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A parallel-plate capacitor with a capacitance of 200 pF is charged to 100 V. Then the battery is removed. A sheet of teflon (K = 2.0) is then slid between the plates. %3D A. By what factor does the charge on the plates change? B. What is the final potential difference between the plates?

A parallel-plate capacitor with a capacitance of 200 pF is charged to 100 V. Then the battery is removed. A sheet of teflon (K = 2.0) is then slid between the plates. %3D A. By what factor does the charge on the plates change? B. What is the final potential difference between the plates?

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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OneNote for Windows 10 Shahid Iqbal Home Insert Draw View Help Class Notebook A Share 6 O Shapes G Ink to Shape a Ink to Text A Ruler 최늘 Math Extra credit Problem A parallel-plate capacitor with a capacitance of 200 pF is charged to 100 V. Then the battery is removed. A sheet of teflon (K = 2.0) is then slid between the plates. A. By what factor does the charge on the plates change? B. What is the final potential difference between the plates? 2:27 PM 2/15/2021 J A A ALBION You 14 others
Transcribed Image Text:OneNote for Windows 10 Shahid Iqbal Home Insert Draw View Help Class Notebook A Share 6 O Shapes G Ink to Shape a Ink to Text A Ruler 최늘 Math Extra credit Problem A parallel-plate capacitor with a capacitance of 200 pF is charged to 100 V. Then the battery is removed. A sheet of teflon (K = 2.0) is then slid between the plates. A. By what factor does the charge on the plates change? B. What is the final potential difference between the plates? 2:27 PM 2/15/2021 J A A ALBION You 14 others
Expert Solution
Step 1

Given data

*The given capacitance of the capacitor is C = 200 pF  = 200 × 10-12 F

*The battery voltage is V = 100 V

*The dielectric constant os teflon is k = 2.0

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