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Chapter 24: Problem 74

An \(8.00-\mu F\) capacitor is fully charged by a \(240 .-V\) battery, which is then disconnected. Next, the capacitor is connected to an initially uncharged capacitor of capacitance \(C,\) and the potential difference across it is found to be \(80.0 \mathrm{~V}\) What is \(C ?\) How much energy ends up being stored in the second capacitor?

Short Answer

Expert verified
Answer: The capacitance of the second capacitor is 16.00 μF, and the energy stored in it is 51.2 mJ.

Step by step solution

01

Determine the initial charge stored in the charged capacitor

Use the capacitor formula: \(Q = C \cdot V\), where \(Q\) represents charge, \(C\) is the capacitance (8.00 μF), and \(V\) is the potential difference (240 V). \(Q_\text{initial} = (8.00 \text{ μF}) \cdot (240 \text{ V})\) \(Q_\text{initial} = 1920 \text{ μC}\)
02

Determine the final charge stored in the charged capacitor

Again, use the capacitor formula, this time using the final potential difference (80.0 V): \(Q_{1_\text{final}} = (8.00 \text{ μF}) \cdot (80.0 \text{ V})\) \(Q_{1_\text{final}} = 640 \text{ μC}\)
03

Determine the charge stored in the second capacitor

By conservation of charge, the charge stored in the second capacitor will be the initial charge minus the final charge of the first capacitor. \(Q_{2_\text{final}} = Q_\text{initial} - Q_{1_\text{final}}\) \(Q_{2_\text{final}} = 1920 \text{ μC} - 640 \text{ μC}\) \(Q_{2_\text{final}} = 1280 \text{ μC}\)
04

Determine the capacitance of the second capacitor

Since both capacitors are connected in parallel, they share the same potential difference. Using the capacitor formula, we can find the capacitance of the second capacitor: \(C = \frac{Q_{2_\text{final}}}{V}\) \(C = \frac{1280 \text{ μC}}{80.0 \text{ V}}\) \(C = 16.00 \text{ μF}\)
05

Determine the energy stored in the second capacitor

To find the energy stored in the second capacitor, we will use the energy formula for a capacitor: \(E = \frac{1}{2} CV^2\). \(E_{2_\text{final}} = \frac{1}{2}(16.00 \text{ μF})(80.0 \text{ V})^2\) \(E_{2_\text{final}} = \frac{1}{2}(16.00 \text{ μF})(6400 \text{ V}^2)\) \(E_{2_\text{final}} = 51200 \text{ μJ} = 51.2 \text{ mJ}\) Thus, the capacitance of the second capacitor is 16.00 μF, and the energy stored in it is 51.2 mJ.

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Most popular questions from this chapter

A parallel plate capacitor has a capacitance of \(120 .\) pF and a plate area of \(100 . \mathrm{cm}^{2}\). The space between the plates is filled with mica whose dielectric constant is \(5.40 .\) The plates of the capacitor are kept at \(50.0 \mathrm{~V}\) a) What is the strength of the electric field in the mica? b) What is the amount of free charge on the plates? c) What is the amount of charge induced on the mica?

The potential difference across two capacitors in series is \(120 . \mathrm{V}\). The capacitances are \(C_{1}=1.00 \cdot 10^{3} \mu \mathrm{F}\) and \(C_{2}=1.50 \cdot 10^{3} \mu \mathrm{F}\) a) What is the total capacitance of this pair of capacitors? b) What is the charge on each capacitor? c) What is the potential difference across each capacitor? d) What is the total energy stored by the capacitors?

The distance between the plates of a parallel plate capacitor is reduced by half and the area of the plates is doubled. What happens to the capacitance? a) It remains unchanged. b) It doubles. c) It quadruples. d) It is reduced by half.

Consider a cylindrical capacitor, with outer radius \(R\) and cylinder separation \(d\). Determine what the capacitance approaches in the limit where \(d \ll R\). (Hint: Express the capacitance in terms of the ratio \(d / R\) and then examine what happens as the ratio \(d / R\) becomes very small compared to \(1 .)\) Explain why the limit on the capacitance makes sense.

A 4.00 -pF parallel plate capacitor has a potential difference of \(10.0 \mathrm{~V}\) across it. The plates are \(3.00 \mathrm{~mm}\) apart, and the space between them contains air. a) What is the charge on the capacitor? b) How much energy is stored in the capacitor? c) What is the area of the plates? d) What would the capacitance of this capacitor be if the space between the plates were filled with polystyrene?
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