Solved Problem on Coulomb's Law
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Three spheres, each of weight W and charged Q, are suspended by insulating strings of
length L attached to the same point. In the equilibrium position, the strings make an angle
θ with the vertical. Calculate the charge Q.
Problem data:
- Weight of each sphere: W;
- Length of strings: L;
- Angle between the string and the vertical: θ;
- Coulomb constant: k0.
Looking at this arrangement of charges from above towards a horizontal plane that contains the charges (Figure 1-A), as all the charges have the same value, they repulse each other, becoming equidistant from each other (Figure 1-B). The charges are at the vertices of an equilateral triangle, the distance between two charges being equal to d, and the angle between the two sides of the triangle is 60°.
The distance of a charge from the center of the distribution is R, and the distance d between two charges, side of the triangle, can be found by applying the Law of Cosines (Figure 1-B)
\[
\begin{gather}
\bbox[#99CCFF,10px]
{c^{2}=a^{2}+b^{2}-2ab\cos \alpha}
\end{gather}
\]
\[
\begin{gather}
d^{2}=R^{2}+R^{2}-2RR\cos 120°
\end{gather}
\]
From the Trigonometry
\( \cos 120°=-{\dfrac{1}{2}} \)
\[
\begin{gather}
d^{2}=2R^{2}-\cancel{2}R^{2}\times\left(-{\frac{1}{\cancel{2}}}\right)\\[5pt]
d^{2}=2R^{2}+R^{2}\\[5pt]
d^{2}=3R^{2}\\[5pt]
d=\sqrt{3R^{2}\;}\\[5pt]
d=R\sqrt{3\;}
\end{gather}
\]
On one of the charges acts the electric force
\( {\vec{F}}_{E} \)
due to the other two charges, using Coulomb's Law
\[
\begin{gather}
\bbox[#99CCFF,10px]
{F_{E}=k_{0}\frac{|Q_{1}||Q_{2}|}{r^{2}}}
\end{gather}
\]
\[
\begin{gather}
F_{E}=k_{0}\frac{Q\;Q}{d^{2}}\\[5pt]
F_{E}=k_{0}\frac{Q^{2}}{\left(R\sqrt{3\;}\right)^{2}}\\[5pt]
F_{E}=k_{0}\frac{Q^{2}}{3R^{2}} \tag{I}
\end{gather}
\]
The angle between these forces is opposite the angle of the triangle where the charge is, so this angle
also measures 60º (Figure 1-B). The net electric force on one of the charges,
\( {\vec{F}}_{ER} \),
is calculated by applying the Law of Cosines
\[
\begin{gather}
F_{ER}^{2}=F_{E}^{2}+F_{E}^{2}+2F_{E}F_{E}\cos 60°
\end{gather}
\]
From the Trigonometry
\( \cos 60°=\dfrac{1}{2} \)
\[
\begin{gather}
F_{ER}^{2}=2F_{E}^{2}+\cancel{2}F_{E}^{2}\frac{1}{\cancel{2}}\\[5pt]
F_{ER}^{2}=2F_{E}^{2}+F_{E}^{2}\\[5pt]
F_{ER}^{2}=3F_{E}^{2}\\[5pt]
F_{ER}=\sqrt{3F_{E}^{2}\;}\\[5pt]
F_{ER}=F_{E}\sqrt{3\;} \tag{II}
\end{gather}
\]
substituting the value of the force between two charges found in expression (I) into expression (II),
we have the resultant on one of the charges
\[
\begin{gather}
F_{E}=k_{0}\frac{Q^{2}}{R^{2}}\frac{\sqrt{3\;}}{3} \tag{III}
\end{gather}
\]
The radius of the circle around which the spheres are distributed can be written in terms of L
and θ given (Figure 2)
\[
\begin{gather}
\sin \theta =\frac{\text{opposite side}}{\text{hpotenuse}}=\frac{R}{L}\\[5pt]
R=L\sin \theta \tag{IV}
\end{gather}
\]
substituting expression (IV) into expression (III)
\[
\begin{gather}
F_{E}=k_{0}\frac{Q^{2}}{L^{2}\sin ^{2}\theta}\frac{\sqrt{3\;}}{3} \tag{V}
\end{gather}
\]
Figure 2
Note: Why in the first Law of Cosines was used the minus sign and in the second
a sum?
However, if side b of the triangle is placed in a position making an angle β with side a, this angle will be the same as the angle between an extension of side a and the original position of side b, these two angles are supplementary (their sum equals to 180°), so the value of β will be
Thus the two expressions coincide, depending only on which angle is considered.
For a triangle of sides a, b, c and angle α, opposite side c
as in the figure, the Law of Cosines is written as
\[
\begin{gather}
c^{2}=a^{2}+b^{2}-2ab\cos \alpha
\end{gather}
\]
as is the first case of the problem.
