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Step 1: The gap between the two plates Is filled with a dlelectric slab of a dlelectric constant of 6.3. The separatlon between the two plates d= 0.054 mm. The area of each plate Is A = 0.037 m2. When the capacitor Is fully charged, the amount of charge on each plate Is Q1 = 9.70 nCnC. inc = 10-9c. Part A - Step 1: what iš the capacitance C1? Part B - Step 1: Find the potential difference between the two plates AVi. Part C- Step 1: What is the magnitude of the electric field Ej in the gap of the capacitor? Part D - Step 1: What is the energy stored in the capacitor? Step 2: The charged capacitor In Step 1 remalns connected to the same charging battery. The dielectric slab Is removed so that the gap between the two plates Is a vacuum. The separation between the two plates Is unchanged d= 0.054 mm. The area of each plate Is unchanged A = 0.037 m2 Part E- Step 2 calculate the magnitude of the electric field in the gap, E2 Part F- Step 2: what is the capacitance C27 Part G- Step 2: what is the amount of charge Q2 on each plate? Step 3 The charged capacitor In step 2 Is DISCONNECTED from the charging battery. The plate area Is unchanged at the original value 0.037 m2. The gap separatlon Is changed to 3/4 of the orlginal d, d3 = 0.0405 m. Part H- Step 3: What is the capacitance C3? Part I- Step 3: What is the potential difference between the two platesAV3.? %3D

Step 1: The gap between the two plates Is filled with a dlelectric slab of a dlelectric constant of 6.3. The separatlon between the two plates d= 0.054 mm. The area of each plate Is A = 0.037 m2. When the capacitor Is fully charged, the amount of charge on each plate Is Q1 = 9.70 nCnC. inc = 10-9c. Part A - Step 1: what iš the capacitance C1? Part B - Step 1: Find the potential difference between the two plates AVi. Part C- Step 1: What is the magnitude of the electric field Ej in the gap of the capacitor? Part D - Step 1: What is the energy stored in the capacitor? Step 2: The charged capacitor In Step 1 remalns connected to the same charging battery. The dielectric slab Is removed so that the gap between the two plates Is a vacuum. The separation between the two plates Is unchanged d= 0.054 mm. The area of each plate Is unchanged A = 0.037 m2 Part E- Step 2 calculate the magnitude of the electric field in the gap, E2 Part F- Step 2: what is the capacitance C27 Part G- Step 2: what is the amount of charge Q2 on each plate? Step 3 The charged capacitor In step 2 Is DISCONNECTED from the charging battery. The plate area Is unchanged at the original value 0.037 m2. The gap separatlon Is changed to 3/4 of the orlginal d, d3 = 0.0405 m. Part H- Step 3: What is the capacitance C3? Part I- Step 3: What is the potential difference between the two platesAV3.? %3D

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 goes through 3 steps shown in +Q, -Q, +Q; -Q, +Q, -Q, Step 1 Step 2 Step 3 step 3 all part do only Step 1: The gap between the two plates Is filled with a dielectric slab of a dlelectric constant of 6.3. The separatlon between the two plates d = 0.054 mm. The area of each plate Is A = 0.037 m2. When the capacitor Is fully charged, the amount of charge on each plate Is Q1 = 9.70 nCnc.inc = 10-9c. Part A - Step 1: what iš the capacitance C1? Part B - Step 1: Find the potential difference between the two plates AV1. Part C- Step 1: What is the magnitude of the electric field Ej in the gap of the capacitor? Part D - Step 1: What is the energy stored in the capacitor? Step 2: The charged capacitor In Step 1 remalns connected to the same charglng battery. The dlelectric slab Is removed so that the gap between the two plates Is a vacuum. The separation between the two plates Is unchanged d= 0.054 mm. The area of each plate Is unchanged A = 0.037 m2 Part E- Step 2: calculate the magnitude of the electric field in the gap, E2 Part F - Step 2: what is the capacitance C2? Part G- Step 2: what is the amount of charge Q2 on each plate? Step 3 The charged capacitor In step 2 Is DISCONNECTED from the charglng battery. The plate area Is unchanged at the original value O.037 m2. The gap separation Is changed to 3/4 of the orlginal d, d3 = 0.0405 m. Part H - Step 3: What is the capacitance C3? Part I- Step 3: What is the potential difference between the two platesAV3.?
Transcribed Image Text:A parallel plate capacitor goes through 3 steps shown in +Q, -Q, +Q; -Q, +Q, -Q, Step 1 Step 2 Step 3 step 3 all part do only Step 1: The gap between the two plates Is filled with a dielectric slab of a dlelectric constant of 6.3. The separatlon between the two plates d = 0.054 mm. The area of each plate Is A = 0.037 m2. When the capacitor Is fully charged, the amount of charge on each plate Is Q1 = 9.70 nCnc.inc = 10-9c. Part A - Step 1: what iš the capacitance C1? Part B - Step 1: Find the potential difference between the two plates AV1. Part C- Step 1: What is the magnitude of the electric field Ej in the gap of the capacitor? Part D - Step 1: What is the energy stored in the capacitor? Step 2: The charged capacitor In Step 1 remalns connected to the same charglng battery. The dlelectric slab Is removed so that the gap between the two plates Is a vacuum. The separation between the two plates Is unchanged d= 0.054 mm. The area of each plate Is unchanged A = 0.037 m2 Part E- Step 2: calculate the magnitude of the electric field in the gap, E2 Part F - Step 2: what is the capacitance C2? Part G- Step 2: what is the amount of charge Q2 on each plate? Step 3 The charged capacitor In step 2 Is DISCONNECTED from the charglng battery. The plate area Is unchanged at the original value O.037 m2. The gap separation Is changed to 3/4 of the orlginal d, d3 = 0.0405 m. Part H - Step 3: What is the capacitance C3? Part I- Step 3: What is the potential difference between the two platesAV3.?
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