Solved Problem on Coulomb's Law

A charge q = 1,0 μC is fixed at a point O in space. A second charge Q = 25,0.10⁻⁸ C and weight W = 2,5.10⁻² N is limited in motion to the vertical passing through O. The charges are in a vacuum. Determine:
a) Is charge Q in equilibrium above or below O?
b) What is the distance between the charges at equilibrium?
c) The type of equilibrium of Q, stable, unstable, or indifferent?

Problem data:

Charge q: q = 1.0 μC = 1.0 × 10⁻⁶ C;
Charge Q: Q = 25.0 × 10⁻⁸ C;
Weight of sphere of charge Q: W = 2.5 × 10⁻² N;
Coulomb constant: \( k_{0}=9\times 10^{9}\;\frac{\text{N.m}^{2}}{\text{C}^{2}} \).

Problem diagram:

We choose a reference frame oriented upwards with the origin at the point where the fixed charge q is placed, the acceleration due to gravity is equal to g oriented downwards (Figure 1). The free charge Q is a distance d from the origin. As the two charges are positive, the electric force \( {\vec{F}}_{E} \) between them is a repulsive force and the weight \( \vec{W} \) acting on the charge Q points downwards in the same direction as the acceleration due to gravity.

Figure 1

Solution

a) For the system to remain in equilibrium, the resultant of the forces acting on the charge Q must be zero.

\[ \begin{gather} \bbox[#99CCFF,10px] {\sum \vec{F}=0} \end{gather} \]

From Figure 1, we can write

\[ \begin{gather} F_{E}-W=0 \tag{I} \end{gather} \]

according to Coulomb's Law, the electric force is given by

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

substituting expression (II) into expression (I)

\[ \begin{gather} k_{0}\frac{|Q||q|}{r^{2}}-P=0\\[5pt] 9\times 10^{9}\times \frac{|25.0\times 10^{-8}\;||1.0\times 10^{-6}\;|}{d^{2}}-2.5\times 10^{-2}=0\\[5pt] 9\times 10^{9}\times \frac{25.0\times 10^{-8}1.0\times 10^{-6}}{d^{2}}=2.5\times 10^{-2}\\[5pt] 9\times 10^{9}\times 25.0\times 10^{-14}=2.5\times 10^{-2}d^{2}\\[5pt] d^{2}=\frac{9\times 25,0.10^{-5}}{2.5\times 10^{-2}}\\[5pt] d^{2}=9\times 10\times 10^{-5}\times 10^{2}\\[5pt] d^{2}=9\times 10^{-2}\\[5pt] d=\sqrt{9\times 10^{-2}\;}\\[5pt] d=\pm 3\times 10^{-1}\;\text{m} \end{gather} \]

For the positive root, the charge is above point O, as the charges are positive, the repulsive force at Q points upwards, and the weight points downwards, these forces cancel each other out, and the charge is in equilibrium (Figure 2-A). For the negative root, the charge is below point O, the repulsive force points downwards, and the weight points downwards, there will be a net force in that direction that will make the charge Q move away from q, the situation is not one of equilibrium.
Charge Q is in equilibrium above point O.

Figure 2

b) At equilibrium, the distance between the charges will be

\[ \begin{gather} \bbox[#FFCCCC,10px] {d=3\times 10^{-1}\;\text{m}} \end{gather} \]

c) Initially, the charge Q is in an equilibrium position at a distance d from the charge q, in this point, the electric force and the weight force cancel each other out (Figure 3-A). As the magnitude of the electric force given by Coulomb's Law is equal to FE if the charge Q is displaced upwards, increasing the distance d from the charge q, the denominator of expression (II) increases as a result, and the electric force decreases. As weight is constant, this gives rise to a net downward force bringing the system back to equilibrium (Figure 3-B). If the charge Q is shifted downwards, decreasing the distance d from the charge q, the denominator of expression (II) decreases, and as a result, the electric force increases. As the weight force is constant, this gives rise to a net upward force which brings the system back to equilibrium (Figure 3-C).
So the equilibrium is stable.

Figure 3

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