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 Ki

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 Ki

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter19: Electric Potential And Electric Field

Section: Chapter Questions

Problem 16CQ: Give the reason why a dielectric material increases capacitance compared with what it would be with...

See similar textbooks

Similar questions

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.)
Calculate the equivalent capacitance between points a and b in Figure P26.77. Notice that this system is not a simple series or parallel combination. Suggestion: Assume a potential difference v between [joints a and b. Write expressions for vab in terms of the charges and capacitances for the various possible pathways from a to b and require conservation of charge for those capacitor plates that are connected to each other.
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.
Give the reason why a dielectric material increases capacitance compared with what it would be with air between the plates of a capacitor. What is the independent reason that a dielectric material also allows a greater voltage to be applied to a capacitor? (The dielectric thus increases C and permits a greater V.)
A parallel-plate capacitor is disconnected from a batter, and the plates are pulled a small distance farther apart. Do the following quantities increase, decrease, or stay the same? (a) C (b) Q (c) E between the plates (d) V (e) PEC
Give a reason why a dielectric material increases capacitance compared with what it would be with air between the plates of a capacitor. How does a dielectric material also allow a greater voltage to be applied to a capacitor? (The dielectric thus increases C and permits a greater V.)
A parallel-plate capacitor of plate separation d is charged to a potential difference V0. A dielectric slab of thickness d and dielectric constant is introduced between the plates while the battery remains connected to the plates. (a) Show that the ratio of energy stored after the dielectric is introduced to the energy stored in the empty capacitor is U/U0 = . (b) Give a physical explanation for this increase in stored energy. (c) What happens to the charge on the capacitor? Note: This situation is not the same as in Example 25.5, in which the battery was removed from the circuit before the dielectric was introduced.
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?
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 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 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