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Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electrical field is E = 3.20 × 10 5 V / m . When the space is filled with dielectric, the electrical field is E = 2.50 × 10 5 V / m . (a) What is the surface charge density on each surface of the dielectric? (b) What is the dielectric constant?

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The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of 1.60 × 10 7 V / m . The capacitor has to have a capacitance of 1.25 nF and must be able to withstand a maximum potential difference 5.5 kV. What is the minimum area the plates of the capacitor may have?

0.0135 m 2

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When a 360-nF air capacitor is connected to a power supply, the energy stored in the capacitor is 18.5 μ J . While the capacitor is connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by 23.2 μ J . (a) What is the potential difference between the capacitor plates? (b) What is the dielectric constant of the slab?

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A parallel-plate capacitor has square plates that are 8.00 cm on each side and 3.80 mm apart. The space between the plates is completely filled with two square slabs of dielectric, each 8.00 cm on a side and 1.90 mm thick. One slab is Pyrex glass and the other slab is polystyrene. If the potential difference between the plates is 86.0 V, find how much electrical energy can be stored in this capacitor.

0.185 μ J

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Additional problems

A capacitor is made from two flat parallel plates placed 0.40 mm apart. When a charge of 0.020 μ C is placed on the plates the potential difference between them is 250 V. (a) What is the capacitance of the plates? (b) What is the area of each plate? (c) What is the charge on the plates when the potential difference between them is 500 V? (d) What maximum potential difference can be applied between the plates so that the magnitude of electrical fields between the plates does not exceed 3.0 MV/m?

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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. 0.277 nF; b. 27.7 nC; c. 50 kV/m

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Suppose that the capacitance of a variable capacitor can be manually changed from 100 to 800 pF by turning a dial connected to one set of plates by a shaft, from 0 ° to 180 ° . With the dial set at 180 ° (corresponding to C = 800 pF ), the capacitor is connected to a 500-V source. After charging, the capacitor is disconnected from the source, and the dial is turned to 0 ° . (a) What is the charge on the capacitor? (b) What is the voltage across the capacitor when the dial is set to 0 ° ?

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Earth can be considered as a spherical capacitor with two plates, where the negative plate is the surface of Earth and the positive plate is the bottom of the ionosphere, which is located at an altitude of approximately 70 km. The potential difference between Earth’s surface and the ionosphere is about 350,000 V. (a) Calculate the capacitance of this system. (b) Find the total charge on this capacitor. (c) Find the energy stored in this system.

a. 0.065 F; b. 23,000 C; c. 4.0 GJ

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Practice Key Terms 5

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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