13 January 2018

Energy Stored in the Capacitor

  • When DC supply is given to capacitor, some energy is stored in the dielectric medium of capacitor. 
  • When capacitor is discharged, this electrostatic energy released.
  • When capacitor is in discharged condition, work done requires transferring charge from one plate to other plate. 
  • Let a charge transfer from positive plate to negative plate is dq when voltage across capacitor is v.
       Therefore the work done dW = vdq…………….. ( 1 )
       Where
       dW = Work done to transfer charge
           v = Voltage across capacitor
         dq = Charge transfer from positive plate to negative plate
  • As q = Cv
        Therefore dq = Cdv ………………..( 2 )
        From equation ( 1 ) and ( 2 )
        dW = Cvdv
        ∫ dW = C ∫ vdv
  • The lower to upper limit for voltage is 0 to V
        W = CV2 / 2
        Energy stored in the capacitor = ½ CV2
  • If capacitance C is given in Farad and V is in voltage, the energy stored is given in Joule.
        E = ½ CVJoule
           = ½ QV  ( As Q = CV )
           = ½ Q2 / C Joule ( As V = Q / C )

Unit of energy stored in the capacitor is Joule

  • E = ½ CV
        The unit of capacitance is Coulomb / voltage
         E = ( Coulomb / voltage ) ( Voltage )2
            = Coulomb × Voltage
            = Coulomb × Joule / coulomb
            = Joule
  • Therefore the unit of E is given in Joule

Energy stored per unit volume

  • The energy stored per unit volume is given by
         E = ½ CV2 / Ad   
         Where
         A = Plate area and
         d = Distance between plates
         E = ½ [ εAV2 / Ad2 ]      ( As C = εA / d )
            = ½ [ ε ( V2 / d2 )]     
            = ½ [ εE2 ]      ( Where electric intensity E = V / d )
            = ½ [ DE ]      ( Where Electrical displacement D = εE )
            = ½ [ D2 / ε ]  in Joule / meter3    ( As E = D / ε )

Unit of Energy stored per unit volume

  • E = ½ [ DE ]   
           = [ Coulomb / meter2 ] [ Voltage / meter ]
           = [ Coulomb – voltage / meter3 ]
           = [Coulomb × Joule / coulomb ] / meter3
           = Joule / meter3
  • E = ½ [ D2 / ε ] 
           = [ Coulomb2 / meter4 ] / [ Farad / meter ]
           = [ Coulomb2 / Farad - meter3 ]
           = [ Coulomb2 - Voltage / Coulomb - meter3 ]  [ As C = Q / V ]
           = [ Coulomb2 - Joule / Coulomb2 - meter3 ]  [ As voltage = Joule / Coulomb ]
           = Joule / meter3