This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 3. Lecture 3
3.1 Dipole and quadrupole
The law of addition of vectors allows us to ﬁnd the electric ﬁeld produced by two
charges. If the two charges are of equal magnitude but opposite sign (and separated by
a distance so that they do not cancel each other) then we have what is called a dipole.
The electric lines can be found in ﬁg.10. In the laboratory practice you will be able to
use a program that draws these lines for you. Try several conﬁgurations. For example
arrange four charges so that they look as two dipoles of opposite orientation and see
what happens. In ﬁg. 11 and 12 we illustrate typical outputs of the program. Besides
the electric lines it also shows equipotential surfaces everywhere perpendicular to the
electric ﬁeld. We discuss what they mean below. − + Figure 10: Electric ﬁeld of a dipole obtained by adding (vectorially) the electric ﬁelds of
each charge. A simple rule for the lines of electric ﬁeld is that they only start from positive
charges and end in negative ones. They can never end “in the middle of the air” so
to say. Before, we mentioned that electric ﬁeld can exist independently of charges, in
that case the lines of electric ﬁeld have to close on themselves since there is nowhere
for them to end. We will see how this works later.
3.2 Electrostatic energy
To motivate the idea of energy we use the Kelvin water dropper (ﬁg.13), a simple – 17 – Figure 11: Typical output from the computer program you use in the lab. In this case
illustrating the electric ﬁeld and equipotentials of a dipole. Figure 12: Illustrating the electric ﬁeld and equipotentials of a quadrupole (meaning four
poles). device that generates electricity. The way it works is illustrated in ﬁg.14. Assuming
that one metallic jar is positively charged and the other negative, opposite charges will
be induced in the streams of water falling through the respective rings. The water
breaks in droplets which then fall into the jars accumulating charge. This is because of
the cross connection between jars and rings which makes the positive droplets fall into
the positively charged jar and the negative ones in the negatively charged one. If the
water were not to break in droplets we would have a conducting circuit from the jars
to the water tank, the charge will be repelled and would not accumulate in the jar. It
is the same eﬀect as when we discussed induction that we needed to cut the connection – 18 – to ground and remove the rod. Ingeniously here the connection breaks simply because
the water stream breaks into droplets. When enough charge is accumulated the electric
ﬁeld is strong enough to push the charge through the bulbs lighting them up which
simultaneously neutralizes the system. Charge will build up again by repeating the
procedure. The initial charge imbalance appears just randomly, the important eﬀect is
that any random charge ﬂuctuation will be ampliﬁed by the device.
The idea we wanted to illustrate is also that we were able to generate electric power
and so the energy has to come from somewhere. It might not be apparent initially where
it is coming from. To ﬁgure that out one has to run the system and see when it will
stop and what I need to replace to make it work again. Obviously many electric devices
have batteries and we know they are the source of energy because when they run out
we need to replace them or the device won’t work anymore. Here it is clear that after
a while the tank runs out of water. So we need to take the water from the jars and
put it back up into the tank. To do that we need to lift water against gravity and
that is where we put energy back into the system. Therefore, gravitational energy is
being converted into electric energy. One way of seeing that is that, since the jars repel
the water droplets, gravity is necessary to push them down into their respective jars
allowing the accumulation of charge.
Having said that let us now look at how we understand energy when there are
electric ﬁelds present.
From Newtonian mechanics we know that forces can be derived from potentials.
Basically a particle feels a force that tries to make it move in the direction where it
can decrease its potential energy faster. The same occurs in the case of electrostatics.
Consider a positive charge Q which is ﬁxed at some point and take another charge q
that we are going to move. Let us say that the other charge is negative so there is an
attractive force. If we want to move the charge q further apart by an amount ∆r we
need to do some work because there is a force that opposes us. The amount of work
needed is
W = F ∆r = ∆U = Uﬁnal − Uinitial (3.1) where ∆U is the change in potential energy of the system. A couple of comments. When
computing the work we should use only the component of the force in the direction of
the motion. Since here we move parallel to the force the formula is OK. If not we need
to multiply by cos θ where θ is the angle between the force and the direction of motion.
