An 8.5-pF parallel-plate capacitor is charged by connectingit to a 12.0-V battery. The battery is then disconnected and aslab of material of dielectric constant k = 3 is inserted into thecapacitor such that it fully fills the space between the plates.The amount (in pJ) by which the energy stored in thecapacitor decreases after the dielectric is inserted isa. 525b. 459O C. 490d. 505O e. 408

given: capacitance=8.5pF potential difference=12V dielectric constant=3 Energy stored = 1/2*C*V2…

An 8.5-pF parallel-plate capacitor is charged by connecting it to a 12.0-V battery. The battery is then disconnected and a slab of material of dielectric constant k = 3 is inserted into the %3D capacitor such that it fully fills the space between the plates. The amount (in pJ) by which the energy stored in the capacitor decreases after the dielectric is inserted is O a. 525 O b. 459 O c. 490 O d. 505 O e. 408

An 8.5-pF parallel-plate capacitor is charged by connecting it to a 12.0-V battery. The battery is then disconnected and a slab of material of dielectric constant k = 3 is inserted into the %3D capacitor such that it fully fills the space between the plates. The amount (in pJ) by which the energy stored in the capacitor decreases after the dielectric is inserted is O a. 525 O b. 459 O c. 490 O d. 505 O e. 408

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter8: Capacitance

Section: Chapter Questions

Problem 70AP: A parallel-plate capacitor with capacitance 5.0F is charged with a 12.0-V battery, after which the...

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