Mastering Physics Solutions: Exercise 16.52

Mastering Physics Solutions: Exercise 16.52

On February 18, 2012, in Chapter 16: Electric Potential, Energy, and Capacitance, by Mastering Physics Solutions

Part A = 2.0 μF

For the arrangement of three capacitors in the figure , what value of C1 will give a total equivalent capacitance of 2.5 μF?

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Mastering Physics Solutions: Exercise 16.39

Mastering Physics Solutions: Exercise 16.39

On February 15, 2012, in Chapter 16: Electric Potential, Energy, and Capacitance, by Mastering Physics Solutions

Part A = 6.31 * 10-9 C Click to use the calculator/solver for this part of the problem
Part B = 3.79 * 10-8 J Click to use the calculator/solver for this part of the problem
Part B = 2700 V/m Click to use the calculator/solver for this part of the problem

A 12.0 V battery remains connected to a parallel plate capacitor with a plate area of 0.264 m2 and a plate separation of 4.44 mm.
What is the charge on the capacitor?
How much energy is stored in the capacitor?
What is the electric field between its plates?

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Mastering Physics Solutions: Exercise 16.10

Mastering Physics Solutions: Exercise 16.10

On February 15, 2012, in Chapter 16: Electric Potential, Energy, and Capacitance, by Mastering Physics Solutions

Part A = -2.3 J Click to use the calculator/solver for this part of the problem

How much work is required to completely separate two charges (each -1.5 μC ) and leave them at rest if they were initially 9.0 mm apart?

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Mastering Physics Solutions: A Simple Network of Capacitors

Mastering Physics Solutions: A Simple Network of Capacitors

On February 14, 2012, in Chapter 16: Electric Potential, Energy, and Capacitance, by Mastering Physics Solutions

Part A = 80.0 μC Click to use the calculator/solver for this part of the problem
Part B = 80.0 μC Click to use the calculator/solver for this part of the problem
Part B = 37.3 V Click to use the calculator/solver for this part of the problem

In the figure are shown three capacitors with capacitances C1 = 6.00 μF, C2 = 3.00 μF, C3 = 5.00 μF. The capacitor network is connected to an applied potential Vab. After the charges on the capacitors have reached their final values, the charge Q2 on the second capacitor is 40.0 μC.
What is the charge Q1 on capacitor C1?
What is the charge on capacitor C3?
What is the applied voltage, Vab?

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Mastering Physics Solutions: The Capacitor as an Energy-Storing Device

Mastering Physics Solutions: The Capacitor as an Energy-Storing Device

On February 12, 2012, in Chapter 16: Electric Potential, Energy, and Capacitance, by Mastering Physics Solutions

Part A = U0 = Aε0V2 / (2d)
Part B = U1 = 3Aε0V2 / (2d)
Part C = U2 = Kε0V2 / (2d)

An air-filled parallel-plate capacitor has plate area A and plate separation d. The capacitor is connected to a battery that creates a constant voltage V.
Find the energy U0 stored in the capacitor.
The capacitor is now disconnected from the battery, and the plates of the capacitor are then slowly pulled apart until the separation reaches 3d. Find the new energy U1 of the capacitor after this process.
The capacitor is now reconnected to the battery, and the plate separation is restored to d. A dielectric plate is slowly moved into the capacitor until the entire space between the plates is filled. Find the energy U2 of the dielectric-filled capacitor. The capacitor remains connected to the battery. The dielectric constant is K.

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Mastering Physics Solutions: Energy in Capacitors and Electric Fields

Mastering Physics Solutions: Energy in Capacitors and Electric Fields

On February 12, 2012, in Chapter 16: Electric Potential, Energy, and Capacitance, by Mastering Physics Solutions

Part A = U = Q2 / (2C)
Part B = U = (CV2) / 2
Part C = U/U0 = 0.5
Part D = U/U0 = 2
Part E = U = (ε0AV2) / 2d
Part F = U = (ε0AdE2) / 2
Part G = u = (ε0E2) / 2

Find the energy U of the capacitor in terms of C and Q by using the definition of capacitance and the formula for the energy in a capacitor.
Find the energy U of the capacitor in terms of C and V by using the definition of capacitance and the formula for the energy in a capacitor.
A parallel-plate capacitor is connected to a battery. The energy of the capacitor is U0. The capacitor remains connected to the battery while the plates are slowly pulled apart until the plate separation doubles. The new energy of the capacitor is U. Find the ratio U/U0.
A parallel-plate capacitor is connected to a battery. The energy of the capacitor is U0. The capacitor is then disconnected from the battery and the plates are slowly pulled apart until the plate separation doubles. The new energy of the capacitor is U. Find the ratio U/U0.
A parallel-plate capacitor has area A and plate separation d, and it is charged to voltage V. Use the formulas from the problem introduction to obtain the formula for the energy U of the capacitor.
A parallel-plate capacitor has area A and plate separation d, and it is charged so that the electric field inside is E. Use the formulas from the problem introduction to find the energy U of the capacitor.
Find the energy density u of the electric field in a parallel-plate capacitor. The magnitude of the electric field inside the capacitor is E.

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