, minus e times integral of dl and cosine of 45 is root 2 over 2, integrated from c to f. This is going to be equal to minus root 2 over 2 and integral of d l, along the path from c to f, is going to give us whatever the length of that path is. In other words if we add all these d l’s to one another, we will end up with the length of this path.

We can easily calculate the length of the path knowing the other two sides of this right triangle by applying Pythagorean theorem. That length is going to be equal to d squared plus d squared in square root which is equal to 2 d squared or root 2 d. So the integral is going to give us root 2 d, which is going to be equal to minus 2 times root 2 is 2, 2 over 2 is 1, so that’s going to be equal to minus ed.

Now in this simple example, we can see that when the charge moves initial to final point, either along a straight line or along this path, first to c and then to f, in both cases, we end up with the same potential difference. The same potential difference implies also the same potential energy difference.

This is also a good example that it is showing us that the work done, because negative of the change in potential energy is the work done, work done is independent of the path. In other words, as the charge moves from initial to final point, it doesn’t make any difference whether it goes along a straight line or through a different path. Work done by the electric field or by the Coulomb force turns out to be always the same.