A parallel plate capacitor of area A and separation d is fully charged so that the charge on the top plate is Q. The battery is disconnected and two dielectric slabs of dielectric constant k are inserted with an air gap between them. Each region takes up one third of the plate separation. (Region 1 has dielectric, 2 has air, and 3 dielectric.) I take ε0 =1 for convenience in writing.
a) Find the electric field in all regions.
b) Find the potential difference across region 1
c) Find potential difference across region 2
d) Find the overall capacitance
e) Find the work done to insert the dielectrics.
D =σ (Gauss Law with dielectrics), ΔV =Ed, C =εA/s, C_eq = (1/C1 +1/C2 + 1/C3)^-1 for capacitors in series,
W=ΔU=ΔC (ΔV)^2/2, since the parallel plate capacitor is an equipotential.
The Attempt at a Solution
A) From Gauss Law the fields are 1: Q/(Aε),2: Q/A, 3: Q/(Aε).
B) Since ΔV=Ed, this is Qd/3(Aε).
C) As in B) this is Qd/3A.
D) The overall capacitance is 3A/d (1/ε +1ε)^-1.
E) From the given expression for work this is A/d ([1/ε +1/ε]^-1 /3 -1) *(ΔV)^2/2.