- 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 CV2 Joule
= 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 CV2
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
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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