Question

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).

Answer 1

Correct Answer: 240

To find the electrostatic energy of the system, we need to consider both the capacitor with and without the dielectric. First, calculate the capacitance without the dielectric. The formula for the capacitance C of a parallel plate capacitor is:

C = (ε₀ * A) / d

where A is the area of the plate and d is the separation between the plates.

The area A of the capacitor is:

A = width * length = 0.04 m * 0.08 m = 0.0032 m²

The separation d is 4 mm = 0.004 m.

Thus, the capacitance C is:

C = (ε₀ * 0.0032) / 0.004 = 0.8 ε₀

Next, consider the dielectric slab. It covers an area A_d (area with the dielectric) and has a thickness equal to the separation.

The area A_d is:

A_d = width * length of dielectric = 0.04 m * 0.01 m = 0.0004 m²

The remaining area without dielectric is:

A_r = total area - A_d = 0.0032 m² - 0.0004 m² = 0.0028 m²

The capacitance with the dielectric C₁ is:

C₁ = (5ε₀ * A_d) / d = (5ε₀ * 0.0004) / 0.004 = 0.5 ε₀

The capacitance without the dielectric C₂ is:

C₂ = (ε₀ * A_r) / d = (ε₀ * 0.0028) / 0.004 = 0.7 ε₀

The total capacitance C_total is the sum:

C_total = C₁ + C₂ = 0.5 ε₀ + 0.7 ε₀ = 1.2 ε₀

The energy U stored in the capacitor is given by:

U = 0.5 * C_total * V²

where V = 20 V. Thus,

U = 0.5 * 1.2 ε₀ * 20² = 240 ε₀

The electrostatic energy of the system is 240 ε₀ J, which is within the given range of 240,240.