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) Numerical answer type. Work it out, then open the answer.

Answer: 240. Slab part and air part are in parallel: C = (ε₀)/(d)[K A₁ + A₂]. A₁ = 1 × 4 = 4 cm², A₂ = 7 × 4 = 28 cm²: C = ε₀ (5 × 4 + 28) × 10⁻⁴4 × 10⁻³ = 1.2ε₀. U = 1/2CV² = 1/2 × 1.2ε₀ × 20² = 240ε₀ J

A parallel plate capacitor with width 4 cm, length 8 cm and… | JEE Main 2022 (27 Jul, Shift 2)

Q70JEE Main 2022 · 27 JUL · SHIFT 2Difficult

Dielectrics

A parallel plate capacitor with width $4\text{ cm}$ , length $8\text{ cm}$ and separation between the plates of $4\text{ mm}$ is connected to a battery of $20\text{ V}$ . A dielectric slab of dielectric constant 5 having length $1\text{ cm}$ , width $4\text{ cm}$ and thickness $4\text{ mm}$ is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be ____________ $\epsilon_0$ J. (Where $\epsilon_0$ is the permittivity of free space)

Numerical answer type. Work it out, then open the answer.

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Answer: 240

Slab part and air part are in parallel: $C = \frac{\epsilon_0}{d}[K A_1 + A_2]$ .
$A_1 = 1 \times 4 = 4\text{ cm}^2$ , $A_2 = 7 \times 4 = 28\text{ cm}^2$ : $C = \frac{\epsilon_0 (5 \times 4 + 28) \times 10^{-4}}{4 \times 10^{-3}} = 1.2\epsilon_0$ .
$U = \frac{1}{2}CV^2 = \frac{1}{2} \times 1.2\epsilon_0 \times 20^2 = 240\epsilon_0\text{ J}$

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