- A dielectric slab is slowly inserted between the plates of a parallel plate capacitor while the capacitor is connected to a battery. As it is being inserted: a. The capacitance and the charge on the plate increase but the potential difference between the plates remains the same b. The capacitance and the charge on the positive plate decrease but the potential difference between the plates remains the same c. The potential difference between the plates increases, the charge on the positive plate decreases, and the capacitance remains the same d. The capacitance, the potential difference between the plates, the charge on the positive plate all decrease

- A dielectric slab is slowly inserted between the plates of a parallel plate capacitor while the capacitor is connected to a battery. As it is being inserted: a. The capacitance and the charge on the plate increase but the potential difference between the plates remains the same b. The capacitance and the charge on the positive plate decrease but the potential difference between the plates remains the same c. The potential difference between the plates increases, the charge on the positive plate decreases, and the capacitance remains the same d. The capacitance, the potential difference between the plates, the charge on the positive plate all decrease

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter8: Capacitance

Section: Chapter Questions

Problem 64AP: A capacitor is made from two flat parallel plates placed 0.40 mm apart. When a charge of 0.020C is...

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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?
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.
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.
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 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 is connected to a battery. What happens to the stored energy if the plate separation is doubled while the capacitor remains connected to the battery? (a) It remains the same, (b) It is doubled. (c) It decreases by a factor of 2. (d) It decreases by a factor of 4. (e) It increases by a factor of 4.
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
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(a) Two spheres have radii a and b, and their centers are a distance d apart. Show that the capacitance of this system is C=401a+1b2d provided d is large compared with a and b. Suggestion: Because the spheres are far apart, assume the potential of each equals the sum of the potentials due to each sphere. (b) Show that as d approaches infinity, the above result reduces to that of two spherical capacitors in series.
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