Solved Problem on Coulomb's Law

Three electric charges, of 1 μC each, are fixed at the vertices of a square of side 1 m, a particle with a charge of 1 μC and mass 1 g is left at rest at the fourth vertex of the square, at this moment begins to act a repulsive force of the other charges. Determine the acceleration of the particle at the moment it is released.

Problem data:

Value of charges: q = 1 μC;
Free charge mass: m = 1 g;
Distance between charges: L = 1 m;
Coulomb constant \( k_{0}=9\times10^{9}\;\frac{\mathrm{N.m}^{2}}{\mathrm{C}^{2}} \).

Problem diagram:

The electric force between two charges is in the direction of the line joining these charges, so \( {\vec{F}}_{1} \) is the electric force between charge q and charge q₁, \( {\vec{F}}_{2} \) is the electric force between charge q and charge q₂ and \( {\vec{F}}_{3} \) is the electric force between charge q and charge q₃ (Figure 1).
The distance between charges q and q₂ will be the diagonal d of the square, using the Pythagorean Theorem

\[ \begin{gather} d^{2}=L^{2}+L^{2}\\[5pt] d^{2}=1^{2}+1^{2}\\[5pt] d^{2}=2\\[5pt] d=\sqrt{2\;}\;\mathrm{m} \end{gather} \]

Figure 1

Solution

First, we convert the unit of mass given in grams to kilograms used in the International System of Units (SI).

\[ \begin{gather} m=1\;\cancel{\mathrm{g}}\times\frac{1\;\mathrm{kg}}{1000\;\cancel{\mathrm{g}}}=\frac{1\;\mathrm{kg}}{10^{3}} =1\times10^{-3}\;\mathrm{kg} \end{gather} \]

According to Coulomb's Law, the electric force is given, in magnitude, by

\[ \begin{gather} \bbox[#99CCFF,10px] {F_{E}=k_{0}\frac{|Q_{1}||Q_{2}|}{r^{2}}} \tag{I} \end{gather} \]

We plot the forces on a Cartesian coordinate system and obtain their components along the x and y directions (Figure 2).

Figure 2

Direction x:

The force \( {\vec{F}}_{1} \) only has a component in the x direction, in magnitude

\[ \begin{gather} F_{1x}=k_{0}\frac{q\;q_{1}}{L^{2}}\\[5pt] F_{1x}=9\times 10^{9}\times\frac{1\times 10^{-6}\times1\times 10^{-6}}{1^{2}}\\[5pt] F_{1x}=9\times 10^{9}\times 1\times 10^{-12}\\[5pt] F_{1x}=9\times 10^{-3} \tag{II} \end{gather} \]

The x component of the force \( {\vec{F}}_{2} \) is given by

\[ \begin{gather} F_{2x}=F_{2}\cos 45° \end{gather} \]

where F₂ is given by the expression (I)

From the Trigonometry \( \cos 45°=\dfrac{\sqrt{2\;}}{2} \)

\[ \begin{gather} F_{2x}=k_{0}\frac{q\;q_{2}}{d^{2}}\cos 45°\\[5pt] F_{2x}=9\times 10^{9}\times \frac{1\times 10^{-6}\times 1\times 10^{-6}}{(\sqrt{2\;})^{2}}\times \frac{\sqrt{2\;}}{2}\\[5pt] F_{2x}=9\times 10^{9}\times \frac{1\times 10^{-12}}{2}\times \frac{\sqrt{2\;}}{2}\\[5pt] F_{2x}=9\times 10^{-3}\times \frac{\sqrt{2\;}}{4} \tag{III} \end{gather} \]

The resultant of the forces along the x direction will be given by the sum of expressions (II) and (III)

\[ \begin{gather} F_{x}=F_{1x}+F_{2x}\\[5pt] F_{x}=9\times 10^{-3}+9\times 10^{-3}\times \frac{\sqrt{2\;}}{4}\\[5pt] F_{x}=9\times 10^{-3}\times \left(1+\frac{\sqrt{2\;}}{4}\right) \tag{IV} \end{gather} \]

