Question

An \(8.00-\mu F\) capacitor is fully charged by a \(240 .-V\) battery, which is then disconnected. Next, the capacitor is connected to an initially uncharged capacitor of capacitance \(C,\) and the potential difference across it is found to be \(80.0 \mathrm{~V}\) What is \(C ?\) How much energy ends up being stored in the second capacitor?

Short answer

Answer: The capacitance of the second capacitor is 16.00 μF, and the energy stored in it is 51.2 mJ.

Step-by-step solution

Determine the initial charge stored in the charged capacitor

Use the capacitor formula: \(Q = C \cdot V\), where \(Q\) represents charge, \(C\) is the capacitance (8.00 μF), and \(V\) is the potential difference (240 V). \(Q_\text{initial} = (8.00 \text{ μF}) \cdot (240 \text{ V})\) \(Q_\text{initial} = 1920 \text{ μC}\)

Determine the final charge stored in the charged capacitor

Again, use the capacitor formula, this time using the final potential difference (80.0 V): \(Q_{1_\text{final}} = (8.00 \text{ μF}) \cdot (80.0 \text{ V})\) \(Q_{1_\text{final}} = 640 \text{ μC}\)

Determine the charge stored in the second capacitor

By conservation of charge, the charge stored in the second capacitor will be the initial charge minus the final charge of the first capacitor. \(Q_{2_\text{final}} = Q_\text{initial} - Q_{1_\text{final}}\) \(Q_{2_\text{final}} = 1920 \text{ μC} - 640 \text{ μC}\) \(Q_{2_\text{final}} = 1280 \text{ μC}\)

Determine the capacitance of the second capacitor

Since both capacitors are connected in parallel, they share the same potential difference. Using the capacitor formula, we can find the capacitance of the second capacitor: \(C = \frac{Q_{2_\text{final}}}{V}\) \(C = \frac{1280 \text{ μC}}{80.0 \text{ V}}\) \(C = 16.00 \text{ μF}\)

Determine the energy stored in the second capacitor

To find the energy stored in the second capacitor, we will use the energy formula for a capacitor: \(E = \frac{1}{2} CV^2\). \(E_{2_\text{final}} = \frac{1}{2}(16.00 \text{ μF})(80.0 \text{ V})^2\) \(E_{2_\text{final}} = \frac{1}{2}(16.00 \text{ μF})(6400 \text{ V}^2)\) \(E_{2_\text{final}} = 51200 \text{ μJ} = 51.2 \text{ mJ}\) Thus, the capacitance of the second capacitor is 16.00 μF, and the energy stored in it is 51.2 mJ.

Key concepts

Capacitance Calculation

The concept of capacitance is central to understanding how capacitors operate. A capacitor stores electrical charge and is characterized by its capacitance, denoted as \( C \). Capacitance is measured in farads (F) but is often seen in smaller units like microfarads (μF) because one farad represents a large capacity.

To calculate the charge stored in a capacitor, we use the formula:

• \( Q = C \cdot V \)

where \( Q \) is the charge in coulombs (C), \( C \) is the capacitance in farads, and \( V \) is the voltage in volts (V).

In the given exercise, we start with an \(8.00 \, \mu \text{F}\) capacitor charged to \(240 \, V\), resulting in an initial charge of \(1920 \, \mu \text{C}\). When this charged capacitor is connected to another uncharged capacitor, charge is redistributed until the potential difference across both capacitors is equalized. By applying the conservation of charge, we find the unknown capacitance \( C \) of the second capacitor to be \(16.00 \, \mu \text{F}\). This outcome is reached using the relation between the final charge and the shared potential.

Energy Storage in Capacitors

Capacitors store energy in the electric field created between their plates when they are charged. The energy \( E \) stored in a capacitor is given by the formula:

• \( E = \frac{1}{2} C V^2 \)

This formula reveals that the energy stored depends on both the capacitance \( C \) and the square of the voltage \( V \).

After charges redistribute between the initial and second capacitor, each will have a known voltage and capacitance. Here, the final voltage is \(80.0 \, V\) and the second capacitor has a capacitance of \(16.00 \, \mu \text{F}\), so the energy stored is calculated as:
\[ E_{2_{\text{final}}} = \frac{1}{2} \cdot (16.00 \, \mu \text{F}) \cdot (80.0 \, V)^2 = 51.2 \, \text{mJ} \]
This equation shows how the stored energy scales with the voltage, explaining that small changes in voltage can significantly affect the energy held.

Charge Conservation

The principle of charge conservation is a key factor when dealing with capacitors in connected systems. Charge conservation tells us that the total electric charge in an isolated system remains constant over time. This principle helps us determine how charge is distributed among components in a circuit.

In our example, initially, the first capacitor is fully charged. When connected with the uncharged capacitor, the charge spreads across both capacitors until equilibrium (same potential difference) is achieved.

By applying the charge conservation formula:

• \( Q_{\text{total initial}} = Q_{1_{\text{final}}} + Q_{2_{\text{final}}} \)

we recognize that the initial charge must equal the sum of the final charges in both capacitors. This insight allows us to find the charge on each capacitor using their respective capacitances and final shared voltage. This exercise beautifully demonstrates how charge conservation aids in solving complex circuit problems.