Part A = 2.0 μF

For the arrangement of three capacitors in the figure , what value of C_{1} will give a total equivalent capacitance of 2.5 μF?

Part A = 6.31 * 10^{-9} C

Part B = 3.79 * 10^{-8} J

Part B = 2700 V/m

A 12.0 V battery remains connected to a parallel plate capacitor with a plate area of 0.264 m^{2} 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?

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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Part A = 80.0 μC

Part B = 80.0 μC

Part B = 37.3 V

In the figure are shown three capacitors with capacitances C_{1} = 6.00 μF, C_{2} = 3.00 μF, C_{3} = 5.00 μF. The capacitor network is connected to an applied potential V_{ab}. After the charges on the capacitors have reached their final values, the charge Q_{2} on the second capacitor is 40.0 μC.

What is the charge Q_{1} on capacitor C_{1}?

What is the charge on capacitor C_{3}?

What is the applied voltage, V_{ab}?

Part A = U_{0} = Aε_{0}V^{2} / (2d)

Part B = U_{1} = 3Aε_{0}V^{2} / (2d)

Part C = U_{2} = Kε_{0}V^{2} / (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 U_{0} 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 U_{1} 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 U_{2} of the dielectric-filled capacitor. The capacitor remains connected to the battery. The dielectric constant is K.

Part A = U = Q^{2} / (2C)

Part B = U = (CV^{2}) / 2

Part C = U/U_{0} = 0.5

Part D = U/U_{0} = 2

Part E = U = (ε_{0}AV^{2}) / 2d

Part F = U = (ε_{0}AdE^{2}) / 2

Part G = u = (ε_{0}E^{2}) / 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 U_{0}. 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/U_{0}.

A parallel-plate capacitor is connected to a battery. The energy of the capacitor is U_{0}. 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/U_{0}.

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.