2.3 Electric Field of an Electric Dipole from Office of Academic Technologies on Vimeo.

**2.3 Electric Field of an Electric Dipole**

Now, let us try to determine the electric field of a system which is called electric dipole. An electric dipole is mainly two point charges with equal magnitudes and opposite signs separated by a small distance from each other.

As an example, let’s try to determine the electric field of a dipole along its axis. Let’s say, z distance from its center. What we have here is two point charges. Let’s say a positive q and a negative q. They are equal in magnitudes, separated from one another by a small distance of d, so d represents the separation distance.

We’re interested with a point along the axis of the dipole. Let’s say somewhere over here, point p, and choosing the center of the dipole as our origin, the location of this point relative to that center is given as z. Therefore, electrical field at point p is the question mark. Since the total distance between the charges is d and this point is the center of the dipole, therefore this distance will be equal to d over 2.

In terms of these distances, the distance between the point of interest and the positive charge is going to be z minus d over 2 and the distance between the point of interest and the negative charge, which is this total distance, will be, considering now this half, z plus d over 2. The positive charge will generate its own electric field at the point of interest in a radially outward direction.

In other words, it will be in the same direction with the Coulomb force which is acting on a positive test charge placed at point p. Therefore it is going to be pointing radially outward direction and let’s call this electrical field as e plus.

In a similar way, the negative charge will generate its own electric field at this location and that’s going to be in a radially inward direction and let’s call that electric field as e minus. The total electric field will be the vector sum of these two fields. The total will therefore be equal to e plus plus e minus.

Now if we choose our positive direction as outward direction, e positive, or e plus is in the positive direction, whereas e minus is in the negative direction, therefore the vector addition of these two fields will be basically the difference of them. Therefore e magnitude will be e plus minus e minus. Of course one can express these fields in terms of the units vectors.

If we choose this outward direction as z axis and call the unit vector in that direction as k, then e plus vector will be e plus magnitude times k and e minus will be e minus magnitude times minus k. Of course when we add them, it will simply be equal to the difference between these two vector quantities.

Now, let’s go ahead and express the magnitude of these electric field vectors using Coulomb’s law. For e plus, we will have Coulomb constant 1 over 4 Pi Epsilon zero times the magnitude of the charge, q divided by the square of the distance between the point of interest and the charge, which is this distance and therefore the square of that will be z minus d over 2 squared.

Similarly e minus will be equal to, again this is the magnitude of the electric field generated by the negative charge q, 1 over 4 Pi Epsilon zero. The magnitude of that charge is again q divided by the square of the distance between that charge and the point of interest is z plus d over 2. Therefore we will have z plus d over 2 squared in the denominator.

Here you have to be careful that the Coulomb’s Law is basically giving us the magnitude of the electric field. Therefore we always use the magnitude of the charges. We have already taken into account their signs while we are establishing the vector diagram and for the positive charge, the electric field goes radially outward and for the negative charge, it goes radially inward direction.

Once we determine the magnitudes therefore, applying a total expression, we will have 1 over 4 Pi Epsilon zero q over z minus d over 2 squared minus, for the e minus, 1 over 4 Pi Epsilon zero q over z plus d over 2 squared.

We can simplify or rewrite this equation by taking the common quantities in the parentheses, which is mainly q and 4 Pi Epsilon zero. So in q over 4 Pi Epsilon zero parentheses, we will have 1 over z minus d over 2 squared, from the first term, and minus 1 over z plus d over 2 quantity squared from the second term.

Therefore this expression is going to give us the magnitude of the electric field generated by this dipole at the point of interest, this point p. Since e plus is greater than e minus, the net direction of the magnetic field will be in the same direction with the e plus and that is in outward direction.

The reason that we can conclude that e plus is larger than e minus, because the electric field is inversely proportional to the square of the distance between the source charge and the point of interest. We can easily see from the diagram that the positive charge is nearer to the point of interest in comparing to the negative charge. Therefore, the strength of the electric field then will generate an upward direction, will be larger than the strength of the electric field generated by the negative charge, which is in the downward direction.

