2. Why do you think the parallel plate capacitor discharged so easily? Why do you think thecylindrical electrolytic capacitor did not discharge as easily?3.

Basically rate of change of capacitor is the capacitor that does not lose its charge at a steady…

Answered: . Why do you think the parallel plate…

University Physics Volume 2

18th Edition

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Author:OpenStax

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Chapter8: Capacitance

Section: Chapter Questions

Problem 75AP: (a) A certain parallel-plate capacitor has plates of area 4.00 m2 , separated by 0.0100 mm of nylon,...

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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) A certain parallel-plate capacitor has plates of area 4.00 m2 , separated by 0.0100 mm of nylon, and stores 0.170 C of charge. What is the applied voltage? (b) What is unreasonable about this result? (c) Which assumptions are responsible or inconsistent?
A parallel-plate capacitor with plates of area LW and plate separation t has the region between its plates filled with wedges of two dielectric materials as shown in Figure P25.48. Assume t is much less than both L and W. (a) Determine its capacitance. (b) Should the capacitance be the same if the labels 1 and 2 are interchanged? Demonstrate that your expression does or does not have this property. (c) Show that if 1 and 2 approach equality to a common value , your result becomes the same as the capacitance of a capacitor containing a single dielectric: C = 0LW/t. Figure P25.48
What total capacitances can you make by connecting a 5.00 F and an 8.00 F capacitor together?
A parallel-plate capacitor is made of two square plates 25 cm on a side and 1.0 mm apart. The capacitor is connected to a 50.0-V battery. With the battery still connected, the plates are pulled apart to a separation of 2.00 mm. What are the energies stored in the capacitor before and after the plates are pulled farther apart? Why does the energy decrease even though work is done in separating the plates?
For a Teflon ™-filled, parallel-plate capacitor, the area of the plate is 50.0 cm and the spacing between the plates is 0.50 mm. If the capacitor is connected to a 200-V battery, find (a) the free charge on the capacitor plates, (b) the electrical field in the dielectric, and (c) the induced charge on the dielectric surfaces.
A 1.00-F capacitor is charged by being connected across a 10.0-V battery. It is then disconnected from the battery and connected across an uncharged 2.00-F capacitor. Determine the resulting charge on each capacitor.
A 40-pF capacitor is charged to a potential difference of 500 V. Its terminals are then connected to those of an uncharged 10-pF capacitor. Calculate: (a) the original charge on the 40-pF capacitor; (b) the charge on each capacitor after the connection is made; and (c) the potential difference across the plates of each capacitor after the connection.
A parallel-plate capacitor has capacitance 3.00 F. (a) How much energy is stored in the capacitor if it is connected to a 6.00-V battery? (b) If the battery is disconnected and the distance between the charged plates doubled, what is the energy stored? (c) The battery is subsequently reattached to the capacitor, but the plate separation remains as in part (b). How much energy is stored? (Answer each part in microjoules.)
Unreasonable Results (a) An 8.00 F capacitor is connected in parallel to another capacitor, producing a total capacitance of 5.00 F. What is the capacitance of the second capacitor? (b) What is unreasonable about this result? (C) Which assumptions are unreasonable or inconsistent?
A parallel-plate capacitor with vacuum between its horizontal plates has a capacitance of 25.0 F. A nonconducting liquid with dielectric constant 6.50 is poured into the space between the plates, filling up a fraction f of its volume. (a) Find the new capacitance as a function of f. (b) What should you expect the capacitance to be when f = 0? Does your expression from part (a) agree with your answer? (c) What capacitance should you expect when f = 1? Does the expression from part (a) agree with your answer?
Show that for a given dielectric material the maximum energy a parallel plate capacitor can store is directly proportional to the volume of dielectric (Volume = A. d). Note that the applied voltage is limited by the dielectric strength.

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. Why do you think the parallel plate capacitor discharged so easily? Why do you think the cylindrical electrolytic capacitor did not discharge as easily?