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Potential Difference Across Capacitor Regions and Work Done

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Homework Statement


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.

Homework Equations


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.
 

Answers and Replies

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Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 

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