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Step-by-Step Solution
Step 1: Understand the Problem
We have a total charge $Q$ divided into two parts: $q$ and $(Q - q)$. These two charges are placed at a distance $r$ apart. We want to determine the value of $q$ (in terms of $Q$) that maximizes the electrostatic repulsion between these two charges.
Step 2: Write the Expression for Electrostatic Force
The electrostatic force between two point charges, according to Coulomb's law, is
$F = \dfrac{k\,q\,(Q - q)}{r^2}$
where $k$ is the Coulomb constant and $r$ is the distance between the charges.
Step 3: Differentiate the Force with Respect to $q$ to Find the Maximum
We treat $F$ as a function of $q$ and set its derivative to zero to locate the maximum:
$F(q) = \dfrac{k}{r^2}\,\bigl(q \, Q - q^2\bigr)$
$\dfrac{dF}{dq} = \dfrac{k}{r^2}\,\bigl(Q - 2q\bigr)$
Set this derivative equal to zero to find the critical point:
$\dfrac{k}{r^2}\,\bigl(Q - 2q\bigr) = 0 \quad \Rightarrow \quad Q - 2q = 0 \quad \Rightarrow \quad Q = 2q$
Step 4: Conclude the Optimal Division of Charges
The condition $Q = 2q$ implies $q = \dfrac{Q}{2}$. So, when $q = \dfrac{Q}{2}$, the force between $q$ and $(Q - q)$ is maximum.
Step 5: Visual Representation (Optional)
The situation can be visualized using the figure:
Final Answer
When the total charge $Q$ is divided as $q = \dfrac{Q}{2}$ and $(Q - q) = \dfrac{Q}{2}$, the electrostatic repulsion is maximum. This corresponds to the condition $Q = 2q.$
