# 26.1 Introduction

**Figure 26.1. System of three charges.**

_{ }

_{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

_{ }

_{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:

_{ }

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

_{ }

_{other}(2) and V

_{other}(3).

### Example: Problem 26.6

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.

**Figure 26.2. Alpha-particle model of**

^{12}C._{ }

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

_{ }

## 26.2 Energy of a System of Conductors

**Figure 26.3. The capacitor.**

_{1}and the potential of plate 2 is V

_{2}. The electrostatic energy of the capacitor is then equal to

_{ }

_{ }

_{1}- V

_{2}between the plates can be obtained by a path integration of the electric field

_{ }

_{ }

_{ }

_{0}

^{.}E

^{2}/2 is called the energy density (potential energy per unit volume).

**Figure 26.4. Field lines at the edge of a capacitor.**

_{ }

_{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

_{ }

_{ }

_{0}:

_{ }

**Figure 26.5. Integration volume discussed in the text.**

_{ }

_{ }

_{ }

_{ }

### Example: Problem 26.27

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.

**Figure 26.6. Two palladium nuclei right after fission of**

^{238}U.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:

_{ }

^{-15}m. Substituting these values into eq.(26.20) we obtain

_{ }

_{Pd}. The total volume of nuclear matter of the system shown in Figure 26.6 is equal to

_{ }

_{ }

_{ }

_{ }

_{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

_{ }

_{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

_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

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