Solved Problem on Coulomb's Law
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Two charged spheres, with charges q1 and q2 of the same sign, are connected by a wire of insulating material of length R and diameter d. Determine the minimum diameter of this wire so that it resists the electric force of repulsion between the charges, knowing that another wire of the same material and diameter D resists at most a tension of magnitude T. Assume the system is in a vacuum.


Problem data:
  • Charge 1:    q1;
  • Charge 2:    q2;
  • Distance between the charges:    R;
  • Known wire diameter:    D;
  • Tension on the known wire:    T;
  • Coulomb constant:    k0.
Problem diagram:

As the charges are of the same sign, there is an electric force of repulsion \( {\vec{F}}_{E} \) between them, this force strains the wire that connects the charges with a force of magnitude \( {\vec{T}}_{E} \) (Figure 1).
Figure 1

Solution

According to Coulomb's Law, the electric force is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {F_{E}=k_{0}\frac{|Q_{1}||Q_{2}|}{r^{2}}} \end{gather} \]
the electric force of repulsion between the spheres will be
\[ \begin{gather} F_{E}=k_{0}\frac{q_{1}\;q_{2}}{R^{2}} \tag{I} \end{gather} \]
Enlarging a cross-section of the wire that joins the spheres (Figure 2), we can see that the tension in the wire can be divided by the unit area of the wire
\[ \begin{gather} T_{u}=\frac{T_{E}}{a} \tag{II} \end{gather} \]
The circular cross-section area of the wire is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {a=\pi r^{2}} \end{gather} \]
the radius is half of the diameter of the wire, \( \left(r=\frac{d}{2}\right) \)
Figure 2
\[ \begin{gather} a=\pi \left(\frac{d}{2}\right)^{2}\\[5pt] a=\pi\frac{d^{2}}{4} \tag{III} \end{gather} \]
since the electric force of repulsion is equal to the tension in the wire, FE = TE, we substitute expressions (I) and (III) into expression (II) to obtain the tension per unit area
\[ \begin{gather} T_{u}=\frac{k_{0}\dfrac{q_{1}\;q_{2}}{R^{2}}}{\pi\dfrac{d^{2}}{4}}\\[5pt] T_{u}=k_{0}\frac{q_{1}\;q_{2}}{R^{2}}\frac{4}{\pi d^{2}} \tag{IV} \end{gather} \]
The problem tells us that another known wire, with diameter D, resists a tension of magnitude T. Using the same argument, we can obtain the tension per unit area in this wire (Figure 3)
\[ \begin{gather} T_{u}=\frac{T}{A} \tag{V} \end{gather} \]
The circular cross-section area of the wire is given by
\[ \begin{gather} A=\pi r^{2} \end{gather} \]
the radius is half of the diameter of the wire, \( \left(r=\frac{D}{2}\right) \)
\[ \begin{gather} A=\pi \left(\frac{D}{2}\right)^{2}\\[5pt] A=\pi\frac{D^{2}}{4} \tag{VI} \end{gather} \]
Figure 3

substituting expression (VI) into expression (V), we obtain the tension per unit area of the known wire
\[ \begin{gather} T_{u}=\frac{T}{\pi \dfrac{D^{2}}{4}}\\[5pt] T_{u}=T\frac{4}{\pi D^{2}} \tag{VII} \end{gather} \]
The thicker wire resists greater tension than the thinner wire, as its area is greater, but the tension per unit area is the same for both wires, equating expressions (IV) and (VII)
\[ \begin{gather} k_{0}\frac{q_{1}\;q_{2}}{R^{2}}\frac{\cancel{4}}{\cancel{\pi} d^{2}}=T\frac{\cancel{4}}{\cancel{\pi} D^{2}}\\[5pt] k_{0}\frac{q_{1}\;q_{2}}{R^{2}}\frac{1}{d^{2}}=T\frac{1}{D^{2}}\\[5pt] k_{0}\frac{q_{1}\;q_{2}}{R^{2}}\frac{D^{2}}{T}=d^{2}\\[5pt] d=\sqrt{k_{0}\frac{q_{1}\;q_{2}}{R^{2}}\frac{D^{2}}{T}} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {d=\frac{D}{R}\sqrt{k_{0}\frac{q_{1}q_{2}}{T}}} \end{gather} \]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .