## 13/01/2018

### 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 )
The 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
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 )
The unit of Energy stored per unit volume is Joule / meter3
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
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