The space between the plates of a parallel plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants K₁=2,K₂=3 and K₃=5 (as shown in figure). If new capacitance is n/3C then the value of n is ____ .

Solution

Physics sir · 4 steps

Read the figure as a circuit: two side-by-side slabs in parallel, and that pair in series with the top slab.

1. Spot the arrangement. K₂ and K₃ sit side by side between the same two surfaces, so they are in parallel. Together they lie one above the other with K₁, so that pair is in series with K₁. Use C = (ε₀ A)/d for the empty capacitor.
2. Each slab. Top slab: C₁ = (2ε₀ A)/(d/2) = 4C. Left half: C₂ = (3ε₀ (A/2))/(d/2) = 3C. Right half: C₃ = (5ε₀ (A/2))/(d/2) = 5C.
3. Parallel pair. C₂ and C₃ add directly: C₂₃ = 3C + 5C = 8C.

   Cₙₑₜ = (C₁ C₂₃)/(C₁ + C₂₃) = (4C × 8C)/(4C + 8C) = 8/3C

4. So. Comparing with n/3C gives n = 8.

The common slip: 4. Using the full gap d instead of d/2 for each slab halves every capacitance and gives 4/3C, so n = 4, not 8.

The idea to keep. Slabs stacked across the gap are in series. Slabs placed side by side across the same gap are in parallel. Use each slab's own area and own thickness in (Kε₀ A)/d.

Answer from the NTA final answer key. Solution by PrepWiser.