Electromagnetism - Capacitor

A parallel-plate capacitor is connected to a battery. V₀ is the potential difference between the plates, Q₀ is the charge on the positive plate, E₀ the magnitude of the electric field, and D₀ the magnitude of the displacement vector. The original vacuum between the plates is filled with a dielectric and then the battery is disconnected. If the corresponding electrical parameters for the final state of the capacitor are denoted by a subscript f, which of the following is true?

A. V_f  > V₀ 
B. V_f  < V₀
C. Q_f  = Q₀
D. E_f  > E₀
E. D_f  > D₀

(GR9277 #88)

Solution:

(A), (B) FALSE
The battery is disconnected after the plates is filled with a dielectric
V_f  = V₀ = constant

(C) FALSE
Charge: Q = CV

V_f  = V₀ and when dielectric is inserted: C_f  = κC₀
Q_f  = C_f V_f  = κC₀V₀ = κQ₀

(D) FALSE
Potential: V = Ed
V_f  = V₀ and d is not changed
E_f  = E₀

(E) TRUE
Displacement vector in empty space: D₀ = ε₀E
with dielectric constant: D_f  = κε₀E
D_f  > D₀

Answer: E