16 - chapter26-question30 A parallel-plate capacitor whose capacitance C is 13.5 pF is charged by a battery to a potential difference V 12.5 V between its plates. The charging battery is now disconnected, and a porcelain slab (K = 6.50) Is slipped between the plates. What is the potential energy (in J) of the capacitor-slab device after the slab is Inserted? Option1 Option2 Option3 Option4 OptionS 6855 x 10-9 1.55 x 10-7 5.5 x 10-6 1.055 x 10-9 1.055 x 10-12

16 - chapter26-question30 A parallel-plate capacitor whose capacitance C is 13.5 pF is charged by a battery to a potential difference V 12.5 V between its plates. The charging battery is now disconnected, and a porcelain slab (K = 6.50) Is slipped between the plates. What is the potential energy (in J) of the capacitor-slab device after the slab is Inserted? Option1 Option2 Option3 Option4 OptionS 6855 x 10-9 1.55 x 10-7 5.5 x 10-6 1.055 x 10-9 1.055 x 10-12

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

1. An air-filled capacitor is made from two flat parallel plates 1.5 mm apart. The inside area of each plate is 7.9 cm². (a) What is the capacitance of this set of plates? F Cair = (b) If the region between the plates is filled with a material whose dielectric constant is 6, what is the new capacitance? C dielectric = F
Please see the image.
100 There is a system with four capacitors connected, as shown in Figure P25.10. (a) Calculate the equivalent capacity between points a and b. (b) Find the charge amount of each capacitor when the potential difference AV between points a and bis 15.0 V. 11 (13). Find the equivalent capacitance a 15.0μF 3.00 μF 120.0 με 16: HH 6.00 με Figure P25.10
Given a 10 pF air-filled capacitor, you are asked to convert it to a capacitor that can store up to 8.3 µJ with a maximum potential difference of 662 V. What must be the dielectric constant of the material that you should you use to fill the gap in the air capacitor if you do not allow for a margin of error? Number Units
1 Lightning Consider the earth and a layer of clouds 800 m above the earth's surface as a parallel plate capacitor. (a) If the cloud layer has an area of 1 km², then what is the capacitance? (b) An electric field strength greater than 3.0 x 106 V/m will generate a lightning strike: The air will break down and conduct electricity. What is the maximum charge that can accumulate on the cloud?
1. Two capacitors with 2.0 F and 3.0 F capacitance, respectively, are connected in parallel. The combination is connected to a 100 V line. Find the (a) total capacitance, (b) charge stored in each capacitor, and (c) potential difference across each capacitor. 2. parallel plate capacitor consists of two rectangular plates, each with an area of 4.5 cm2 and are separated from each other by a 2.00 mm thick dielectric with a dielectric constant of 5.26. The capacitor is connected to a 12.0 V battery. How much energy is stored in the capacitor?
You connect three capacitors as shown in the diagram. C1 = C3 = 2.5 μF, and C2 = 5.0 μF. A potential difference of 9.0 V is maintained between the terminals A and B. The magnitude of the charge on capacitor C3 is approximately Choose the correct answer
6
Question 13 The plates of a 1.50 pF parallel plate capacitor are separated by 1.00x10-4 m. The space between the plates is filled with a material with a dielectric constant of 2.26. What is the area of the plates? 1.00x10-6 m² 4.35x10-6 m² 7.50x10-6 m² 9.00x10-6 m²
A parallel-plate capacitor has plates 2.0cm by 3.0cm. The plates are separated by a 1.0mm thickness of paper (k = 3.7). Determine the capacitance of the device. Write your answer in E notation. Example, Your answer: 1.50x10-10 or 1.50x1010 E Notation: 1.50E-10 or 1,50E10
An air-filled parallel plate capacitor is connected to a battery so the difference in potential between the plates is AV and the magnitude of the electric field between the plates is E. If the battery is disconnected, and then the distance between the plates is increased, O AV increases, E remains the same O AV remains the same, E remains the same O AV increases, E increases O AV decreases, E decreases O AV decreases, E remains the same O AV remains the same, E increases O AV decreases, E increases O AV increases, E decreases O AV remains the same, E decreases
8. 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