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Copper slab inserted in capacitor

Short answer

Initially C₀ = ε₀A/d. With the slab, only air gaps of total thickness d−b sustain field, so C = ε₀A/(d−b). The change ΔC = ε₀A b / [(d−b)d].

A copper slab of thickness b is inserted into a parallel plate capacitor exactly half way between the plates. If the separation of the plates is d and the area of each plate is A, show that the change in capacitance is equal to ε0Ab/((d − b)d).

Parallel plate capacitor: plate area A, separation d. Without the slab, C₀ = ε0A/d.

   ______________________  + plate
        d1 (air)
   ======================  copper slab, thickness b
        d2 (air)
   ______________________  - plate
   d1 + d2 = d - b ; here d1 = d2 = (d-b)/2

With the slab

  1. Charge +q on the top plate induces −q on the upper face of the slab and +q on the lower face. The field inside copper is zero.
  2. Field exists only in the air gaps: E = q/(ε0 A) in each gap (Gauss's law).
  3. Potential difference between the plates = field × total air thickness:

V = E (d₁ + d₂) = (q/ε0A)(d − b)

  1. New capacitance:

C = q/V = ε0A / (d − b)

The result depends only on the total air gap (d − b), not on where the slab sits. The two gaps behave as two capacitors in series: 1/C = d₁/ε0A + d₂/ε0A = (d − b)/ε0A.

Change in capacitance

ΔC = C − C₀ = ε0A/(d − b) − ε0A/d = ε0A [ d − (d − b) ] / [ d (d − b) ]

ΔC = ε0 A b / [ (d − b) d ]

Shown. The capacitance increases, since the effective separation shrinks from d to d − b.

✓ Answer checked by Claude Opus 5.5 (high effort)

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