Animated Solution for Physics - Electric Charges and Fields: A certain charge $Q$ is divided into two parts $q$ and $(Q-q)$. How should the charges $Q$ and $q$ be divided, so that $q$ and $(Q-q)$ placed at a certain distance apart experience maximum electrostatic repulsion ?

Have you ever wondered how nature inherently prefers balance and symmetry? This classic problem from electrostatics is a beautiful demonstration of that exact principle. We are given a total charge $Q$ and asked to divide it into two parts, $q$ and $(Q-q)$, such that when they are placed at a fixed distance apart, the electrostatic repulsion between them is maximized.

At first glance, you might think that keeping one charge very large and the other very small would do the trick. But as we will see, physics and mathematics conspire to show us that equality is the key to maximum impact. Let's dive into the mechanics of this problem and uncover the elegant math behind it.

Imagine you have a lump of charge, $Q$. You split it into two distinct pieces. If the first piece takes a charge $q$, the law of conservation of charge dictates that the second piece must have whatever is left over, which is $(Q-q)$.

We then take these two charges and pin them down at a specific, constant distance $r$ from each other. Because both pieces originated from the same initial charge $Q$ (assuming it's positive), they will both carry the same sign. And as we know from the fundamental rules of electrostatics, like charges repel.

Our goal is to find the exact value of $q$ that makes this repulsive force as large as physically possible.

To quantify this repulsion, we reach for our trusty tool: Coulomb's Law. Coulomb's Law states that the electrostatic force $F$ between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.

Mathematically, this is expressed as:

Here, $k$ is Coulomb's constant, $q_1$ and $q_2$ are our two charges, and $r$ is the separation distance.

Let's substitute our specific charges into this master equation. We plug in $q_1=q$ and $q_2=(Q-q)$. The distance $r$ remains constant.

This equation is the heart of our problem. It tells us exactly how the force behaves as we vary the amount of charge $q$ we allocate to the first piece.

To find the maximum force, we need to look at our force equation as a mathematical function of $q$. Let's expand the numerator to make it easier to work with:

Notice that $k$, $r$, and $Q$ are all constants. The only variable changing is $q$. This function is a downward-opening parabola, which means it definitely has a peak—a maximum value.

How do we find the peak of a function? We use the power of differential calculus! The derivative of a function gives us its slope or rate of change. At the exact peak of a curve, the slope is perfectly flat, meaning the derivative is zero.

So, to find the maximum force, we must take the first derivative of $F$ with respect to $q$ and set it equal to zero:

Let's execute the differentiation. We apply the derivative operator to our force function:

Since $\frac{k}{r^2}$ is a constant multiplier, it simply passes through the derivative:

Now, we differentiate the terms inside the parenthesis. The derivative of $qQ$ with respect to $q$ is simply $Q$. The derivative of $q^2$ with respect to $q$ is $2q$, using the power rule.

For this entire expression to equal zero, the term inside the parenthesis must be zero, because the constant $\frac{k}{r^2}$ cannot be zero.

Solving for $Q$, we get our final, elegant result:

Or, equivalently:

What does this mean physically? It means that to achieve the maximum possible electrostatic repulsion, you must divide the original charge $Q$ exactly in half!

This is a profound yet intuitive result. It aligns perfectly with a famous mathematical concept known as the AM-GM inequality (Arithmetic Mean-Geometric Mean inequality), which states that for a given sum of two numbers, their product is maximized when the two numbers are equal. Since the force depends on the product $q(Q-q)$, making the two parts equal maximizes that product.

So, the next time you need to maximize an interaction between two parts of a whole, remember this problem. Nature loves symmetry, and the math proves it!