Solved Problem on Coulomb's Law
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We want to distribute a charge Q between two bodies. One of the bodies receives a charge q1 and the other a charge q2. The distribution of the charges is done in such a way that q1+q2=Q. Determine the ratio between the charges so that the Coulomb's force of repulsion between q1 and q2 is maximum for any distance between the charges.
Solution:
The problem gives us the condition that the sum of the distributed charges equals the total charge
\[
\begin{gather}
Q=q_1+q_2 \tag{I}
\end{gather}
\]
The electric force Fel is given by Coulomb's Law
\[
\begin{gather}
\bbox[#99CCFF,10px]
{F_{el}=\frac{1}{4\pi\varepsilon_0}\frac{|q_{\small A}||q_{\small B}|}{r^2}}
\end{gather}
\]
\[
\begin{gather}
F_{el}=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2} \tag{II}
\end{gather}
\]
From the condition (I), we can express the charge q2 as a function of charge
q1
\[
\begin{gather}
q_2=Q-q_1 \tag{III}
\end{gather}
\]
defining the constant term as
\( k_e=\frac{1}{4\pi\epsilon_0} \)
and substituting the equation (III) into equation (II)
\[
\begin{gather}
F_{el}=k_e\frac{q_1(Q-q_1)}{r^2}\\[5pt]
F_{el}=\frac{k_e}{r^2}(Qq_1-q_1^2) \tag{IV}
\end{gather}
\]
To find the maximum point of equation (IV), we take the derivative of the function with respect to
q1 and equal to zero.
\[
\begin{gather}
\frac{dF_{el}}{dq_1}=0\\[5pt]
\frac{d}{dq_1}\left[\frac{k_e}{r^2}(Qq_1-q_1^2)\right]=0
\end{gather}
\]
Differentiation of \( \displaystyle F_{el}=\frac{k_e}{r^2}(Qq_1-q_1^2) \)
the term \( \frac{k_e}{r^2} \) is moved outside of the derivative, and the derivative of the difference is the difference of the derivatives.
the term \( \frac{k_e}{r^2} \) is moved outside of the derivative, and the derivative of the difference is the difference of the derivatives.
\[
\begin{gather}
\frac{dF_{el}}{dq_1}=\frac{k_e}{r^2}\left[\frac{d}{dq_1}(Qq_1)-\frac{d}{dq_1}(q_1^2)\right]\\[5pt]
\frac{dF_{el}}{dq_1}=\frac{k_e}{r^2}\left[Q-2q_1\right]
\end{gather}
\]
\[
\begin{gather}
\frac{k_e}{r^2}(Q-2q_1)=0\\[5pt]
Q-2q_1=0\times \frac{r^2}{k_e}\\[5pt]
Q-2q_1=0\\[5pt]
2q_1=Q\\[5pt]
q_1=\frac{Q}{2}
\end{gather}
\]
substituting this result into equation (III), we obtain q2
\[
\begin{gather}
q_2=Q-\frac{Q}{2}\\[5pt]
q_2=\frac{2Q-Q}{2}\\[5pt]
q_2=\frac{Q}{2}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{q_1=q_2=\frac{Q}{2}}
\end{gather}
\]
The result is independent of the distance r between the charges, and the force is maximum when the
total charges are equally distributed between the bodies.
Note 1: the equation (IV) represents a quadratic function with the term of highest-degree
term negative, q1 < 0, it is a parabola that opens downward, thus the function has
a maximum point.
Note 2: the same result would be obtained if we wrote q1 as a function of q2, \( q_1=Q-q_2 \), and took the derivative of the force with respect to q2, \( \left(\frac{dF_{el}}{dq_{2}}=0\right) \).
Note 2: the same result would be obtained if we wrote q1 as a function of q2, \( q_1=Q-q_2 \), and took the derivative of the force with respect to q2, \( \left(\frac{dF_{el}}{dq_{2}}=0\right) \).
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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 .