6.5.23∗. Uniformly charged faces of a regular tetrahedron have the same charge. To put two faces of a tetrahedron together, you need to do work $A$. What kind of work do you need to do to put all the faces of a tetrahedron in one pile?
Solution
Suppose the potential energy of an isolated face of the tetrahedron is $U_1$. Then, the potential energy of two faces of the tetrahedron pressed together is $4U_1$, because the charge is doubled in the same space. Let the potential energy of two faces of the tetrahedron making a dihedral angle with each other be $2U_1+U_2$. Then the work done in pressing the two faces together is $A=4U_1-(2U_1+U_2)=2U_1-U_2$.
Originally, the potential energy of the tetrahedron is $4U_1+6U_2$, because there are $6$ possible pairs out of $4$ faces. When the four faces are pressed together, the potential energy is $16U_1$, because the charge is quadrupled in the same space. Thus, the work done in collapsing the tetrahedron is $16U_1-(4U_1+6U_2)=12U_1-6U_2=6A$.
