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Answer :
The capacitance [tex]\( C_3 \)[/tex] must be [tex]\( 8.00 \, \text{pF} \) for the network to store \( 2.70 \times 10^{-3}[/tex] J of electrical energy.
To find the capacitance [tex]\( C_3 \) that allows the network to store \( 2.70 \times 10^{-3} \)[/tex] J of electrical energy, we can use the formula for the energy stored in a capacitor:
[tex]\[E = \frac{1}{2} C V^2\][/tex]
Where:
E is the electrical energy stored,
C is the capacitance, and
V is the voltage across the capacitor.
Given that [tex]\( C_1 = C_2 = 4.00 \, \text{pF} \), \( C_4 = 8.00 \, \text{pF} \), and \( E = 2.70 \times 10^{-3} \, \text{J} \)[/tex], we'll first need to determine the equivalent capacitance of the circuit, and then we'll calculate the voltage across the circuit.
1. Equivalent Capacitance [tex](\( C_{\text{eq}} \))[/tex]
The capacitors [tex]\( C_1 \) and \( C_2 \)[/tex] are in parallel, so their equivalent capacitance is the sum of their individual capacitances:
[tex]\[C_{\text{eq}} = C_1 + C_2 = 4.00 \, \text{pF} + 4.00 \, \text{pF} = 8.00 \, \text{pF}\][/tex]
The capacitors [tex]\( C_{\text{eq}} \) and \( C_3 \) are in series, and this combination is in parallel with \( C_4 \)[/tex]. The formula for calculating the equivalent capacitance of two capacitors in series is:
[tex]\[\frac{1}{C_{\text{series}}} = \frac{1}{C_{\text{eq}}} + \frac{1}{C_3}\][/tex]
Given that [tex]\( C_{\text{eq}} = 8.00 \, \text{pF} \) and \( C_4 = 8.00 \, \text{pF} \)[/tex], the equivalent capacitance of the series combination[tex]\( C_{\text{series}} \) will be \( 4.00 \, \text{pF} \).[/tex] Thus:
[tex]\[\frac{1}{4.00 \, \text{pF}} = \frac{1}{8.00 \, \text{pF}} + \frac{1}{C_3}\]\[\frac{1}{C_3} = \frac{1}{4.00 \, \text{pF}} - \frac{1}{8.00 \, \text{pF}}\]\[\frac{1}{C_3} = \frac{1}{4.00 \, \text{pF}} - \frac{1}{8.00 \, \text{pF}} = \frac{1}{8.00 \, \text{pF}}\]\[\frac{1}{C_3} = \frac{1}{8.00 \, \text{pF}} = 0.125 \, \text{pF}^{-1}\]\[C_3 = \frac{1}{0.125 \, \text{pF}^{-1}} = 8.00 \, \text{pF}\][/tex]
So, the capacitance [tex]\( C_3 \)[/tex] must be [tex]\( 8.00 \, \text{pF} \) for the network to store \( 2.70 \times 10^{-3}[/tex]J of electrical energy.
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