Solved Problem on Coulomb's Law and Electric Field
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We want to share a charge Q between two bodies. One of the bodies receives a charge q1 and the other a charge q2. The splitting of the charges is made in such a way that if you have 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 shared charges is the total charge
\[
\begin{gather}
Q=q_{1}+q_{2} \tag{I}
\end{gather}
\]
The electric force Fel is given by Coulomb's Law
\[ \bbox[#99CCFF,10px]
{F_{el}=\frac{1}{4\pi \varepsilon_{0}}\frac{|q_{A}||q_{B}|}{r^{2}}}
\]
\[
\begin{gather}
F_{el}=\frac{1}{4\pi \varepsilon _{0}}\frac{q_{1}q_{2}}{r^{2}} \tag{II}
\end{gather}
\]
From the condition (I), we can write 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
\( k_{0}=\frac{1}{4\pi \epsilon _{0}} \)
and substituting the expression (III) into expression (II)
\[
\begin{gather}
F_{el}=k_{0}\frac{q_{1}(Q-q_{1})}{r^{2}}\\
F_{el}=\frac{k_{0}}{r^{2}}(Qq_{1}-q_{1}^{2}) \tag{IV}
\end{gather}
\]
To find the maximum point of the expression (IV), we take the derivative of the function relative to
q1 and equal to zero.
\[
\begin{gather}
\frac{dF_{el}}{dq_{1}}=0\\
\frac{d}{dq_{1}}\left[\frac{k_{0}}{r^{2}}(Qq_{1}-q_{1}^{2})\right]=0
\end{gather}
\]
Differentiation of \( \displaystyle F_{el}=\frac{k_{0}}{r^{2}}(Qq_{1}-q_{1}^{2}) \)
the term \( \frac{k_{0}}{r^{2}} \) is moved outside of the derivative, and the derivative of a sum is equal to the sum of the derivatives
the term \( \frac{k_{0}}{r^{2}} \) is moved outside of the derivative, and the derivative of a sum is equal to the sum of the derivatives
\[
\begin{gather}
\frac{dF_{el}}{dq_{1}}=\frac{k_{0}}{r^{2}}\left[\frac{d}{dq_{1}}(Qq_{1})-\frac{d}{dq_{1}}(q_{1}^{2})\right]\\
\frac{dF_{el}}{dq_{1}}=\frac{k_{0}}{r^{2}}\left[Q-2q_{1}\right]
\end{gather}
\]
\[
\begin{gather}
\frac{k_{0}}{r^{2}}(Q-2q_{1})=0\\
Q-2q_{1}=0.\frac{r^{2}}{k_{0}}\\
Q-2q_{1}=0\\
2q_{1}=Q\\q_{1}=\frac{Q}{2}
\end{gather}
\]
substituting this result in the expression (III), we obtain q2
\[
\begin{gather}
q_{2}=Q-\frac{Q}{2}\\
q_{2}=\frac{2Q-Q}{2}\\
q_{2}=\frac{Q}{2}
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{q_{1}=q_{2}=\frac{Q}{2}}
\]
The result is independent of the distance r between the charges and the force is maximum when the
total load is equally divided between the bodies.
Note 1: The expression (IV) represents a quadratic function with the term of highest-degree
negative, q1 < 0, it is a parabola that opens down, so 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 take the derivative relative 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 take the derivative relative 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 .