The capacitor in the figure has a capacitance of 21 µF and is initially uncharged. The battery provides a potential difference of 160 V. After switch S is closed, how much charge will pass through it?

College Physics
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Author:Raymond A. Serway, Chris Vuille
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Chapter16: Electrical Energy And Capacitance
Section: Chapter Questions
Problem 42P: Consider the combination of capacitors in Figure P16.42. (a) Find the equivalent single capacitance...
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### Capacitors and Charge Calculation

**Problem Statement:**

The capacitor in the figure has a capacitance of 21 µF (microfarads) and is initially uncharged. The battery provides a potential difference of 160 V (volts). After the switch \( S \) is closed, how much charge will pass through it?

**Illustration Description:**

- The figure illustrates a simple electric circuit consisting of a battery, a switch (\( S \)), and a capacitor (\( C \)).
- The battery is connected in series with the switch and the uncharged capacitor.
- Once the switch (\( S \)) is closed, the circuit is completed, and the capacitor will start to charge.

**Calculation:**

When the switch is closed, the charge (\( Q \)) on the capacitor can be calculated using the formula:

\[ Q = C \times V \]

where:
- \( C \) is the capacitance of the capacitor (21 µF).
- \( V \) is the potential difference provided by the battery (160 V).

**Provide your answer:**

- Number: [    ] Units: [   ]

**Explanation:**

To find the charge, multiply the capacitance by the potential difference:

\[ Q = 21 \, \mu F \times 160 \, V \]

After performing the multiplication, you will obtain the value of the charge \( Q \) in microcoulombs (µC).

### Notes:

- **Capacitors** are electronic components that store electric charge.
- **Capacitance** (C) is a measure of a capacitor's ability to store charge, and it's expressed in farads (F).
- **Voltage** (V) is the potential difference that drives the charge through the capacitor.

Feel free to complete the calculation and input your answer in the provided fields.
Transcribed Image Text:### Capacitors and Charge Calculation **Problem Statement:** The capacitor in the figure has a capacitance of 21 µF (microfarads) and is initially uncharged. The battery provides a potential difference of 160 V (volts). After the switch \( S \) is closed, how much charge will pass through it? **Illustration Description:** - The figure illustrates a simple electric circuit consisting of a battery, a switch (\( S \)), and a capacitor (\( C \)). - The battery is connected in series with the switch and the uncharged capacitor. - Once the switch (\( S \)) is closed, the circuit is completed, and the capacitor will start to charge. **Calculation:** When the switch is closed, the charge (\( Q \)) on the capacitor can be calculated using the formula: \[ Q = C \times V \] where: - \( C \) is the capacitance of the capacitor (21 µF). - \( V \) is the potential difference provided by the battery (160 V). **Provide your answer:** - Number: [ ] Units: [ ] **Explanation:** To find the charge, multiply the capacitance by the potential difference: \[ Q = 21 \, \mu F \times 160 \, V \] After performing the multiplication, you will obtain the value of the charge \( Q \) in microcoulombs (µC). ### Notes: - **Capacitors** are electronic components that store electric charge. - **Capacitance** (C) is a measure of a capacitor's ability to store charge, and it's expressed in farads (F). - **Voltage** (V) is the potential difference that drives the charge through the capacitor. Feel free to complete the calculation and input your answer in the provided fields.
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