Question

The plates of a parallel plate capacitor has an area of 100cm2 each and are separated by 3mm the capacitor is charged by connecting it to a 400v supply. Calculate 1.energy stored in the capacitor 2.energy per unit volume 3.if dielectric constant 20.5 is introduced between the plated of the capacitor then find the energy stored and also change in the energy stored.

Answer 1

Change in energy : 46.8 x 10⁻⁶ J

Explanation:

Area of plates A = 100 cm²  = 10⁻² m²

distance between plates d = 3 mm = 3 x 10⁻³ m³

Potential applied V = 400 V

The capacity of parallel plate capacitor = ε₀A/d = 8.85 x 10⁻¹² x 10⁻² /3 x 10⁻³

= 2.95 x 10⁻¹¹ Farad

The energy of capacitor E = 1/2 C V² = 1/2 x 2.95 x 10⁻¹¹ x ( 400 )²

= 24 x 10⁻⁷ J

2. Energy per unit volume = E/Ad = 24 x 10⁻⁷/ 10⁻²x 3 x 10⁻³ = 8 x 10⁻² Farad/m³

3. When dielectric is placed in between the plates

The capacity  C₀ =  ε₀K A/d = 2.95 x 10⁻¹¹ x 20.5 = 6.15 x 10⁻¹⁰ Farad

Energy E₀ = 1/2 C₀ V² = 1/2 x 6.15 x 10⁻¹⁰ x ( 400 )² = 49.2 x 10⁻⁶ J

The change in energy = 49.2 x 10⁻⁶ - 24 x 10⁻⁷ = 49.2 x 10⁻⁶ - 2.4 x 10⁻⁶

= 46.8 x 10⁻⁶ J