Second comment is that the formula is valid for constant force. Since the force depends
on r we are going to consider ∆r r and therefore to consider the force constant is a – 19 – Figure 13: Kelvin invented this device to generate electricity. I ﬁnd it of great interest
since it is basically just water dropping from a tank and nevertheless it is able to generate
electricity enough to light a few neon bulbs. Essentially illustrates how, by understanding
how nature works one can create something interesting out of almost nothing. good approximation. With this in mind we compute
1 Qq
∆U = Uﬁnal − Uinitial = −
δr
4πε0 r2 (3.2) The minus sign is because we should have ∆U > 0 as discussed but the product of
the charges is negative since they have opposite sign. We claim now that the potential
energy is
1 Qq
U=
(3.3)
4πε0 r
Indeed the diﬀerence in potential energy between the initial and ﬁnal situation is
1
Qq
1 Qq
1
Qq
∆U = Uﬁnal − Uinitial =
−
=−
∆r
4πε0 r + ∆r 4πε0 r
4πε0 r(r + ∆r) (3.4) If we use now that ∆r r we ﬁnd that
∆U = − 1 Qq
δr
4πε0 r2 (3.5) as we computed from the force. This veriﬁes our expression (3.3) for the potential
energy. – 20 – water tank +
+ −−
−−
−−
−− +
+ −
− ++
++
++
+ − −
−
−
− rings +
−
−
−
−
−
−
−
− +
− + −
−
−
−
−
−
−−− −− − − +
+
+
+
+ − +
+
+
+
+
+
+
+ +++++ metallic jars neon lamps
Figure 14: The key is the cross connection of the rings and the jars. This ampliﬁes any
charge diﬀerence accumulating opposite charges in the two metallic jars. Another case in which we can compute the potential energy is that of a constant
uniform electric ﬁeld. Namely E is independent of the position, that is a probe charge
feels the same force no matter where it is located. Of course this is an idealized situation
but in many cases is a good approximation for the electric ﬁeld in certain regions where
it does not change very much. In such case, let us say that the electric ﬁeld points in
the direction x. If we move a charge in directions, y or z , perpendicular to x, the force
ˆ
ˆ
ˆ
ˆ
does not oppose or help the motion so we do not need to do any work. If we move it – 21 – in direction x however, since we move in the direction of the force we extract work, or,
ˆ
we make negative work:
W = −F ∆x = −q E ∆x = δ U = Uﬁnal − Uinitial = U (x + ∆x) − U (x) (3.6) The ﬁnal energy of the system is therefore smaller than the initial one. We see that
the equation is satisﬁed if the potential energy is simply given by
U = −q E x (3.7) Another way to ﬁgure out the sign is to notice that the charge will move in the direction
of decreasing energy. So, if q > 0 then U has to decrease toward the right and therefore
the minus sign.
3.3 Electrostatic potential
We see, as a consequence of the force being proportional to the charge, that so is the
potential energy. For that reason we deﬁne the electrostatic potential V (x, y, z ) such
that is we put a charge at position (x, y, z ) the energy of the system changes by
U = qV (x, y, z ) (3.8) For a charge Q the electrostatic potential is given by
V= 1Q
,
4πε0 r where, as always, r=
x2 + y 2 + z 2 (3.9) On the other hand, for a constant electric ﬁeld in direction x it is
ˆ
V = −E x (3.10) By analogy with the equation that determines the work done in terms of the potential
energy we see that if we move by a distance ∆x in a direction parallel to the electric
ﬁeld the change in electrostatic potential is
∆ V = −E ∆ x (3.11)
Equivalently the component of E in direction x is
Ex = − ∆V
∆x (3.12) So, the electric ﬁeld can be computed from the potential by moving in a direction x
by an amount ∆x. The change in potential ∆V determines the electric ﬁeld through – 22 – eq.(3.12). This actually works if we move in any direction, it always give the component
of E in that particular direction (or equivalently the projection of E along the direction
of motion). For example, if we move in a direction perpendicular to E then there is no
change in the potential. This gives rise to the notion of equipotential surfaces. This
means surfaces of the same (equal) potential, that is surfaces where V has a constant
value. From what we just said, such surfaces should be perpendicular to the electric
ﬁeld. For example, for a single charge they are spheres concentric with the charge. For
a constant electric ﬁeld they are planes perpendicular to E . In other cases, for example
for the dipole, they are more complicated but one can have some idea by drawing the
electric ﬁeld and then drawing surfaces perpendicular to it (see ﬁg. 11).
An important property of equipotential surfaces is that if we move a charge along
such surface we do not do any work.
Another important property is that to be in a static situation, namely with charges
not moving, the surface of a conductor has to be an equipotential surface. This is
because on the surface of the conductor the electric ﬁeld is perpendicular to it, otherwise
if there were a component parallel to the surface, charges would move until they cancel
the electric ﬁeld. Similarly, if charges are not moving then inside a conductor the electric
ﬁeld is zero, implying that all the conductor has the same value of the potential. We
should emphasize that this refers to the static situation. If charges are moving then
the electric ﬁeld inside a conductor need not be zero and the conductor need not be all
at the same potential.
The principle of superposition also applies to the electrostatic potential. So if we
have several charges, the total electrostatic potential is the sum of the electrostatic
potentials due to each of them. It becomes apparent the usefulness of the potential,
since the potential add as numbers as opposed to the electric ﬁeld which add as vectors.
Finally the potential is measured in N m/C since the electric ﬁeld is measured in
N/C and ∆x in meters. Since this is a very common unit it has received the special
name of Volt (V). So we have
Nm
1V = 1
(3.13)
C
The potential diﬀerence many times is referred simply as “voltage”
Another demo illustrates the existence of this potential by measuring directly the
potential of a charged sphere and another by showing how putting a ﬂuorescent light
in the electric ﬁeld produces a discharge because of the potential diﬀerence between
the two ends of the tube (ﬁg.15). – 23 – Figure 15: A voltmeter and a ﬂuorescent bulb make manifest the existence of a potential
diﬀerence (and electric ﬁeld). – 24 – ...
View Full
Document
 Fall '11
 NA
 Charge

Click to edit the document details