Now, let’s look at an interesting special case for our problem. That is, let’s consider the points of interest such that the distance z, their location, relative to the dipole separation, d, is such that z is much greater than d. Under this condition, then we can easily say that d over z will be much smaller than 1.

So we will try to take advantage of this ratio in our expression and obtain an approximate value for such a specific case. Of course, in the crude approximation, we will compare d over z with 1 and whenever we have such a comparison, in the first crude approximation we will try to neglect this ratio in comparing to one.

Since we are able to have such a ratio, let’s try to rearrange our final expression which was e is equal to q over 4 Pi Epsilon zero. To be able to have d over z ratio, I will take the z in these power brackets outside of the power bracket. If I do that, since z is in a square bracket, it’s going to come out as z squared and inside of the bracket, we will have 1 minus d over 2z squared.

For the second term, we can do a similar type of analysis. Take z out and it’s going to come out as z squared and inside we will have 1 plus d over 2z quantity squared. Now again we can rewrite this expression since z squared in the denominator is common.

We can write this expression by taking the first and second term in z squared common parentheses, therefore we are left 1 over 1 minus d over 2z squared for the first term and minus 1 over 1 plus d over 2z squared from the second term.

Since our condition says that d over z is much smaller than 1, and as we can see we have d over 2z in our power brackets next to 1, d over 2z will be even smaller than 1. Then in the first crude approximation we can neglect d over 2z in comparing to 1. In terms of numerical values, this is something like 1 minus 0 point 00000 something and this is 1 plus 0 point 0000 something, so since that ratio is much smaller, in the first crude approximation we can neglect in comparing to 1. If we do that, then we will have 1 left in the power bracket, so the square of 1 will be 1 and the first term will give us just 1.

Similarly, the second term, again we will just have 1 in the bracket, square of 1 is 1, 1 over 1 will give us 1, and what we are going to end up is, that the approximation will have just 1 for the first term and minus 1 for the second term and when we add them, they cancel and we end up with zero.

Although this is a correct result, this is basically the answer when the point of interest is infinite distance away from the dipole. We already know that if we go infinite distance away from the dipole, the electric field will go to zero. In other words, it will not give us something new.

What we are interested in is that these are our charges, our dipole charges with plus q and minus q and there is d and we’re at a point, p, such that this distance z, from the center of the dipole, is much greater than d. We’d like to obtain an electrical field expression which is non-zero, because we know if we go infinite distance away or approach infinity, then the associated electric field will drop to zero. Therefore we need some other tool to be able to obtain such an approximated value.

To do that, we will take advantage of the mathematical series expansion called binomial expansion. According to this expansion, if you consider a polynomial of the type 1 plus x to the power n, this polynomial can be expanded in terms of binomial expansion under the condition such that x is much smaller than one as 1 plus an x over 1 factorial plus n times n minus 1 x square over 2 factorial and plus third and higher order terms.

Looking at our quantities, or out brackets over here, we have a similar type of situation. We have a term, which is much smaller than 1, which will correspond to the x term in this polynomial expression and the power of the bracket is in the second power so n will be equal to 2 and since d over 2z, which corresponds to x, is much smaller than 1, then we are able to expand these brackets in binomial expansion.

The procedure is such that first we will expand up to the first term, in other words, 1 plus nx over 1 factorial term, and then we will neglect second and higher order terms. Once we do that, if we end up with a non-zero result, then our approximation is done. That expression will give us the final result.

If we still end up with a value of zero, like in the previous case, once we neglected d over 2z in comparing to one, we ended up with zero. So in the first order approximation, neglecting the second and higher order terms, if we still end up with zero, then we’ll go back and include the second order term in our expression. If we end up with non-zero result, everything is done. If we end up with zero, then we go and include the third order term and so on and so forth. That’s the procedure.

