A parallel plate capacitor with width 4 cm, length 8 cm and separation between the plates of 4 mm is connected to a battery of 20 V. A dielectric slab of dielectric constant 5 having length 1 cm, width 4 cm and thickness 4 mm is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be ____________ ε₀ J. (Where ε₀ is the permittivity of free space)

Solution

Physics sir · 4 steps

The slab covers only 1 cm of the 8 cm length, so you have two capacitors side by side, in parallel.

1. Split the plates. The slab fills the full gap of 4 mm but covers only a strip of area 1 × 4 = 4 cm². The remaining 7 × 4 = 28 cm² has air. Both parts share the same plates and the same voltage, so they are in parallel.

   C = ε₀/d(K A₁ + A₂) = ε₀/(4 × 10⁻³)(5 × 4 × 10⁻⁴ + 28 × 10⁻⁴)

2. Find C. The bracket is 48 × 10⁻⁴ m², so C = (48 × 10⁻⁴)/(4 × 10⁻³) ε₀ = 1.2 ε₀ (in farads, with ε₀ as a number).
3. Find the energy. The battery holds V = 20 V, so U = 1/2CV² = 1/2 × 1.2 ε₀ × 400.
4. So. U = 240 ε₀ J, so the answer is 240.

The common slip: 800. Using K = 5 for the whole plate area, as if the slab filled the capacitor, gives C = 4 ε₀ and an energy of 800, not 240.

The idea to keep. Slab fills the gap but covers part of the area: parallel capacitors, so add KA and A. Slab covers the full area but only part of the gap: series capacitors.

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