A capacitor of 10 μ F capacitance whose plates are separated by 10 mm through air and each plate has area 4 cm² is now filled equally with two dielectric media of K₁=2, K₂=3 respectively as shown in figure. If new force between the plates is 8 N. The supply voltage is ________ V.

Solution

Physics sir · 5 steps

Side-by-side dielectrics add in parallel, and the plate force is the stored energy divided by the gap.

1. Parallel halves. Each dielectric covers half the plate area across the full gap d, so the two halves act as capacitors in parallel. Each half has capacitance (Kᵢ C₀)/2, where C₀ = 10 μF is the air value.
2. New capacitance. C = (K₁ + K₂)C₀/2 = ((2+3) × 10)/2 = 25 μF. Once C is known, the plate area is not needed.
3. Force between plates. At fixed charge, U = Q²/2C. Since C ∝ 1/d, U grows in proportion to d, so the attraction is F = U/d.

   F = CV²/2d ⇒ V² = 2Fd/C

4. Put in numbers. V² = (2 × 8 × 0.01)/(25 × 10⁻⁶) = 0.16/(25 × 10⁻⁶) = 6400 V².
5. So. V = √6400 = 80 V, so the supply voltage is 80.

The idea to keep. Dielectrics side by side across the gap add in parallel; layers stacked one above the other add in series. The pull between the plates is F = CV²/2d, the stored energy divided by the gap.

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