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9.5 Determination of Electrostatic Forces from Energy
We saw that we can find electric forces between charged bodies only if we know
the charge distribution on them, which is rarely the case. Moreover, the previously
discussed method cannot be used to determine forces on polarized bodies except in
a few simple cases.
For example, suppose that a parallel-plate capacitor is partially dipped in a liq-
uid dielectric, as in Fig. 9.3a. If the capacitor is charged, polarization charges exist
only on the two vertical sides of the dielectric inside the capacitor. The electric force
acting on them has only a horizontal component, if any. Yet experiment tells us that
when we charge the capacitor, there is a small but noticeable rise in the dielectric level
between the plates. How can we explain this phenomenon?
The answer lies in what happens not at the top of the dielectric but near the
bottom edge of the capacitor. In that region, the dipoles in the dielectric orient them-
selves as shown in Fig. 9.3b. The net force on the dipoles points essentially upward
and pushes the dielectric up between the plates. Although we can explain the nature
of this force, based on what we have learned so far we have no idea how to calculate
it. The method described next enables us to determine the electric forces in this and
many other cases where the direct method fails. In addition, conceptually the same
method is used for the more important determination of magnetic forces in practical
Figure 9.3 (a) When a parallel-plate capacitor dipped in a liquid
dielectric is charged, the level between the plates rises due to
electric forces acting on dipoles in the dielectric in the region
around the edge of the capacitor, where the field is not uniform.
(b) Enlarged domain of the capacitor fringing field in the
dielectric, indicating the force on a dipole in a nonuniform field.
ENERGY, FORCES, AND PRESSURE IN THE ELECTROSTATIC FIELD 129
Figure 9.4 A body in an electrostatic system
moved a small distance dx by the electric
Consider an arbitrary electrostatic system consisting of a number of charged
conducting and polarized dielectric bodies. We know that there are forces acting
on all these bodies. Let us concentrate on one of the bodies, for example the one
in Fig. 9.4, that may be either a conductor or a dielectric. Let the unknown electric
force on the body be F, as indicated in the figure.
Suppose we let the electric force move the body by a small distance dx in the
direction of the x axis indicated in the figure. The electric force would in this case do
work equal to
where F, is the projection of the force F on the x axis.
At first glance we seem to have gained nothing by this discussion: we do not
know the force F, so we do not know the work dAel,fo,c, either. However, we will now
show that if we know how the electric energy of the system depends on the coordi-
nate x, we can determine the work dAel,f,,,,, and then from Eq. (9.10), the component
F, of the force F. In this process, either (1) the charges on all the bodies of the system
can remain unchanged or (2) the potentials of all the conducting bodies can remain
Let us consider case (1) first. The charges can remain unchanged in spite of the
change in the system geometry only if none of the conducting bodies is connected to a
source that could change its charge (for example, a battery). Therefore, by conservation
of energy, the work in moving the body can be done only at the expense of the electric
energy contained in the system.
Let the system energy as a function of the coordinate x of the body, We(x), be
known. The increment in energy after the displacement, dW,(x), is negative because
some of the energy has been used for doing the work. Since work has to be a positive
number, we have in this case dA,l,fo,,, = -dW,(x). Combining this expression with
Eq. (9.10), the component F, of the electric force on the body is
F - -.-------
(charges kept constant).
Figure 9.5 Determination of the force on the
electrodes of a parallel-plate capacitor using
Example 9.4-Force acting on one plate of a parallel-plate capacitor. In this example,
we will find the electric force acting on one plate of a parallel-plate capacitor. The dielectric is
homogel~eous, of permittivity E, the area of the plates is S, and the distance between them is
x. One plate is charged with Q and the other with -Q (Fig. 9.5). Let the electric force move the
right plate by a small distance dx. The energy in the capacitor is given by W,(x) = Q2
Q2x/(2cS), SO the force that tends to irlcrease the distance between the plates is
This is the same result as in Example 9.2, except for the sign. The minus sign tells us that the
force tends to decrease the coordinate x, i.e., that it is attractive.
Example 9.5-Force per unit length acting on a conductor of a two-wire line. The wires
of a two-wire line of radii a are x apart, and are charged with charges Q' and -Q'. The energy
per unit length of the line is
using C' as calculated in problem P8.13. From Eq. (9.11) we obtain the force per unit length on
the right conductor, tending to increase the distance between them, as
This is the same as in Example 9.3, except for the minus sign. We know that this means only
that the force tends to decrease the distance x between the wires, i.e., that it is attractive.
Example 9.6-Force acting on a dielectric partly inserted into a parallel-plate capacitor.
Let us find the electric force acting on the dielectric in Fig. 9.6. Equation (9.11) allows us to do
this in a simple way. The capacitance of a capacitor such as this one is given by
ENERGY, FORCES, AND PRESSURE IN THE ELECTROSTATIC FIELD 131
Figure 9.6 Determination of the force on the
dielectric partly inserted between the electrodes of
a parallel-plate capacitor using Eq. (9.11)
(see problem P8.8). The energy in the capacitor is
We(x) = - = Q2 -
2C 2(C1+ C2) 2b[cx + to(a - x)] '
The derivative dW,(x)/dx in this case is a bit more complicated to calculate, and it is left as an
exercise. The force is found to be
Note that this force is always positive because t > to. This means that the forces tend to pull the
dielectric further in between the plates.
Example 9.7-Rise of level of liquid dielectric partly filling a parallel-plate capacitor.
As a final example of the application of Eq. (9.11), let us determine the force that raises the level
of the liquid dielectric between the plates of the capacitor in Fig. 9.3. Assume the dielectric is
distilled water with t-, = 81, the width of the plates is b, their distance is d = 1 cm, and the
capacitor was charged by being connected to V = 1000 V. The electric forces will raise the level
of the water between the plates until the weight of the water between the plates becomes equal
to this force. The weight is equal to
where p,, is the mass density of water and g = 9.81 m/s2
. By equating this force to the force
that we found in Example 9.6, we get
So far, we have discussed examples of case (I), where the charges in a system
were kept constant. Case (2) is finding forces from energy when the voltage, not the
charge, of the n conducting bodies of the system is kept constant (for example, we
connect the system to a battery). When a body is moved by electric forces again by
dx, some changes must occur in the charges on the conducting bodies, due to electro-
static induction. These changes are made at the expense of the energy in the sources
(battery). So we would expect the energy contained in the electric field to increase in
this case. It can be shown in a relatively straightforward way that the expression for
the component F, of the electric force on the body in this case is
Fs = +--- We(x) (potentials kept constant).
Of course, this formula in all cases leads to the same result for the force as
Eq. (9.11), but in some cases it is easier to calculate dW,/dx for constant potentials
than for constant charges, and conversely.
Example 9.8-Example 9.6 revisited. Let us compute the force from Example 9.6 using
Eq. (9.12) instead of Eq. (9.11), which we used in Example 9.6. Now we assume the potential
of the two plates to be constant, and therefore express the system energy in the form
The result is easier to obtain than in Example 9.6.
Questions and puoblems: P9.17 to P9.20
Ultimo aggiornamento: 2014-04-01
Ultimo aggiornamento: 2014-03-28
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