An air-filled parallel plate capacitor with each plate having an area of 50 ??2 and the separation of the two plates is 4.0 mm. The capacitor is connected to a 25.0 V battery. a. Calculate the capacitance of the capacitor. b. Calculate the charge stored in the capacitor. c. Calculate the electric field between the plates.

Four capacitors are connected as shown in Figure P25.11. (a) Find the equivalent capacitance between points a and b. (b) Calculate the charge on each capacitor, taking Vab = 15.0 V. Figure P25.11

(a) How much charge can be placed 011 a capacitor with air between the plates before it breaks down if the area of each plate is 5.00 cm2? (b) What If? Find the maximum charge if polystyrene is used between the plates instead of air.

a parallel-plate capacitor with area 0.200 m2 and plate separation of 3.00 mm is connected to a 6.00-V battery. (a) What is the capacitance? (b) How much charge is stored on the plates? (c) What is the electric field between the plates? (d) Find the magnitude of the charge density on each plate. (e) Without disconnecting the battery, the plates are moved farther apart. Qualitatively, what happens to each of the previous answers?

(a) How much charge can be placed on a capacitor with air between the plates before it breaks down if the area of each plate is 5.00 cm2? (b) Find the maximum charge if polystyrene is used between the plates instead of air. Assume the dielectric strength of air is 3.00 106 V/m and that of polystyrene is 24.0 106 V/m.

A parallel-plate capacitor with only air between its plates is charged by connecting the capacitor to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates, (a) A voltmeter reads 45.0 V when placed across the capacitor. When a dielectric is inserted between die plates, completely filling the space, the voltmeter reads 11.5 V. What is the dielectric constant of the material? (b) What will the voltmeter read if the dielectric is now pulled away out so it fills only one-third of the space between the plates?

An air-filled (empty) parallel-plate capacitor is made from two square plates that are 25 cm on each side and 1.0 mm apart. The capacitor is connected to a 50-V battery and fully charged. It is then disconnected from the battery and its plates are pulled apart to a separation of 2.00 mm. (a) What is the capacitance of this new capacitor? (b) What is the charge on each plate? (c) What is the electrical field between the plates?

(a) When a 9.00-V battery is connected to the plates of a capacitor, it stores a charge of 27.0 C. What is the value of the capacitance? (b) If the same capacitor is connected to a 12.0-V battery, what charge is stored?

True or False? (a) From the definition of capacitance C = Q/V it follows that an uncharged capacitor has a capacitance of zero. (b) As described by the definition of capacitance, the potential difference across an uncharged capacitor is zero.

A parallel-plate capacitor has plates of area A = 7.00 102 m2 separated by distance d = 2.00 104 m. (a) Calculate the capacitance if the space between the plates is filled with air. What is the capacitance if the space is filled half with air and half with a dielectric of constant = 3.70 as in (b) Figure P16.56a, and (c) Figure P16.56b? (Hint: In (b) and (c), one of the capacitors is a parallel combination and the other is a series combination.) Figure P16.56