Direction y:

The force \( {\vec{F}}_{3} \) only has a component in the y direction, in magnitude

\[ \begin{gather} F_{3y}=k_{0}\frac{q\;q_{3}}{L^{2}}\\[5pt] F_{3y}=9\times 10^{9}\times \frac{1\times 10^{-6}\times 1\times 10^{-6}}{1^{2}}\\[5pt] F_{3y}=9\times 10^{9}\times 1\times 10^{-12}\\[5pt] F_{3y}=9\times 10^{-3} \tag{V} \end{gather} \]

The y component of the force \( {\vec{F}}_{2} \) is given by

\[ \begin{gather} F_{2y}=F_{2}\sin 45° \end{gather} \]

where F₂ is given by the expression (I)

From the Trigonometry \( \sin 45°=\dfrac{\sqrt{2\;}}{2} \)

\[ \begin{gather} F_{2y}=k_{0}\frac{q\;q_{2}}{d^{2}}\sin 45°\\[5pt] F_{2y}=9\times 10^{9}\times \frac{1\times 10^{-6}\times 1\times 10^{-6}}{(\sqrt{2\;})^{2}}\times \frac{\sqrt{2\;}}{2}\\[5pt] F_{2y}=9\times 10^{9}\times \frac{1\times 10^{-12}}{2}\times \frac{\sqrt{2\;}}{2}\\[5pt] F_{2y}=9\times 10^{-3}\times \frac{\sqrt{2\;}}{4} \tag{VI} \end{gather} \]

The resultant of the forces along the y direction will be given by the sum of (V) and (VI)

\[ \begin{gather} F_{y}=F_{3y}+F_{2y}\\[5pt] F_{y}=9\times 10^{-3}+9\times 10^{-3}\times \frac{\sqrt{2\;}}{4}\\[5pt] F_{y}=9\times 10^{-3}\times \left(1+\frac{\sqrt{2\;}}{4}\right) \tag{VII} \end{gather} \]

The magnitude of the resultant electric force F_E will be obtained using the Pythagorean Theorem using expressions (IV) and (VII)

\[ \begin{gather} F_{E}^{2}=F_{x}^{2}+F_{y}^{2}\\[5pt] F_{E}^{2}=\left[9\times 10^{-3}\times \left(1+\frac{\sqrt{2\;}}{4}\right)\right]^{2}+\left[9\times 10^{-3}\times \left(1+\frac{\sqrt{2\;}}{4}\right)\right]^{2}\\[5pt] F_{E}^{2}=2\times \left[9\times 10^{-3}\times \left(1+\frac{\sqrt{2\;}}{4}\right)\right]^{2}\\[5pt] F_{E}=\sqrt{2\times \left[9\times 10^{-3}\times \left(1+\frac{\sqrt{2\;}}{4}\right)\right]^{2}\;}\\[5pt] F_{E}=9\times 10^{-3}\times \left(1+\frac{\sqrt{2\;}}{4}\right)\times \sqrt{2\;}\\[5pt] F_{E}=9\times 10^{-3}\times \left(\sqrt{2\;}+\frac{\sqrt{2\;}\times \sqrt{2\;}}{4}\right)\\[5pt] F_{E}=9\times 10^{-3}\times \left(\sqrt{2\;}+\frac{2}{4}\right)\\[5pt] F_{E}=9\times 10^{-3}\times \left(\sqrt{2\;}+\frac{1}{2}\right)\\[5pt] F_{E}=1.72\times 10^{-2}\;\mathrm{N} \end{gather} \]

Using Newton's Second Law

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

the only force acting on the charge is the electric force F_E, so the acceleration will be in the same direction as the resultant electric force

Figure 3

\[ \begin{gather} F_{E}=ma\\[5pt] a=\frac{F_{E}}{m}\\[5pt] a=\frac{1.72\times 10^{-2}}{1\times 10^{-3}}\\[5pt] a=1.72\times 10^{-2}\times 10^{3}\\[5pt] a=1.72\times 10 \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {a=17.2\;\mathrm{m/s}^{2}} \end{gather} \]

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