A composite parallel plate capacitor is made up of two different dielectric materials of different thicknesses, the two layers being separated by a thin conducting foil F. The upper layer has εᵣ₁=3 and t₁=0.5 mm, and the lower layer has εᵣ₂=4 and t₂=1 mm. The top plate is held at 100 V and the bott…

Question

A composite parallel plate capacitor is made up of two different dielectric materials of different thicknesses, the two layers being separated by a thin conducting foil F. The upper layer has εᵣ₁=3 and t₁=0.5 mm, and the lower layer has εᵣ₂=4 and t₂=1 mm. The top plate is held at 100 V and the bottom plate at 0 V. The voltage of the conducting foil is V.

Accepted Answer

60. The foil splits the capacitor into two capacitors in series, so they carry the same charge q. C₁=(εᵣ₁ε₀ A)/(t₁)=(3ε₀ A)/(0.5)=6ε₀ A and C₂=(4ε₀ A)/1=4ε₀ A (thicknesses in mm). Equal charge gives (V₁)/(V₂)=(C₂)/(C₁)=4/6=2/3, with V₁+V₂=100 V. V₁=40 V across the upper layer and V₂=60 V across the lower layer. The foil sits above the bottom plate, which is at 0 V, so its potential is V₂=60 V.

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