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**Example 1- Potential of a point charge**

Now let’s calculate the potential of a point charge. Let’s assume that we have a positive point charge, *q*, sitting over here, and now we know that it generates electric field in radially outward direction, filling the whole space surrounding the charge and going from charge to the infinity in radially out direction. The potential, by choosing the 0 potential at infinity, was defined as minus integral of **E** dot *d***r**, integrated from infinity to the point of interest in space.

Let’s assume that the point that we’re interested is over here and it is *r* distance away from the source. *d***r** is the incremental displacement vector in radial direction and recall that electric field is *q* over 4*π* *ε**0* *r*2 for a point charge. It is of course radially outward direction for a positive charge. Then the potential of this charge becomes equal to minus magnitude of the first vector, and that is *q* over 4*π* *ε**0* *r*2, magnitude of the second vector *dr*, and again, *dr* is an incremental displacement vector in radial direction. An electric field is also in radial direction at this point. Therefore the angle between these two vectors is 0 degrees, so we have here then cosine of 0 as a result of this dot product.

This quantity will be integrated from infinity to the point of interest, which is located *r* distance away from the charge. Cosine of 0 is 1 and *q* over 4*π* *ε**0* constant can be taken outside of the integral and potential *V*, therefore becomes equal to –*q* over 4*π* *ε**0* times integral of *dr* over *r*2 integrated from infinity to *r*. Integral of *dr* over *r*2 is -1 over *r*, so *V* is equal to minus *q* over 4*π* *ε**0* times -1 over *r* evaluated at infinity and *r*.

This minus and that minus will make a positive and if you substitute *r* for the little *r* in the denominator we will have *q* over 4*π* *ε**0 **r.* If we substitute infinity for little *r*, then the quantity will go to 0 because any number divided by infinity goes to 0. Therefore our result is going to be that the potential of a point charge is equal to charge divided by 4*π* *ε**0* times *r*.

Here *r* is the distance between the point charge and the point of interest. Here we should also make an important note, as you recall that the potential was electrical potential energy *U* per unit charge. Here, energy is a scalar quantity, charge is also a scalar quantity, and whenever we divide any scalar by a scalar, we end up also with a scalar quantity. Therefore potential does not have any directional properties. It is not a vector, and that makes also dealing with potential much easier than dealing with the electric field, because we don’t have to worry about any directional properties for this case.

If we have more than one point charge in our region of interest, then since we are dealing with scalar quantities, we can calculate the potential of the specific point simply by calculating the potentials generated by each individual point charges at the location of interest and then simply adding them. If we make a note of that over here, for more than one point charge, for example if I have *q**1* and *q**2* and *q**3* and so on and so forth, and if I am interested with the potential at this point, I look at the distances of these charges to the point of interest and calculate their potentials. *V**1* will be equal to *q**1* over 4*π* *ε**0* *r**1* over and let’s give some sign to these charges also.

Let’s say that this is positive, this is negative, this is positive. *V**1* will be *q**1* over 4*π* *ε**0* *r**1*. *V**2* is going to be equal to, again this is positive charge *q**2* over 4*π* *ε**0* *r**2*, and *V**3* is going to be equal to, since it is negative, –*q**3* over 4*π* *ε**0* *r**3*. Once we determine their potentials relative to this point, then the total potential will be equal to *V**1* plus *V**2* plus *V**3*, or is going to simply be equal to 1 over 4*π* *ε**0* will be common, *q**1* over *r**1* plus *q**2* over *r**2* minus *q**3* over *r**3*. Therefore the total potential that this system of charges generates at this point *P* is going to be equal to this quantity.