Question

Two capacitors, with capacitances \(C_{1}\) and \(C_{2},\) are connected in series. A potential difference, \(V_{0}\), is applied across the combination of capacitors. Find the potential differences \(V_{1}\) and \(V_{2}\) across the individual capacitors, in terms of \(V_{0}\), \(C_{1},\) and \(C_{2}\).

Short answer

Question: Given two capacitors with capacitances \(C_1\) and \(C_2\) connected in series and a total potential difference \(V_0\), express the potential differences \(V_1\) and \(V_2\) across the individual capacitors in terms of \(V_0\), \(C_1\), and \(C_2\). Answer: The potential differences across the individual capacitors can be expressed as: \(V_1 = \frac{C_2}{C_1 + C_2} \cdot V_0\) \(V_2 = \frac{C_1}{C_1 + C_2} \cdot V_0\)

Step-by-step solution

Find the equivalent capacitance of capacitors in series

To find the potential difference across each capacitor, we first need to determine the equivalent capacitance (\(C_{eq}\)) of two capacitors connected in series. The formula for the equivalent capacitance of capacitors in series is: $$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}.$$

Calculate the total charge stored in the combination of capacitors

We can find total charge stored in the combination of capacitors (\(Q_{total}\)) using the equivalent capacitance and the given potential difference \(V_0\): $$Q_{total} = C_{eq} \cdot V_0.$$

Determine charge on each capacitor

Since the capacitors are connected in series, the charge stored on each capacitor is the same as the total charge stored: $$Q_1 = Q_2 = Q_{total}.$$

Find the potential difference across each capacitor

Now, we can find the potential difference across each capacitor using their charge and capacitance: $$V_1 = \frac{Q_1}{C_1} = \frac{Q_{total}}{C_1}$$ and $$V_2 = \frac{Q_2}{C_2} = \frac{Q_{total}}{C_2}.$$

Express the potential differences in terms of \(V_0\), \(C_1\), and \(C_2\)

To express the potential differences in terms of \(V_0\), \(C_1\), and \(C_2\), we can substitute the expression for \(Q_{total}\) from step 2: $$V_1 = \frac{Q_{total}}{C_1} = \frac{C_{eq} \cdot V_0}{C_1}$$ and $$V_2 = \frac{Q_{total}}{C_2} = \frac{C_{eq} \cdot V_0}{C_2}.$$ Finally, substitute the expression for \(C_{eq}\) from step 1: $$V_1 = \frac{\frac{1}{\frac{1}{C_1} + \frac{1}{C_2}} \cdot V_0}{C_1} = \frac{C_2}{C_1 + C_2} \cdot V_0$$ and $$V_2 = \frac{\frac{1}{\frac{1}{C_1} + \frac{1}{C_2}} \cdot V_0}{C_2} = \frac{C_1}{C_1 + C_2} \cdot V_0.$$

Key concepts

Equivalent Capacitance

When combining capacitors in our circuits, we need to understand how they collectively affect the storage of electric charge. This combined effect is captured by a concept called 'equivalent capacitance.' Suppose we connect two capacitors, with capacities labeled as \(C_{1}\) and \(C_{2}\), in a series in our circuit. To determine their equivalent capacitance, \(C_{eq}\), we use a special formula:
\[\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}\].
What this formula tells us is that the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances of the capacitors connected in series. One can notice that the equivalent capacitance is always less than the smallest individual capacitor’s capacitance in the series. Understanding how to calculate the equivalent capacitance is vital in predicting how the system will behave when a voltage is applied.

Potential Difference

The potential difference across a component in a circuit, such as a capacitor, is indicative of the energy difference experienced by charge as it moves across that component. When we have two capacitors in series, like \(C_{1}\) and \(C_{2}\), both experiencing a total potential difference of \(V_{0}\), they share this voltage in proportion to their capacitances.
The formula to find the voltage across each capacitor is simply:
\[V = \frac{Q}{C}\].
Solving for the potential difference across individual capacitors when in series requires us to first understand the charge distribution and then apply this understanding to the formula. Since they share the charge, each capacitor will have a different voltage drop depending on its capacity.

Charge Conservation

A fundamental principle in circuits is the conservation of charge. When dealing with series circuits, this principle takes on a special implication: all components of the series circuit share the same charge. For our two capacitors in series, \(C_{1}\) and \(C_{2}\), regardless of their individual capacitances, the total charge \(Q\) stored must be the same on both.
So when we calculate the total charge using the formula:
\[Q_{total} = C_{eq} \cdot V_0\],
we can assert that:\[Q_1 = Q_2 = Q_{total}\].
This sharing of charge is what ultimately leads us to understand the differing potential differences across each capacitor, as the same charge is spread over different capacities.

Capacitance Formula

To delve deeper into the calculation and implication of capacitance, let's consider the fundamental formula:\[C = \frac{Q}{V}\].
This tells us that the capacitance \(C\) of a capacitor is the amount of charge \(Q\) it can store per unit of potential difference \(V\) across it. This formula becomes essential when we have already determined the total charge on capacitors in series and want to find the voltage across each one.
Understanding this formula allows us to manipulate the known values in our problem—total charge and individual capacitances—to find out how much voltage each capacitor will drop:\[V_1 = \frac{Q_{total}}{C_1}\] and \[V_2 = \frac{Q_{total}}{C_2}\].
It’s a bridge connecting the quantity of electric charge and the electric potential, fundamental in the study of capacitors.