Q3:A parallel plate capacitor of area 0.64 cm? , when the plate in vacuum, the capacitance of the capacitor is 4.9 pf. 1- Calculate the value of the capacitance if the space between the plates is filled with nylon (k=3.4) 2- What maximum the potential difference that can be applied to the plates without causing discharge (Ema=14 x10° v/m)

Q3:A parallel plate capacitor of area 0.64 cm? , when the plate in vacuum, the capacitance of the capacitor is 4.9 pf. 1- Calculate the value of the capacitance if the space between the plates is filled with nylon (k=3.4) 2- What maximum the potential difference that can be applied to the plates without causing discharge (Ema=14 x10° v/m)

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) 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 capacitor is made from two flat parallel plates placed 0.40 mm apart. When a charge of 0.020C is placed oil the plates the potential difference between them is 250 V. (a) What is the capacitance of the plates? (b) What is the area of each plate? (c) What is the charge on the plates when the potential difference between them is 500 V? (d) What maximum potential difference can be applied between the plates so that the magnitude of electrical fields between the plates does not exceed 3.0 MV/m?
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 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 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 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
Review. A storm cloud and the ground represent the plates of a capacitor. During a storm, the capacitor has a potential difference of 1.00 x 108 V between its plates and a charge of 50.0 C. A lightning strike delivers 1.00% of the energy of the capacitor to a tree on the ground. How much sap in the tree can be boiled away? Model the sap as water initially at 30.0C. Water has a specific heat of 4 186 J/kg C, a boiling point of 100C, and a latent heat of vaporization of 2.26 X 106 J/kg.
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.
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?
Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500-F capacitors and an 800-V charging source? (a) 500 V (b) 8.00 kV (c) 400 kV (d) 800 V (e) 0
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 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