Solved Problem on Coulomb's Law

Four positive charges equal to q are located at the vertices of a regular tetrahedron with edges equal to d. Find the magnitude of the electric force due to the three charges at the base of the tetrahedron on the charge located at point P above the base.

Problem data:

Value of electrical charges: +q;
Distance between charges: d.

Problem diagram:

As all charges have the same sign, the charges at the base of the tetrahedron will repel the charge located at point P, and as the values of the charges are equal and the distance between them is the same, the magnitude of the repulsive force \( {\vec{F}}_{\small E} \) will be the same (Figure 1).

Figure 1

Solution

Coulomb's Law is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {F_{\small E}=k_{0}\frac{|q||Q|}{r^{2}}} \end{gather} \]

Looking at a vertical plane passing through one of the base charges and the charge at point P (Figure 2), applying Coulomb's Law, the repulsive electric force will be

\[ \begin{gather} F_{\small E}=k_{0}\frac{|q||q|}{d^{2}}\\[5pt] F_{\small E}=k_{0}\frac{q^{2}}{d^{2}} \tag{I} \end{gather} \]

This electric force is decomposed in the directions parallel to the plane of the base of the tetrahedron \( {\vec{F}}_{\small{EP}} \) and the normal to it \( {\vec{F}}_{\small{EN}} \). We draw the electric force in a Cartesian coordinate system and we have its components (Figure 3)

Figure 2

\[ \begin{gather} F_{\small{EP}}=F_{\small E}\sin \theta \tag{II-a} \end{gather} \]

\[ \begin{gather} F_{\small{EN}}=F_{\small E}\cos \theta \tag{II-b} \end{gather} \]

where the angle θ is measured between the vector force \( {\vec{F}}_{\small E} \) and the normal component \( {\vec{F}}_{\small{EN}} \) to the plane, it is the same angle measured between the edge d of the tetrahedron and its height h (these are opposite angles).

Figure 3

Forces parallel to the plane, \( {\vec{F}}_{\small{EP}} \):

By symmetry of the problem, each charge at the base will interact in the same way with the charge at P, so we have three parallel components \( {\vec{F}}_{\small{EP}} \) acting at that point (Figure 4-A). Looking from above (Figure 4-B) we see these forces are equally distributed around the charge at P. By the Polygon Method for adding vectors (Figure 4-C) we have that these forces form a closed polygonal, therefore the resultant of the parallel forces due to base charges is zero.

\[ \begin{gather} F_{\small{EP}}+F_{\small{EP}}+F_{\small{EP}}=0 \end{gather} \]

Figure 4

Forces normal to the plane, \( {\vec{F}}_{\small{EN}} \):

To find the value of the normal component to the \( {\vec{F}}_{\small{EN}} \) plane, we find the cosine of the angle θ as a function of the distance d between the charges. The cosine θ is calculated by

\[ \begin{gather} \cos \theta =\frac{h}{d} \tag{III} \end{gather} \]

As the tetrahedron is regular, the lateral surfaces are equilateral triangles (they have three equal sides), and the height a of one of the surfaces can be found using the Pythagorean Theorem (Figure 5)

\[ \begin{gather} d^{2}=a^{2}+\left(\frac{d}{2}\right)^{2}\\[5pt] d^{2}=a^{2}+\frac{d^{2}}{4}\\[5pt] a^{2}=d^{2}-\frac{d^{2}}{4} \end{gather} \]

multiplying and dividing by 4 the first term on the right-hand side of the equation

Figure 5

\[ \begin{gather} a^{2}=\frac{4}{4}\times d^{2}-\frac{d^{2}}{4}\\[5pt] a^{2}=\frac{4d^{2}-d^{2}}{4}\\[5pt] a^{2}=\frac{3d^{2}}{4}\\[5pt] a=\sqrt{\frac{3d^{2}}{4}}\\[5pt] a=\frac{d\sqrt{3}}{2} \tag{IV} \end{gather} \]

The height h of the tetrahedron divides the base into two segments of sizes m and n (Figure 6), from which we can write the following relations, the sum of m and n is the height a of the base triangle

\[ \begin{gather} a=m+n \end{gather} \]

substituting the value of a found in (IV)

\[ \begin{gather} \frac{d\sqrt{3}}{2}=m+n \tag{V} \end{gather} \]

Figure 6

applying the Pythagorean Theorem to the right triangle with legs m and h and hypotenuse a

\[ \begin{gather} a^{2}=m^{2}+h^{2} \end{gather} \]

substituting the value of a found in (IV)

\[ \begin{gather} \left(\frac{d\sqrt{3}}{2}\right)^{2}=m^{2}+h^{2}\\[5pt] \frac{3d^{2}}{4}=m^{2}+h^{2} \tag{VI} \end{gather} \]

applying the Pythagorean Theorem to the right triangle at left with legs n and h and hypotenuse d

\[ \begin{gather} d^{2}=n^{2}+h^{2} \tag{VII} \end{gather} \]

