Question

A parallel plate capacitor without a dielectric is charged up so that it stores potential energy Uo, and it is then disconnected so that its charge remains the same. A dielectric with constant k = 2 is then inserted between the plates. What is the new potential energy stored in the capacitor with the dielectric? a) 400 b) 200 c) U. d) 14U

Answer 1

`(U_(o))/(2)`

Step-by-step explanation:

Q = Amount of charge stored by parallel plate capacitor

C₀ = Capacitance of the capacitor

U₀ = Amount of electric potential energy stored

Amount of electric potential energy stored is given as

`U_(o) = (Q^(2))/(2C_(o))`

k = dielectric constant of dielectric = 2

New capacitance of the capacitor is given as

C = k C₀

New amount of electric potential energy stored is given as

`U = (Q^(2))/(2C)`

`U = (Q^(2))/(2 k C_(o))`

`U = (U_(o))/(k)`

`U = (U_(o))/(2)`