## English Issue, December 2002 | ||||

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## Solutions of problems for physics P

**P. 3432.** *Identical metal spheres are placed into the vertices of a regular tetrahedron. The spheres do not touch. When a single sphere *(*A*)* is given a charge of *20 nC* it reaches the same potential as when A** and another sphere are given *15 nC* each. What equal charge should be given to A** and to two other spheres, and what equal charge to all four spheres so that the potential of sphere A** is always the same?*

(*6 points*)

Submitted by: *Bihary Zsolt,* Irvine, California

**Solution.** If, in a certain tetrahedral arrangement, the electric charges are in equilibrium on the metal spheres, then increasing the charges proportionally (say \(\displaystyle \lambda\) times) the equilibrium will not change but the original values of the field strengths and potentials will increase \(\displaystyle \lambda\) times. It is also true that if we add together two equilibrium arrangements then the result is also an equilibrium arrangement in which the field strengths and the potentials at every point are the sum of the original vectors or scalars. The two above-mentioned characteristics can together be described as the *principle of superposition.*

Let us term the first arrangement (with only one charged sphere) I, and the second (where *A* and an other sphere is charged) II. Turn by 120^{o} arrangement II around a line connecting the centre of sphere *A* and the centre of the tetrahedron (let us call this arrangement III), and turn it by -120^{o} (let us call this arrangement IV).

After this let us take the superposition of the \(\displaystyle \lambda\)_{1}-fold value of arrangement I and the \(\displaystyle \lambda\)_{2}-fold values of arrangements II and III and let us choose the coefficients in a way that there are three spheres of the same charge and the potential of sphere *A* is exactly the same (*U*) as in arrangement I. These requirements are fulfilled if

20\(\displaystyle \lambda\)_{1}+15\(\displaystyle \lambda\)_{2}+15\(\displaystyle \lambda\)_{2} = 15\(\displaystyle \lambda\)_{2}, and \(\displaystyle \lambda\)_{1}*U*+\(\displaystyle \lambda\)_{2}*U*+\(\displaystyle \lambda\)_{2}*U* = *U*.

The solution of the above equation system is \(\displaystyle \lambda_1=-\frac{3}{5}\), \(\displaystyle \lambda_2=\frac{4}{5}\) and, accordingly, the three charged spheres will have 12 nC charge each.

Similarly, taking the proper superposition of arrangement I and arrangement *II*+*III*+*IV*, an arrangement can be obtained where all four spheres have the same charge and the potential of *A* is *U*. In this case the charge of the spheres are 10 nC each.

Based on the paper of *Pápai Tivadar* (11th form student of Dráva Völgye Secondary School, Barcs)

*Note.* If the distance of the spheres were much greater than the radii of the spheres, their electrostatic field could be obtained in the point charge approximation. In this case (with the numbers given) this is not feasible, the size of the spheres and their distances are commeasurable and therefore the spheres polarize each other. The calculation of the charge distributions and the electric field is very difficult, but fortunately they are not needed - if we follow the above considerations.