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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

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.

Two capacitors with capacitances \(C_{1}\) and \(C_{2}\) are connected in series. Show that, no matter what the values of \(C_{1}\) and \(C_{2}\) are, the equivalent capacitance is always less than the smaller of the two capacitances.

The space between the plates of an isolated parallel plate capacitor is filled with a slab of dielectric material. The magnitude of the charge \(Q\) on each plate is kept constant. If the dielectric material is removed from between the plates, the energy stored in the capacitor a) increases. c) decreases. b) stays the same. d) may increase or decrease.

Does it take more work to separate the plates of a charged parallel plate capacitor while it remains connected to the charging battery or after it has been disconnected from the charging battery?

A dielectric slab with thickness \(d\) and dielectric constant \(\kappa=2.31\) is inserted in a parallel place capacitor that has been charged by a \(110 .-\mathrm{V}\) battery and having area \(A=\) \(100, \mathrm{~cm}^{2},\) and separation distance \(d=2.50 \mathrm{~cm}\). a) Find the capacitance, \(C,\) the potential difference, \(V,\) the electric field, \(E,\) the total charge stored on the capacitor \(Q\), and electric potential energy stored in the capacitor, \(U\), before the dielectric material is inserted. b) Find \(C, V, E, Q,\) and \(U\) when the dielectric slab has been inserted and the battery is still connected. c) Find \(C, V, E, Q\) and \(U\) when the dielectric slab is in place and the battery is disconnected.
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