Subtracting equation (VI) from the equation (VII)

\[ \begin{gather} \frac{ \begin{matrix} \quad d^{2}=n^{2}+h^{2}\\ (-)\quad \dfrac{3d^{2}}{4}=m^{2}+h^{2}\quad\;\; \end{matrix}} {d^{2}-\dfrac{3d^{2}}{4}=n^{2}+h^{2}-(m^{2}+h^{2})}\\[5pt] d^{2}-\frac{3d^{2}}{4}=n^{2}+h^{2}-m^{2}-h^{2}\\[5pt] d^{2}-\frac{3d^{2}}{4}=n^{2}-m^{2} \end{gather} \]

From the Special Binomials Products \( x^{2}-y^{2}=(x+y)(x-y) \)

\[ x^{2}-y^{2}=(x+y)(x-y) \]

on the left-hand side multiplying and dividing by 4 the first term of the equation

\[ \begin{gather} \frac{4}{4}\times d^{2}-\frac{3d^{2}}{4}=(n+m)(n-m)\\[5pt] \frac{4d^{2}-3d^{2}}{4}=(n+m)(n-m)\\[5pt] \frac{d^{2}}{4}=(n+m)(n-m) \end{gather} \]

substituting the term n+m by the value given in (V)

\[ \begin{gather} \frac{d^{2}}{4}=\frac{d\sqrt{3\;}}{2}(n-m)\\[5pt] n-m=\frac{d^{2}}{\cancelto{2}{4}}\times\frac{\cancel{2}}{d\sqrt{3\;}}\\[5pt] n-m=\frac{d}{2\sqrt{3\;}} \end{gather} \]

multiplying the numerator and denominator by \( \sqrt{3\;} \) on the right side of the equality

\[ \begin{gather} n-m=\frac{d}{2\sqrt{3\;}}\times\frac{\sqrt{3\;}}{\sqrt{3\;}}\\[5pt] n-m=\frac{d\sqrt{3\;}}{2\times 3}\\[5pt] n-m=\frac{d\sqrt{3\;}}{6} \tag{VIII} \end{gather} \]

Adding expressions (V) and (VIII)

\[ \begin{gather} \frac{ \begin{matrix} \quad n+m=\dfrac{d\sqrt{3\;}}{2}\\ (\text{+})\quad n-m=\dfrac{d\sqrt{3\;}}{6}\quad\; \end{matrix}} {2n=\dfrac{d\sqrt{3\;}}{2}+\dfrac{d\sqrt{3\;}}{6}} \end{gather} \]

multiplying the numerator and denominator by 3 on the right side of the equation

\[ \begin{gather} 2n=\frac{3}{3}\times\frac{d\sqrt{3\;}}{2}+\frac{d\sqrt{3\;}}{6}\\[5pt] 2n=\frac{3d\sqrt{3\;}+d\sqrt{3\;}}{6}\\[5pt] n=\frac{4d\sqrt{3\;}}{2\times 6}\\[5pt] n=\frac{d\sqrt{3\;}}{3} \tag{IX} \end{gather} \]

Using equation (VII), we can determine h in terms of d

\[ \begin{gather} d^{2}=\left(\frac{d\sqrt{3\;}}{3}\right)^{2}+h^{2}\\[5pt] d^{2}=\frac{3d^{2}}{9}+h^{2}\\[5pt] h^{2}=d^{2}-\frac{3d^{2}}{9} \end{gather} \]

multiplying and dividing by 9 the first term on the right-hand side of the equation

\[ \begin{gather} h^{2}=\frac{9}{9}\times d^{2}-\frac{3d^{2}}{9}\\[5pt] h^{2}=\frac{9d^{2}-3d^{2}}{9}\\[5pt] h^{2}=\frac{6d^{2}}{9}\\[5pt] h=\sqrt{\frac{6d^{2}}{9}\;}\\[5pt] h=\frac{d\sqrt{6}}{3} \tag{X} \end{gather} \]

substituting expression (X) for h into expression (III) for the cosine of θ

\[ \begin{gather} \cos \theta =\frac{\dfrac{d\sqrt{6}}{3}}{d}\\[5pt] \cos \theta=\frac{\cancel{d}\sqrt{6}}{3}\times\frac{1}{\cancel{d}}\\[5pt] \cos \theta=\frac{\sqrt{6}}{3} \end{gather} \]

Using expressions (I), (II), and the cosine, calculated above, the normal electric force for the interaction between one of the base charges and the charge at point P will be

\[ \begin{gather} F_{\small{EN}}=k_{0}\frac{q^{2}}{d^{2}}\times\frac{\sqrt{6}}{3} \end{gather} \]

By symmetry, the charge at point P interacts equally with the other two charges on the base, so the net electric force on the charge at P will be (Figure 7)

\[ \begin{gather} F_{\small{ER}}=3F_{\small{EN}}\\[5pt] F_{\small{ER}}=\cancel{3}\times k_{0}\frac{q^{2}}{d^{2}}\times\frac{\sqrt{6}}{\cancel{3}} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {F_{\small{ER}}=k_{0}\frac{q^{2}\sqrt{6}}{d^{2}}} \end{gather} \]

Figure 7

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .