_{
}

where q_{1} and q_{2} are the electric charges of the two
objects, and r is their separation distance. The electric potential energy of
a system of three point charges (see Figure 26.1) can be calculated in a
similar manner

_{
}

where q_{1}, q_{2}, and q_{3} are the electric charges
of the three objects, and r_{12}, r_{13}, and r_{23}
are their separation distances (see Figure 26.1). The potential energy in
eq.(26.2) is the energy required to assemble the system of charges from an
initial situation in which all charges are infinitely far apart. Equation
(26.2) can be written in terms of the electrostatic potentials V:

_{
}

where V_{other}(1) is the electric potential at the position of charge
1 produced by all other charges

_{
}

and similarly for V_{other}(2) and V_{other}(3).

According to the alpha-particle model of the nucleus some nuclei
consist of a regular geometric arrangement of alpha particles. For instance,
the nucleus of ^{12}C consists of three alpha particles on an
equilateral triangle (see Figure 26.2). Assuming that the distance between
pairs of alpha particles is 3 x 10^{-15} m, what is the electric energy
of this arrangement of alpha particles ? Treat the alpha particles as
pointlike.

The electric potential at the location of each alpha particle is equal to

_{
}

where d = 3.0 x 10^{-15} m. The electric energy of this configuration
can be calculated by combining eq.(26.5) and eq.(26.3):

_{
}

_{
}

The electric field E between the plates is a function of the charge density [sigma]

_{
}

The potential difference V_{1} - V_{2} between the plates can
be obtained by a path integration of the electric field

_{
}

Combining eq.(26.9) and eq.(26.7) we can calculate the electrostatic energy of the system:

_{
}

This equation shows that electrostatic energy can be stored in a capacitor. Equation (26.10) can be rewritten as

_{
}

where Volume is the volume between the capacitor plates. The quantity
[epsilon]_{0} ^{.} E^{2}/2 is called the energy density
(potential energy per unit volume).

_{
}

where V_{1} and V_{2} are the electrostatic potential of the
top and bottom plate, respectively. The potential difference, V_{1} -
V_{2}, is related to the electric field between the plates

_{
}

The electric field E(l) can be related to the charges on the small segments of the capacitor plates via Gauss' law. Consider a volume with its sides parallel to the field lines (see Figure 26.5). The electric flux through its surface is equal to

_{
}

where E(l) is the strength of the electric field at a distance l from the
bottom capacitor plate (see Figure 26.5) and dS(l) is the area of the top of
the integration volume. The flux is negative since the field lines are
entering the integration volume. The flux through the sides of the integration
volume is zero since the sides are chosen to be parallel to the field lines.
The flux through the bottom of the integration volume is also zero, since the
electric field in any conductor is zero. Gauss' law requires that the flux
through the surface of any volume is equal to the charge enclosed by that
volume divided by [epsilon]_{0}:

_{
}

_{
}

Equations (26.12), (26.13) and (26.16) can be combined to give

_{
}

This calculation can be generalized to objects of arbitrary shapes, and the electrostatic energy of any system can be expressed as the volume integral of the energy density u which is defined as

_{
}

Thus

_{
}

where the volume integration extends over all regions where there is an electric field.

In symmetric fission, the nucleus of uranium (^{238}U) splits
into two nuclei of palladium (^{119}Pd). The uranium nucleus is
spherical with a radius of 7.4 x 10^{-15} m. Assume that the two
palladium nuclei adopt a spherical shape immediately after fission; at this
instant, the configuration is as shown in Figure 26.6. The size of the nuclei
in Figure 26.6 can be calculated from the size of the uranium nucleus because
nuclear material maintains a constant density.

b) Calculate the total electric energy of the palladium nuclei in the configuration shown in Figure 26.6, immediately after fission. Take into account the mutual electric potential energy of the two nuclei and also the individual electric energy of the two palladium nuclei by themselves.

c) Calculate the total electric energy a long time after fission when the two palladium nuclei have moved apart by a very large distance.

d) Ultimately, how much electric energy is released into other forms of energy in the complete fission process ?

e) If 1 kg of uranium undergoes fission, how much electric energy is released ?

a) The electric energy of the uranium nucleus before fission can be calculated using the equations derived in Example 26.4 in Ohanian:

_{
}

For the uranium nucleus q = 92e and R = 7.4 x 10^{-15} m. Substituting
these values into eq.(26.20) we obtain

_{
}

b) Suppose the radius of a palladium nucleus is R_{Pd}. The total
volume of nuclear matter of the system shown in Figure 26.6 is equal to

_{
}

Since the density of nuclear matter is constant, the volume in eq.(26.22) must be equal to the volume of the original uranium nucleus

_{
}

Combining eq.(26.23) and (26.22) we obtain the following equation for the radius of the palladium nucleus:

_{
}

The electrostatic energy of each palladium nucleus is equal to

_{
}

where we have used the radius calculated in eq.(26.24) and a charge
q_{Pd} = 46e. Besides the internal energy of the palladium nuclei, the
electric energy of the configuration must also be included in the calculation
of the total electric potential energy of the nuclear system

_{
}

where q_{Pd} is the charge of the palladium nucleus (q_{Pd} =
26e) and R_{int} is the distance between the centers of the two nuclei
(R_{int} = 2 R_{Pd} = 11.7 x 10^{-15} m). Substituting
these values into eq.(26.26) we obtain

_{
}

The total electric energy of the system at fission is therefore

_{
}

c) Due to the electric repulsion between the positively charge palladium nuclei, they will separate and move to infinity. At this point, the electric energy of the system is just the sum of the electric energies of the two palladium nuclei:

_{
}

d) The total release of energy is equal to the difference in the electric energy of the system before fission (eq.(26.21)) and long after fission (eq.(26.29)):

_{
}

e) Equation (26.30) gives the energy released when 1 uranium nucleus fissions. The number of uranium nuclei in 1 kg of uranium is equal to

_{
}

The total release of energy is equal to

_{
}

To get a feeling for the amount of energy released when uranium fissions, we can compare the energy in eq.(26.32) with the energy released by falling water. Suppose 1 kg of water falls 100 m. The energy released is equal to the change in the potential energy of the water:

_{
}

The mass of water needed to generate an amount of energy equal to that released in the fission of 1 kg uranium is

_{
}

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