Question

In a circular parallel plate capacitor radius of each plate is 5cm and they are separated by a distance of 2mm.Calculate the capacitance and the energy stored when it is charged by connecting the battery of 200v?​

Answer 1

Radius of each plate = 5 cm

Area = Pie * Radius² = 3.14 * 5² = 78.5 cm² = 0.00785 m²

distance between Plates = 2mm = 0.2cm = 0.002 m

C = ∈ A / d

∈ = 8.85 pF/m  (p = pico = 10⁻¹²)

C = 8.85 * 0.00785 /0.002  pF

=> C = 34.736 pF

Capacitance = 34.736 pF = 34.736 × 10⁻¹² F

Energy Stored = (1/2)CV²

=  (1/2) × 34.736 × (200)²  × 10⁻¹²  J

= 200 × 100 × 34.736 × 10⁻¹²  J

= 0.69472 × 10⁻⁶  J

= 0.695 × 10⁻⁶  J

Energy Stored = 0.695 × 10⁻⁶  J   = 0.695 μJ