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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 the parallel plate capacitor. The electrostatic energy of this system will be ______ ε₀ J, where ε₀ is the permittivity of free space.

Asked in JEE Main 27th July 2nd Shift 2022 · Dielectric slab

Answer: 240

Step-by-step solution

The slab is as thick as the gap, so it fills the full separation over only part of the plate area: the capacitor becomes two capacitors side by side, in parallel.

Slab-covered area A₁=1 cm×4 cm=4×10⁻⁴ m²; remaining air area A₂=7 cm×4 cm=28×10⁻⁴ m².

C=(ε₀)/d(A₂+KA₁)=(ε₀)/(4×10⁻³)(28×10⁻⁴+5×4×10⁻⁴)=(48×10⁻⁴)/(4×10⁻³)ε₀=1.2ε₀.

U=1/2CV²=1/2×1.2ε₀×(20)²=240ε₀ J.

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