However, if side b of the triangle is placed in a position making an angle β with side a, this angle will be the same as the angle between an extension of side a and the original position of side b, these two angles are supplementary (their sum equals to 180°), so the value of β will be
\[
\begin{gather}
\alpha +\beta =180°\\[5pt]
\beta=180°-\alpha
\end{gather}
\]
Applying the Law of Cosines to this case
\[
\begin{gather}
c^{2}=a^{2}+b^{2}+2ab\cos \beta\\[5pt]
c^{2}=a^{2}+b^{2}+2ab\cos (180°-\alpha )
\end{gather}
\]
using the cosine identity of the difference of arcs
\[
\begin{gather}
\cos (x-y)=\cos x\cos y+\sin x\sin y
\end{gather}
\]
\[
\begin{gather}
c^{2}=a^{2}+b^{2}+2ab(\cos 180°\cos \alpha+\sin 180°\sin \alpha)\\[5pt]
c^{2}=a^{2}+b^{2}+2ab(-1.\cos \alpha +0.\sin \alpha)\\[5pt]
c^{2}=a^{2}+b^{2}-2ab\cos \alpha
\end{gather}
\]
the second case of the problem is equivalent to the first.Thus the two expressions coincide, depending only on which angle is considered.
Looking toward a vertical plane that contains a charge and the string supporting it (Figure 5-A). The
weight
\( \vec{W} \),
the tension in the string
\( \vec{T} \),
and the electric force act on the charge
\( {\vec{F}}_{E} \)
due to the other charges (Figure 5-B). The angle of the string, where the charge is attached, and the
vertical is given as θ this is also the angle between the string and the vertical passing
through the charge, are alternate angles.
We plot the forces on a Cartesian coordinate system and decompose the forces in the x and y directions (Figure 6). Applying Newton's Second Law
We plot the forces on a Cartesian coordinate system and decompose the forces in the x and y directions (Figure 6). Applying Newton's Second Law
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\vec{F}=m\vec{a}} \tag{VI}
\end{gather}
\]
- y direction:
\[
\begin{gather}
W-T_{y}=0\\[5pt]
W=T_{y} \tag{VII}
\end{gather}
\]
The angle θ is measured between the tension force and the y-axis, so the component
of the tension force in the y direction is given by
\[
\begin{gather}
T_{y}=T\cos \theta \tag{VIII}
\end{gather}
\]
substituting expression (VIII) into expression (VII)
\[
\begin{gather}
W=T\cos \theta \tag{IX}
\end{gather}
\]
- x direction:
\[
\begin{gather}
T_{x}-F_{E}=0\\[5pt]
T_{x}=F_{E} \tag{X}
\end{gather}
\]
the component of the tension force in the x direction is given by
\[
\begin{gather}
T_{x}=T\sin \theta \tag{XI}
\end{gather}
\]
substituting expression (XI) into expression (X)
\[
\begin{gather}
F_{E}=T\sin \theta \tag{XII}
\end{gather}
\]
Dividing expression (XI) by expression (IX)
\[
\begin{gather}
\frac{F_{E}}{P}=\frac{\cancel{T}\sin \theta }{\cancel{T}\cos \theta }
\end{gather}
\]
From the Trigonometry
\( \dfrac{\sin \theta }{\cos \theta }=\tan \theta \)
\[
\begin{gather}
\frac{F_{E}}{W}=\tan \theta\\[5pt]
F_{E}=P\tan \theta
\end{gather}
\]
substituting the electric force by the value found in the expression (V)
\[
\begin{gather}
k_{0}\frac{Q^{2}}{L^{2}\sin ^{2}\theta}\frac{\sqrt{3\;}}{3}=W\tan \theta\\[5pt]
Q^{2}=\frac{3}{\sqrt{3\;}}\frac{W\tan \theta}{k_{0}}L^{2}\sin ^{2}\theta
\end{gather}
\]
multiplying the term on the right-hand side of the equation by
\( \frac{\sqrt{3\;}}{\sqrt{3\;}} \)
\[
\begin{gather}
Q^{2}=\frac{3}{\sqrt{3\;}}\frac{\sqrt{3\;}}{\sqrt{3\;}}\frac{W\tan \theta}{k_{0}}L^{2}\sin ^{2}\theta\\[5pt]
Q^{2}=\frac{\cancel{3}\sqrt{3\;}}{\cancel{3}}\frac{W\tan \theta}{k_{0}}L^{2}\sin ^{2}\theta\\[5pt]
Q^{2}=\frac{\sqrt{3\;}W\tan \theta}{k_{0}}L^{2}\sin ^{2}\theta\\[5pt]
Q=\sqrt{\frac{\sqrt{3\;}W\tan \theta}{k_{0}}L^{2}\sin ^{2}\theta \;}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{Q=L\sin \theta\sqrt{\frac{\sqrt{3\;}W\tan \theta }{k_{0}}\;}}
\end{gather}
\]
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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 .