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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.](https://content.bartleby.com/qna-images/question/af1ef7c6-a145-4dbc-a50b-84dc7d2e578d/98e787d2-5510-4a92-a6bf-1751f881c672/8aixm9h_processed.jpeg)
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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