To be able to apply the binomial expansion to our case, let’s try to put our parentheses exactly in a form such that we can directly apply this expansion. To do that, I will rewrite the electric field expression. Q over 4 Pi Epsilon zero z square. By moving the terms in the denominator to the numerators, then the first term will be 1 minus d over 2z to the power minus 2 minus, for the second one, we will have 1 plus d over 2z to the power minus 2 and close parentheses. We will look at each one of these terms and apply binomial expansion.

For the first term, we have 1 minus d over 2z to the power minus 2. Again, this term will correspond to x and the power of the bracket will correspond to n term in the binomial expansion formula. This will be approximately equal to, since we will neglect the second and higher order terms, 1 plus n is minus 2 and x d over 2z with its sign, which is minus d over 2z, and divided by 1 factorial, which is equal to 1. We will neglect second and higher order terms.

Therefore, 1 minus d over 2z to the power minus 2 approximately becomes equal to 1 minus minus, we’ll make plus, and this 2 and the 2 in the denominator will cancel and we are going to end up 1 plus d over z from the first term.

Similarly, for 1 plus d over 2x to the power minus two, again, d over 2z is the x term and minus 2 is the n term in the binomial expansion formula. We will have 1 plus nx, again n is minus 2, x is d over 2z in this case with a positive sign. Again we have 1 factorial in the denominator and again we will neglect second and higher order terms. Therefore 1 plus d over 2z to the power minus 2 will be approximately equal to, the 2’s will cancel, 1 minus d over z, since 1 factorial is just 1.

Now we will go back to our original expression, which electrical field was equal to q over 4 Pi Epsilon zero z squared times the first term here 1 minus d over 2z to the power minus 2. Will be approximated 1 plus d over z by applying binomial expansion, that is the first term. Minus the second term, which is 1 minus d over 2z to the power minus 2. Is going to be approximated 1 minus d over z using the same expansion.

If we continue, we will have q over 4 Pi Epsilon zero z squared, 1 plus d over z from the first term, minus 1 plus d over z from the second term. Plus 1 and minus 1 will cancel, therefore the electric field is going to be equal to d over z plus d over z, will give us 2d over z. q over 4 Pi Epsilon zero squared times 2d over z. We can cancel this 2 with our 4 in the denominator and z times z squared will give us z cubed.

Therefore, our electric field vector is going to be equal to q d divided by 2 Pi Epsilon zero z cubed. For the electric field of a dipole along its axis, z distance from its center such that z is much greater than d.

Again, if we’d like to write this down in vector form, we can multiply this by the unit vector pointing in upward direction and let’s call that as k in three dimensional rectangular coordinates. If this is x, this y, and this is z, then we call the associated unit vectors as i, j, and k.

If we look at this expression, which we have the product of the magnitude of the charge of the dipole q, and also the separation distance of the dipole. Of course, the quantity z in the denominator is the distance of the point of interest from the center of the dipole. Both q and d are unique properties of a given dipole. In other words, every dipole will have a unique charge magnitude and also will have a unique separation distance.

So the product of these two quantities will also be a unique value for a given dipole, which we will have a special name for that product and we’ll denote that by p and it is called magnitude of electric dipole moment vector. Therefore this quantity is actually a vector quantity and the direction of electric dipole moment vector is such that it points from negative charge, minus q, to positive charge, plus q.

Therefore, if we redraw the picture, and if this is our dipole, with plus q here and minus q here, and the separation distance d between them, electric dipole moment vector, p, is such that it points from negative charge to positive charge and the magnitude of p is equal to magnitude of q times d. That’s the definition of magnetic dipole moment vector.

Now, why did we consider the condition of this point z much greater than d in the beginning anyway? When we look at the molecular structure of some materials, we see that as the atoms come together, to make a specific molecule, they show some electric dipole characteristics.

For such molecules, therefore, our observation distance, which is z, and that is the distance between the point that we look at that molecule relative to the center of that electric dipole will be much greater than the separation distance of those charges; in other words, the size of the molecule.

**Therefore, for those type of cases, it becomes a very reasonable approximation to be able to estimate the electric field of such a molecule at the location of our observation point.**