Правка разделов «Statement», «Answer»
en/6.5.23.md
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| ### Statement | |||
| − | $6.5.23.$ [Insert the problem statement] | ||
| + | $6.5.23.$ 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 $16U_1-(4U_1+6U_2)=12U_1-6U_2=6A$. | |||
| @@ -10,4 +10,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $6A$ | ||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $6.5.23.$ [Insert the problem statement] | $6.5.23.$ 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 | ### 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 $A=4U_1-(2U_1+U_2)=2U_1-U_2$. | 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 $16U_1-(4U_1+6U_2)=12U_1-6U_2=6A$. | 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 $16U_1-(4U_1+6U_2)=12U_1-6U_2=6A$. | ||
| @@ -10,4 +10,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $6A$ | ||