A parallel plate air capacitor has a capacitance of 5.0 F. What is the capacitance if a dielectric of relative permeability of 2 inserted between the plates of the original or capacitor? a. 1.25F c. 2.5F b. 5.0F d. 10.0F

A parallel plate air capacitor has a capacitance of 5.0 F. What is the capacitance if a dielectric of relative permeability of 2 inserted between the plates of the original or capacitor? a. 1.25F c. 2.5F b. 5.0F d. 10.0F

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter8: Capacitance

Section: Chapter Questions

Problem 82CP: A parallel-plate capacitor is filled with two dielectrics, as shown below. When the plate area is A...

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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?
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.
If you have ever tried to hang a picture or a mirror, you know it can be difficult to locate a wooden stud in which to anchor your nail or screw. A carpenters stud finder is a capacitor with its plates arranged side by side instead of facing each other as shown in Figure 20.32. When the device is moved over a stud, does the capacitance (a) increase or (b) decrease?
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 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 is filled with two dielectrics, as shown below. When the plate area is A and separation between plates is d, show that the capacitance is given by C=0Adk1+k22 C=0Adk1+k22 C=0Adk1+k22
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 capacitor is constructed from two square, metallic plates of sides and separation d. Charges +Q and Q are placed on the plates, and the power supply is then removed. A material of dielectric constant is inserted a distance x into the capacitor as shown in Figure P25.49 (page 690). Assume d is much smaller than x. (a) Find the equivalent capacitance of the device. (b) Calculate the energy stored in the capacitor. (c) Find the direction and magnitude of the force exerted by the plates on the dielectric. (d) Obtain a numerical value for the force when x = /2, assuming = 5.00 cm, d = 2.0 mm, the dielectric is glass ( = 4.50), and the capacitor was charged to 2.00 103 V before the dielectric was inserted. Suggestion: The system can be considered as two capacitors connected in parallel. Figure P25.49
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?
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.)
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?