I just finished compsing a new handout The negative sign in the
dV=-E*dx confuses many students. I am addressing this aspect.
I know that many are teaching electricity now, otherwise I would
wait a little before posting this message.
Comments and corrections will be appreciated. Feel free to use
my draft in any way you wish. I will post it on my web site later.
Ludwik Kowalski
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DRAFT #1 (Ludwik Kowalski, February 7, 2002)
The easiest way to introduce the concept of electrical potential, V, is to consider
a uniform electric field, such as that produced by a pair of parallel planes of
charge, for example, as illustrated below. We already know that E=sigma/esp_o
(where sigma is the magnitude of the charge surface density on each plane) and
that the electric field lines are directed from the positive plane to the negative
plane. The field lines are horizontal in this illustration because the planes of
charge are vertical; in general they can be oriented in any way we want (by
changing the orientation of planes of + and - charges).
Consider a positive probe charge, q. The force acting on it is q*E at every point;
this is like weight (m*g) near the earth surface, where the gravitational field, g, is
constant. The direction of this force, for a positive q, is the same as the direction
of field lines. Note the similarity between the vectors E and g; the unit of E is N/C
while the unit of g and N/kg. Can a probe charge be moved horizontally with a
constant velocity? Yes, but only if the net force acting on it is zero, as required by
Newtonís second law. In other words, an external force F, equal and opposite to
q*E, must be constantly acting on it to prevent the acceleration along the field lines.
How much work is done by this force (F=-q*E) when the probe charge is moved
horizontally from a point A to a point B? In this simple case the work is W=F*d =
-q*E*d, where d is the displacement from A to B. Is W positive or negative?
The sign of the probe charge is positive and for that reason the sign of W depends
only on the sign of the displacement d. The work is negative when d is positive and
it is positive when d negative. Note that W is not the work done by the electric force,
q*E, it is the work done by an external force F (an agent, if you wish). The external
force is equal and opposite to q*E. This is not different from what we do in mechanics;
recall that gravitational potential energy (m*g*d) is defined by the work done by a
force equal and opposite to weigh, m*g, for example, when a box is moved to a higher
or lower elevation. An agent moving a probe charge in the direction opposite to E is
increasing the electric potential energy; an agent moving it along E (preventing
acceleration) is decreasing the electric potential energy. Our textbook defines the
electric potential, V, as W/q. That is why the unit of V is J/C. Note that the unit of
energy was named after Joule while the unit of charge was named after Coulomb.
The derived unit J/C is most frequently called volt, after Volta, the inventor of the
first electric battery.
Referring to the above illustration, we can declare, that the potential energy is zero
everywhere on the left plane. This is not different, in principle, from declaring that
the gravitational potential energy is zero at the sea level, or at any other convenient
location. Then what happens to the potential energy when the probe charge is moved
toward the right plane? Without the external force the charge would accelerate to the right
and the work done by the electric field would be positive. But potential is not defined in
terms of the work done by the field, it is defined in terms of the work done by the agent
working against the field. For our choice of the reference, the work done by the agent is
negative and the potential becomes more and more negative as one moves away from
the positive plane toward the negative plane.
Since the choice of zero is arbitrary, let us make a new declaration; let us say that V is
zero on the negatively charged plane. How does the potential change when the probe
charge is moved from the negative plane on the right toward the positive plane on the
left? Please convince yourself that the potential becomes more and more positive. The
rule is simple to remember; the electric potential decreases when a positive charge is
moved along the field lines and it increases when that charge moves in the opposite
direction. That is why we say that electric field lines point downhill, electrically
speaking. Also convince yourself that the opposite would be true if the probe charge
were negative. Keep in mind, however, that a convention according to which a probe
charge is positive is well established. In the next chapter we will learn that the direction
of an electric current is always defined as a direction along which a hypothetical positive
charge would move along the field lines. Suppose a wire is used to connect a positive
terminal of an electric battery with its negative terminal. We will say that a current
outside a battery flows from its positive terminal toward its negative terminal. It flows
along the electric field lines established inside the wire by the battery.
Note that potential is a scalar quantity, like temperature. We already know that a
unique value of E (a vector quantity) can be assigned to any point in space when
locations of electric charges are specified. In this chapter we started learning how to
assign V do different locations. This handout shows how to do this in a uniform field.
Referring to the above illustration, and assuming that V=0 on the left plane, we can say
that V at a point situated d meters away from the left plane is V=-E*d = -(sigma/eps_o)*d.
For example, if the magnitude of sigma is 10^-8 C/m^2 then V, at d=0.1 meters, must
be ?113 volts. Likewise V=-226 volts at d=0.2 meters away from the reference. If
the negative side were declared as our zero reference (instead of the positive side) then
the potential, at any given point situated d meters away from the reference, would be
+(sigma/eps_o)*d. These formulas are not worth memorizing; they can easily be
reconstructed by calculating the amount of work needed to bring a probe charge from
the reference location to a given location, and dividing by q (because V=W/q).
Our textbook shows that the potential V(r), at a distance r from a point charge Q can
be calculated as k*Q/r. Thus V is positive when Q is positive and negative when Q
is negative. Be aware that the reference V=0 is at "infinity." In physics infinity is
usually interpreted as a "very large r"; so large that k*Q/r is practically zero. Note
that any r could have been chosen as a zero reference but assigning zero V to infinity
makes the V(r) formula less cumbersome. Also note that the potential V, due to a single
point chare, is inversely proportional to r while the magnitudes of E, due to a single
point charge, is inversely proportional to the square of r. The principle of superposition
is often used to calculate V due to several point charges, as illustrated in our textbook.