Question

An air filled capacitor is made from two flat parallel plate 1mm apart. The magnitude of the charge on each plate is 0.005×10^-6 when the potential difference is 500v calculate a,the energy stored. b, the area each plate. C,the energy density

Answer 1

Given, Charge stored in the plates, Q= 0.005×10⁻⁶C

Separation between the plates, d= 1mm = 0.001m

Potential difference V =500v

We know that, Q= C.V

   C= Q/V  = (0.005×10⁻⁶)/ 500

                 = 0.001×10⁻⁸ F

                = 0.01 ×10⁻⁹F= 0.01 μF

 

(a) Energy stored in the capacitor,

U= (1/2)CV²

 = (1/2) 0.01×10⁻⁹×(500)²

U = 1.25 μJ

(b) We know that, capacitance in a parallel plate capacitor,

C= ε₀A/d

A= C.d/ε₀

  = (0.01×10⁻⁹×0.001)/ (8.85×10⁻¹²)                

  = 1.13×10⁻³ m²

(c) Energy density (energy per unit volume) =  U/Ad

or

Energy density = (1/2)ε₀(V/d)²

                         = (1/2)×8.85×10⁻⁹×(500/0.001)²

                         = 1.11 J/m³