15. If half of the space between the plates are filled with a dielectric material of dielectric constant K 1 and rest of the half space is filled with a dielectric material of dielectric constant K 2, then 1. Find the equivalent dielectric constant. 2. Find the capacitance of this configuration. 3. Ifk 1=2 K 2, calculate the total stored energy in the capacitor. 4. Following the conditions and answers found for 3. , calculate the percentage of energy stored in the space of the capacitor filled with the dielectric constant K 2. Separation Plate Area

15. If half of the space between the plates are filled with a dielectric material of dielectric constant K 1 and rest of the half space is filled with a dielectric material of dielectric constant K 2, then 1. Find the equivalent dielectric constant. 2. Find the capacitance of this configuration. 3. Ifk 1=2 K 2, calculate the total stored energy in the capacitor. 4. Following the conditions and answers found for 3. , calculate the percentage of energy stored in the space of the capacitor filled with the dielectric constant K 2. Separation Plate Area

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter8: Capacitance

Section: Chapter Questions

Problem 62P: When a 360-nF air capacitor is connected to a power supply, the energy stored in the capacitor is...

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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 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.)
An air-filled capacitor is made from two flat parallel plates 1.0 mm apart. The inside area of each plate is 8.0cm2. (a) What is the capacitance of this set of plates? (b) If the region between the plates is filled with a material whose dielectric constant is 6.0, what is the new capacitance?
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?
(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?
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
(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?
(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?
Unreasonable Results (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
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?
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