A parallel plate capacitor with plate area A and plate separation d = 2 m has a capacitance of 4 μF. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant K = 3 (as shown in figure) will be :

Solution

Physics sir · 3 steps

The slab fills half the gap across the full plate area, so split it into two capacitors in series.

1. Read the figure. The slab covers the whole plate area but only half the separation, d/2. A field line from one plate to the other crosses the air layer first and then the dielectric layer, so the two layers act as two capacitors in series.
2. Each layer. With C₀ = (ε₀ A)/d = 4 μF, the air layer gives C₁ = (ε₀ A)/(d/2) = 2C₀ = 8 μF. The dielectric layer gives C₂ = (Kε₀ A)/(d/2) = 2KC₀ = 24 μF.

   C = (C₁ C₂)/(C₁ + C₂) = (8 × 24)/(8 + 24) μF = 6 μF

3. So. The new capacitance is 3/2C₀ = 6 μF, which is option (c). The value d = 2 m is never needed, because everything scales from C₀.

The idea to keep. Slab parallel to the plates, filling part of the gap: layers in series. Slab filling part of the area across the full gap: pieces in parallel. Look at which way the slab cuts the field.

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