Capacitors \(C_{1}\) and \(C_{2}\) are in series, with voltage \(V\) across the combination. Show that the voltages across the individual capacitors are \(V_{1}=C_{2} V /\left(C_{1}+C_{2}\right)\) and \(V_{2}=C_{1} V /\left(C_{1}+C_{2}\right)\).

When capacitors are connected in series, they are lined up in a single path for the electric current to flow. Here’s what happens in a series circuit:

Each capacitor in a series connection will store the same amount of electrical charge ('Q'). This is because there is only one path for the current, and all of the charge passing through the first capacitor has to pass through the subsequent ones. Think of it like a single-file line of people passing a ball; the ball represents the charge, and it's the same ball (same charge) being passed along.

The total capacitance of the series (\(C_{total}\) can be calculated using the formula \(1/C_{total} = 1/C_{1} + 1/C_{2} + ... \), where \(C_{1}\), \(C_{2}\), etc., are the individual capacitances. The result is always less than the smallest capacitance in the series because the electric field lines have to squeeze through the smaller space available, making it harder for them to pass through.

The voltage distribution in series capacitors is quite different from the distribution of charge. Since the same charge 'Q' is stored on all capacitors in series, the voltage across each capacitor depends on its individual capacitance. The capacity to store charge is like a container's ability to hold water; some containers (capacitors) are bigger and can store more. A larger capacitance means that the container can hold more charge for a given voltage.

Now, if you imagine the total voltage 'V' applied across a series of capacitors as water pressure, it becomes clear that capacitors with smaller capacitance will have a higher voltage drop across them, while those with larger capacitance will have a lower voltage drop. That's why, if you use the formula derived in the problem, \(V_{1} = C_{2} V /(C_{1}+C_{2})\) and \(V_{2} = C_{1} V /(C_{1}+C_{2})\), the capacitor with the smaller capacitance gets a larger share of the total voltage.

Let's explore the concept of charge on capacitors. A capacitor's charge capacity is directly related to its capacitance 'C' and the voltage 'V' applied to it. The fundamental relationship is given by \(Q = CV\), where 'Q' is the charge stored in the capacitor.

In a series circuit, as shown in the exercise, the same charge is present on all capacitors. This is a crucial point because it underlines the principle of charge conservation within an isolated system. No matter the capacitance of each capacitor in the series, the charge remains consistent across them, much like the quantity of water remaining the same regardless of how many times you pour it from one glass to another of different sizes. This constraint of equal charge is foundational in determining the voltage drop across each capacitor in series arrangements.