Sample
Problem 25-6
A parallel-plate capacitor whose capacitance C
is 13.5 pF is charged by a battery to a
potential difference V = 12.5 V between its plates. The charging battery is
now disconnected, and a porcelain slab
(κ = 6.50) is slipped between
the plates. What is the potential energy
(1) of the capacitor before the slab is inserted and (2) of the capacitor-slab device after the slab is inserted?
Solution:
When the slab
is introduced, the potential energy decreases by a factor of K.
The
"missing" energy, in principle, would be apparent to the person who
introduced the slab. The capacitor would
exerts tiny tug on the slab and would do work on it, in amount: