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Required information A parallel plate capacitor of capacitance 5.19 µF has the space between the plates filled with a slab of glass with K = 3.00. The capacitor is charged by connecting it to a 1.20-V battery. After the capacitor is disconnected from the battery, the dielectric slab is removed. XV (1) Glass X = 1.20 V. Find the energy stored in the capacitor after the glass is removed.

Required information A parallel plate capacitor of capacitance 5.19 µF has the space between the plates filled with a slab of glass with K = 3.00. The capacitor is charged by connecting it to a 1.20-V battery. After the capacitor is disconnected from the battery, the dielectric slab is removed. XV (1) Glass X = 1.20 V. Find the energy stored in the capacitor after the glass is removed.

College Physics
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)...
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A parallel plate capacitor of capacitance 5.19 μF has the space between the plates filled with a slab of glass with κ = 3.00. The capacitor is charged by connecting it to a 1.20-V battery. After the capacitor is disconnected from the battery, the dielectric slab is removed.

 

 

 
 X = 1.20 V.

 

 

 

 

Find the energy stored in the capacitor after the glass is removed.

 

 μJ

Required information A parallel plate capacitor of capacitance 5.19 µF has the space between the plates filled with a slab of glass with K = 3.00. The capacitor is charged by connecting it to a 1.20-V battery. After the capacitor is disconnected from the battery, the dielectric slab is removed. XV (1) Glass X = 1.20 V. Find the energy stored in the capacitor after the glass is removed.
Transcribed Image Text:Required information A parallel plate capacitor of capacitance 5.19 µF has the space between the plates filled with a slab of glass with K = 3.00. The capacitor is charged by connecting it to a 1.20-V battery. After the capacitor is disconnected from the battery, the dielectric slab is removed. XV (1) Glass X = 1.20 V. Find the energy stored in the capacitor after the glass is removed.
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