plates of a capacitor are 4.0 mm apart i. C/m?. Calculate the energy density in betwe 5. A capacitor circuit is shown in figure. Calculate the equivalent capacitance of the circuit. b. Calculate the charge on capacitor C1. a. C2= 2 mF 5.0 V C4= 3 mF ic field C1= 1mF C3=2 mF

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**Educational Content on Capacitor Circuits**

**Question 5:**

- The parallel plates of a capacitor are **4.0 mm** apart in a medium with a permittivity of **\(5.6 \times 10^{-11} F/m\)**. Calculate the energy density between the plates.

**Question 6:**

A capacitor circuit is shown in the figure.

a. Calculate the equivalent capacitance of the circuit.

b. Calculate the charge on capacitor \(C_1\).

**Diagram Explanation:**

The diagram shows a circuit with a voltage source of **5.0 V** connected to a combination of capacitors:

- \(C_1 = 1 \, \text{mF}\)
- \(C_2 = 2 \, \text{mF}\)
- \(C_3 = 2 \, \text{mF}\)
- \(C_4 = 3 \, \text{mF}\)

Capacitors \(C_2\) and \(C_3\) are connected in parallel. The combination of \(C_2\) and \(C_3\) is in series with \(C_4\). This entire arrangement is connected in parallel with \(C_1\).

To solve this:

1. Calculate the equivalent capacitance of \(C_2\) and \(C_3\) which are in parallel:
   \[
   C_{23} = C_2 + C_3
   \]

2. Calculate the equivalent capacitance of \(C_{23}\) in series with \(C_4\):
   \[
   C_{234} = \left( \frac{1}{C_{23}} + \frac{1}{C_4} \right)^{-1}
   \]

3. Finally, add \(C_{234}\) in parallel with \(C_1\) to find the total equivalent capacitance \(C_{\text{eq}}\):
   \[
   C_{\text{eq}} = C_{234} + C_1
   \]

For the charge on \(C_1\), use:
\[
Q_1 = C_1 \times \text{Voltage across } C_1
\] 

The voltage across \(C_1\) is the same as the source voltage since they are in parallel.
Transcribed Image Text:**Educational Content on Capacitor Circuits** **Question 5:** - The parallel plates of a capacitor are **4.0 mm** apart in a medium with a permittivity of **\(5.6 \times 10^{-11} F/m\)**. Calculate the energy density between the plates. **Question 6:** A capacitor circuit is shown in the figure. a. Calculate the equivalent capacitance of the circuit. b. Calculate the charge on capacitor \(C_1\). **Diagram Explanation:** The diagram shows a circuit with a voltage source of **5.0 V** connected to a combination of capacitors: - \(C_1 = 1 \, \text{mF}\) - \(C_2 = 2 \, \text{mF}\) - \(C_3 = 2 \, \text{mF}\) - \(C_4 = 3 \, \text{mF}\) Capacitors \(C_2\) and \(C_3\) are connected in parallel. The combination of \(C_2\) and \(C_3\) is in series with \(C_4\). This entire arrangement is connected in parallel with \(C_1\). To solve this: 1. Calculate the equivalent capacitance of \(C_2\) and \(C_3\) which are in parallel: \[ C_{23} = C_2 + C_3 \] 2. Calculate the equivalent capacitance of \(C_{23}\) in series with \(C_4\): \[ C_{234} = \left( \frac{1}{C_{23}} + \frac{1}{C_4} \right)^{-1} \] 3. Finally, add \(C_{234}\) in parallel with \(C_1\) to find the total equivalent capacitance \(C_{\text{eq}}\): \[ C_{\text{eq}} = C_{234} + C_1 \] For the charge on \(C_1\), use: \[ Q_1 = C_1 \times \text{Voltage across } C_1 \] The voltage across \(C_1\) is the same as the source voltage since they are in parallel.
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