For the circuit shown in the figure below, the switch S is initially open and the voltage across the capacitor is 60 V. The switch is then closed at time t = 0 s. What is the charge on the capacitor when the current in the circuit is 20 µA? 22 µF 1.5 MQ Ο 140 μC Ο 290 μC Ο 310 μ Ο 970 μC Ο 660 μ ww

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## Useful Constants

- Coulomb's constant (\( k \)):  
  \[
  k = 9.00 \times 10^9 \, \frac{N \cdot m^2}{C^2}
  \]

- Permittivity of free space (\( \varepsilon_0 \)):  
  \[
  \varepsilon_0 = 8.85 \times 10^{-12} \, \frac{C^2}{N \cdot m^2}
  \]

- Elementary charge (\( e \)):  
  \[
  e = 1.6 \times 10^{-19} \, C
  \]

- Electron mass (\( m_e \)):  
  \[
  m_e = 9.11 \times 10^{-31} \, kg
  \]

- Proton mass (\( m_p \)):  
  \[
  m_p = 1.67 \times 10^{-27} \, kg
  \]

- Neutron mass (\( m_n \)):  
  \[
  m_n = 1.68 \times 10^{-27} \, kg
  \]

These constants are fundamental in physics, particularly in the fields of electromagnetism and particle physics.
Transcribed Image Text:## Useful Constants - Coulomb's constant (\( k \)): \[ k = 9.00 \times 10^9 \, \frac{N \cdot m^2}{C^2} \] - Permittivity of free space (\( \varepsilon_0 \)): \[ \varepsilon_0 = 8.85 \times 10^{-12} \, \frac{C^2}{N \cdot m^2} \] - Elementary charge (\( e \)): \[ e = 1.6 \times 10^{-19} \, C \] - Electron mass (\( m_e \)): \[ m_e = 9.11 \times 10^{-31} \, kg \] - Proton mass (\( m_p \)): \[ m_p = 1.67 \times 10^{-27} \, kg \] - Neutron mass (\( m_n \)): \[ m_n = 1.68 \times 10^{-27} \, kg \] These constants are fundamental in physics, particularly in the fields of electromagnetism and particle physics.
**Circuit Analysis Problem**

For the circuit shown in the diagram, the switch \( S \) is initially open and the voltage across the capacitor is \( 60 \, \text{V} \). The switch is then closed at time \( t = 0 \, \text{s} \). What is the charge on the capacitor when the current in the circuit is \( 20 \, \mu \text{A} \)?

**Circuit Description:**

- Capacitor: \( 22 \, \mu \text{F} \)
- Resistor: \( 1.5 \, \text{M}\Omega \)

**Multiple Choice Answers:**

- \( 140 \, \mu \text{C} \)
- \( 290 \, \mu \text{C} \)
- \( 310 \, \mu \text{C} \)
- \( 970 \, \mu \text{C} \)
- \( 660 \, \mu \text{C} \)

**Solution Explanation:**

To solve for the charge on the capacitor when the current is \( 20 \, \mu \text{A} \), we use the formula for the charging process of an RC circuit:

1. Implement the time constant \(\tau = R \times C\)

2. Use the relationship for charge \( Q \) at a given time with current \( I \):

   \[
   Q = C \times V
   \]

   And since the current \( I = \frac{dQ}{dt} \), integrate as needed.

**Note:** This theoretical approach involves understanding the exponential nature of capacitor charging and the application of Ohm's Law (\( V = I \times R \)) in transient analysis.
Transcribed Image Text:**Circuit Analysis Problem** For the circuit shown in the diagram, the switch \( S \) is initially open and the voltage across the capacitor is \( 60 \, \text{V} \). The switch is then closed at time \( t = 0 \, \text{s} \). What is the charge on the capacitor when the current in the circuit is \( 20 \, \mu \text{A} \)? **Circuit Description:** - Capacitor: \( 22 \, \mu \text{F} \) - Resistor: \( 1.5 \, \text{M}\Omega \) **Multiple Choice Answers:** - \( 140 \, \mu \text{C} \) - \( 290 \, \mu \text{C} \) - \( 310 \, \mu \text{C} \) - \( 970 \, \mu \text{C} \) - \( 660 \, \mu \text{C} \) **Solution Explanation:** To solve for the charge on the capacitor when the current is \( 20 \, \mu \text{A} \), we use the formula for the charging process of an RC circuit: 1. Implement the time constant \(\tau = R \times C\) 2. Use the relationship for charge \( Q \) at a given time with current \( I \): \[ Q = C \times V \] And since the current \( I = \frac{dQ}{dt} \), integrate as needed. **Note:** This theoretical approach involves understanding the exponential nature of capacitor charging and the application of Ohm's Law (\( V = I \times R \)) in transient analysis.
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