Problem 40 - Physics

Question

A capacitor stores \(5.3 \times 10^{-5} \mathrm{C}\) of charge when connected to a 6.0 -V battery. How much charge does the capacitor store when connected to a \(9.0-\mathrm{V}\) battery?

Step-by-Step Solution

Answer

The capacitor stores \( 7.95 \times 10^{-5} \) C when connected to a 9.0 V battery.

Step 1: Identify Key Variables

Let's identify the given values: the initial voltage, \( V_1 \), is \( 6.0 \) V, and the initial charge, \( Q_1 \), is \( 5.3 \times 10^{-5} \) C. The final voltage, \( V_2 \), is \( 9.0 \) V.

Step 2: Understand the Charge-Voltage Relationship

The charge \( Q \) stored in a capacitor is directly proportional to the voltage \( V \) across it. This means \( \frac{Q_1}{V_1} = \frac{Q_2}{V_2} \), where \( Q_2 \) is the final charge at voltage \( V_2 \).

Step 3: Set Up the Proportion

Based on the proportional relationship, we can write the equation: \[ \frac{5.3 \times 10^{-5}}{6.0} = \frac{Q_2}{9.0} \]

Step 4: Solve for Final Charge, Q2

Solve for \( Q_2 \) by cross-multiplying: \[ Q_2 = \frac{5.3 \times 10^{-5} \times 9.0}{6.0} \] Perform the multiplication first: \( 5.3 \times 9.0 = 47.7 \times 10^{-5} \). Now divide: \[ Q_2 = \frac{47.7 \times 10^{-5}}{6.0} = 7.95 \times 10^{-5} \] C.

Key Concepts

Charge-Voltage Relationship

When studying capacitors, one of the fundamental concepts is the relationship between charge and voltage. Capacitors are devices that store electric charge, and their ability to do so is influenced by the voltage applied across their plates. The relationship can be expressed by the equation \( Q = CV \), where \( Q \) is the charge stored, \( C \) is the capacitance (a measure of the capacitor's ability to store charge), and \( V \) is the voltage applied.

• The equation shows that the amount of charge stored \( Q \) is directly proportional to the applied voltage \( V \).
• As the voltage increases, so does the charge stored.

This relationship is crucial for understanding how changes in voltage affect the amount of charge a capacitor can store. For a given capacitor with constant capacitance, doubling the applied voltage would double the stored charge according to this linear relationship.

Proportionality in Capacitors

The concept of proportionality in capacitors is essential for solving problems involving changes in voltage and charge. Given the charge-voltage relationship \( Q = CV \), we directly observe proportionality between \( Q \) and \( V \), assuming the capacitance \( C \) remains constant.

• The formula \( \frac{Q_1}{V_1} = \frac{Q_2}{V_2} \) is used to maintain this proportionality when the voltage changes.
• This equation allows us to calculate unknown charges based on changes in voltage by keeping the ratio constant.

For example, if a capacitor stores \(5.3 \times 10^{-5} \) C at \(6.0\) V, the proportionality equation helps us find the new charge when the voltage is increased to \(9.0\) V by setting up the proportion \( \frac{5.3 \times 10^{-5}}{6.0} = \frac{Q_2}{9.0} \). By solving this, we determine that the new charge is \(7.95 \times 10^{-5} \) C.

Electric Charge Storage

Capacitors are designed to store electric charge, which is crucial for many electronic applications. When a capacitor is connected to a power source, it accumulates charge until it reaches its storage capacity determined by its capacitance and the voltage applied.

• The stored charge is proportional to the electric field across its plates, which induces separation of charge and creates potential difference.
• Charge storage in capacitors helps in various applications such as filtering signals, tuning radios, and powering electronic devices briefly during power off situations.
• Capacitors can also release stored energy quickly when needed, making them essential components in circuits requiring quick energy discharge.

In essence, understanding how capacitors store charge, how this is influenced by the voltage, and the significance of proportionality helps in designing and using capacitors effectively across different electronic applications.