A parallel-plate capacitor of plate separation d is charged to a potential difference AV. A dielectric slab of thickness d and dielectric constant k is introduced between the plates while the battery remains con- nected to the plates. (a) Show that the ratio of energy stored after the dielectric is introduced to the energy stored in the empty capacitor is U/U, = K. (b) Give a physical explanation for this increase in stored energy. (c) What happens to the charge on the capacitor? Note: This situation is not the same as in Example 26.5, in which the battery was removed from the circuit before the dielectric was introduced.

A parallel-plate capacitor of plate separation d is charged to a potential difference AV. A dielectric slab of thickness d and dielectric constant k is introduced between the plates while the battery remains con- nected to the plates. (a) Show that the ratio of energy stored after the dielectric is introduced to the energy stored in the empty capacitor is U/U, = K. (b) Give a physical explanation for this increase in stored energy. (c) What happens to the charge on the capacitor? Note: This situation is not the same as in Example 26.5, in which the battery was removed from the circuit before the dielectric was introduced.

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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