Chapter 3
The Electric Potential

3.1  Definition of Electric Potential

3.1.1  Motivation

We are still operating under the assumption that the electric field provides a convenient "mechanism" for dealing with electrical phenomena. Even though electrical forces are what we observe, the electric field is assumed to the be the "agent" by which the force is manifested on charged particles. However, just as we found with Newton's Laws, it becomes increasingly clear, with some experience, that we would prefer, for practical reasons, to adopt a way of describing the electric field in terms of scalar quantities rather than through the vector nature of the E field itself. With Newton's Laws of Motion, it turned out to be very useful to define the concepts of work and kinetic energy. These definitions, namely

W = ó õ ® F   ·ds K = 1 2 mv2 (3.1.1.1)

relate work to an integral over a path (which is broken into infinitesimal increments of displacement, ds) and the kinetic energy to the mass and velocity of a point particle. We note that these two are related according to the work-energy theorem, W = K_f - K₀. Furthermore, for work done by conservative forces, we can define a potential energy,

U(x) = ó õ F(x) dx

(3.1.1.2)

The generalized work-energy theorem relates changes in potential energy to work and hence to changes in kinetic energy for the case of conservative forces

W = -DU = DK ÞK0 + U0 = Kf + Uf

(3.1.1.3)

The potential energy and kinetic energy allowed us to work with scalar quantities dealing with motion and the change in motions caused by forces. We can develop an analogous scalar quantity to the electric field. It is called the electric potential and is usually represented as V (due, as we will see, to the common name for the unit of electric potential, the volt). The relationships we seek then look, schematically, as follows

Figure 3.1: Relationships between scalar and vector quantities related to static electric fields.

There's nothing especially tricky here. V becomes the scalar quantity of choice to use for problems involving the electric field. The idea is that the scalar nature of V will be generally easier, and therefore more practical, for uses than E just as potential energy is more useful for many problems than force, F. However, for V to be useful, we need a definition for it which allows us to consistently realize the pictorial relationship between force, potential energy, electric field, and electric potential that is drawn above. The definition which works is

U(x) = - ó õ F(x) dx = - ó õ q0Ex dx DV º - W q0 = DU q0 Þ V = U q0 (3.1.1.4)

That is, we state that the work per unit charge is the change in potential. This is entirely consistent with thinking about V in the same way we think about height for the gravitational field. The higher the value of V with respect to some fixed reference, the more potential energy (for a positive charge) there is in the electric field, just as increasing the height of a mass raises the potential energy in the gravitational field. One consequence of this definition is the relationship between electric potential and electric field. From the generalized work-energy theorem and the definition of electric potential, we have

DV = - W q0 = DU q0 = - 1 q0 ó õ ® F   ·d ® s   = - 1 q0 ó õ q ® E   ·d ® s   DV = - ó õ ® E   ·d ® s   (3.1.1.5)

The s refers to a path along a electric field line. V becomes the scalar quantity of choice to use for problems involving the electric field. The idea is that the scalar nature of V will be generally easier, and therefore more practical, for uses than E just as potential energy is more useful for many problems than force, F.

3.1.2  Calculating the Electric Potential

We will find many uses for the electric potential in the near future. For now, we note that it is indeed easier to work with than the electric field because of its scalar nature.

The first way to calculate the potential is through the use of integral as defined above. We can apply this to a case for which we know the electric field, namely a finite length charged, rod as shown below. We wish to find the electric potential at a point P which is a distance h above the midpoint of the rod of length L.

Figure 3.2: Finding the electric potential over the midpoint of a finite length, charged rod.

You may remember that we got an E field whose direction is vertically up away from the rod and with a magnitude

E = 2kQ h   _______ ÖL2 + 4h2

(3.1.2.6)

If we wish to find the potential at point P relative to infinity, then we should integrate the dot product of E with ds over a path from infinity in to point P. The easiest path to integrate is a straight-line path as the E field is always vertically upward and opposite to a straight line path from infinity down to point P. So, noting that y is the integration variable and goes from the line charge out to infinity (and therefore is opposite the ds direction), we find

VP = - ó õ ® E   ·d ® s   = - ó õ h ¥  -E ds = - ó õ h ¥  2kQ y   _______ ÖL2 + 4y2    dy = - ó õ h ¥  kQ y   _________ Ö(L/2)2 + y2     dy = -kQ -1 L/2 ln é ê ê ê ë L/2 +   _________ Ö(L/2)2 +y2   y ù ú ú ú û h ¥  VP = 2kQ L ln é ê ê ê ë L/2 +   _________ Ö(L/2)2 +h2   h ù ú ú ú û (3.1.2.7)

We can also consider doing this calculation another way. If we make use of the techniques of calculus, we can consider dividing the rod into infinitesimal lengths of charge, dq, then considering the potential at point P for each. Adding up the potential for all the dq's by integration should yield the same result as above. Since the infinitesimal lengths can be considered as point charges, we need to know the potential at point P, relative to infinity, for a point charge dq. First, consider any point charge, Q, in space. Then the electric potential at a point a distance r away can be calculated using y as the integration variable as follows. The electric field due to a point charge is radial, so we can integrate over a straight line path along a radius from infinity to r (note that E and s are in the same direction for this calculation).

V(r) = - ó õ ® E   ·d ® s   = - ó õ r ¥  E dy = - ó õ r ¥  kq y2  dy = é ê ë kq y ù ú û r ¥  V(r) = kq r (3.1.2.8)

Now we can apply this to the present problem. For infinitesimal lengths, dx, with charge, dq, we can do the integration by noting that the distance from dq to point P is

r =   ______ Öx2 + h2

(3.1.2.9)

with x being the distance of dq from the midpoint of the line segment charge. Even though we are not dealing with vector components, we can still take advantage of symmetry in that we see that the right and left halfs of the line segment (i.e. to the right or left of the midpoint of the line segment) are completely symmetrical: they give equal contributions to the potential.

V = ó õ dV = ó õ L/2 0  2k dQ r = 2k ó õ L/2 0  (Q/L)   ______ Öh2 + x2  dx = 2kQ L ln é ë x +   ______ Öx2 + h2   ù û L/2 0  VP = 2kQ L ln é ê ê ê ë L/2 +   __________ Ö(L/2)2 + h2   h ù ú ú ú û (3.1.2.10)

As expected, this gives the same result. This example brings up something of a lesson here. If you wish to find the potential of a charge distribution of finite size, you can always break the distribution into infinitesimal pieces, treat each as a point charge for getting the potential at the desired point, then integrate over the whole distribution to get the net potential. If you know the electric field distribution of a charge distribution or can easily calculate it, you can find the potential relative to some reference point by integration of the electric field over some path. The fact that you can use any path means you always want to choose the path that minimizes the difficulty of the integral.

Send comments to larryg@upenn5.hep.upenn.edu.
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