Animated Solution for Physics - Electrostatics: In the given circuit, charge $Q_2$ on the $2\mu \text{F}$ capacitor changes as $C$ is varied from $1\mu \text{F}$ to $3\mu \text{F}$. $Q_2$ as a function of $C$ is given properly by (figures are drawn schematically and are not to scale)

Let's embark on a journey to decode the charge dynamics of this intriguing circuit. Imagine the circuit as a network of water pipes, where the battery is the pump and the capacitors are storage tanks. We have a battery providing an electromotive force (EMF) $E$. This battery is connected in series with a variable capacitor $C$.

Following this variable capacitor, the circuit branches out into a parallel combination of two fixed capacitors: one with a capacitance of $1\mu \text{F}$ and another with $2\mu \text{F}$. Our primary goal is to understand how the charge $Q_2$ on the $2\mu \text{F}$ capacitor behaves as we tweak the value of the variable capacitor $C$ from $1\mu \text{F}$ to $3\mu \text{F}$.

To find the charge on any specific component, we first need to understand the circuit as a whole. We start by simplifying the parallel branch. For capacitors connected in parallel, their equivalent capacitance is simply the sum of their individual capacitances.

Therefore, the equivalent capacitance of the parallel section, let's call it $C_p$, is:

Now, we can visualize the circuit as just two capacitors in series: the variable capacitor $C$ and our new equivalent capacitor $C_p=3\mu \text{F}$. The total equivalent capacitance $C_{eq}$ of two capacitors in series is given by the product over their sum:

With the total equivalent capacitance in hand, we can determine the total charge $Q$ drawn from the battery. Using the fundamental relation $Q=CV$, we find:

This total charge $Q$ flows entirely through the first capacitor $C$. However, when it reaches the parallel junction, it must split. In a parallel combination, the voltage across each branch is identical, which means the charge distributes itself in direct proportion to the capacitance of each branch.

The $2\mu \text{F}$ capacitor will take a fraction of the total charge equal to its capacitance divided by the total capacitance of the parallel branch. Thus, the charge $Q_2$ is:

Substituting our expression for the total charge $Q$ into this equation yields:

We now have a beautiful mathematical function for $Q_2$ in terms of $C$. To understand how this function behaves, let's rewrite it by dividing both the numerator and the denominator by $C$:

Let's analyze this form. As the variable capacitance $C$ increases, the term $\frac{3}{C}$ strictly decreases. Consequently, the entire denominator $\left(1+\frac{3}{C}\right)$ decreases. Dividing a constant numerator ($2E$) by a decreasing denominator results in an increasing overall value. Therefore, we can confidently say that as $C$ increases, $Q_2$ must also increase.

Knowing the function is increasing isn't enough to pick the right graph; we need to know how it increases. Does it shoot up exponentially, or does it level off? To find out, we turn to calculus and examine the slope of the function by taking its derivative with respect to $C$.

Using the quotient rule on our original expression $Q_2=\frac{2CE}{C+3}$:

Since $E$ is positive and the denominator is squared, the derivative $\frac{d{Q}_2}{dC}$ is always positive, confirming our earlier conclusion that the function is strictly increasing.

However, the critical insight lies in the denominator. As $C$ increases, the term $(C+3{)}^2$ grows larger. This means the overall value of the derivative—the slope of the curve—is continuously decreasing.

A curve that is increasing but has a decreasing slope is geometrically described as concave downwards. It rises, but at a progressively slower rate, eventually approaching a horizontal asymptote (which, if we take the limit as $C\to \infty$, is $2E$). Looking at the given options, the graph that perfectly depicts an increasing, concave-down curve is option (b).