Question

A dielectric-filled parallel-plate capacitor has plate area A = 15.0cm2 , plate separation d = 5.00mm and dielectric constant k = 5.00. The capacitor is connected to a battery that creates a constant voltage V = 10.0V . Throughout the problem, use ?0 = 8.85

Answer 1

a.The energy sored in the capacitor is U=1/2*C*V²

The capacitance of a parallel plate capacitor with a dielectric is

Thus U=1/2CV² = 1/2 () V²

b. The siuatio is having two capacitors i parallel one with k' =l(air) and one with k'=k. Each has area of A/2

C1= (A/2)1/d C₂ = k (A/2)1/d

C_eq = C₁+C₂ =1/2 (1+k)A/D

U=1/2C_eqV² = 1/4(1+k) A/D V²

c. The charge on the plate remains constant through out the process

Initially U₂ = 1/2QV =1/2(2U₂/V)² 1/C

The capacitance is just the capacitance of the dielectric field capacitor C=A/d

U₃ =1/2Q²/C = 1/2(2U₂ /V)² *1/C

C=A/D

U₃ =1/2(2U₂/V)² (d/A)= 2U₂² d/ AV²

d. By energy coservation he work done is difference of energy between initial and final states of the capacitor

U₃=U₂+W