Capacitor Part – 4
Energy stored in a capacitor:
Let us consider initially the plates of a capacitor are uncharged and small positive charges are repeatedly transferred from one plate to the other plate. Let at any instant q be the total quantity of charge transferred and v be the potential difference between the plates. Then, q = Cv (where C is the capacitance). If additional dq charge is transferred from the negative plate to the positive plate then the small work done dw = v.dq = 
dq.
The total work done in transferring charge Q [and potential of plate is V] is W = 
dW = 
dq = 
|
= 
= 
[As Q = CV].
Energy density in parallel plate capacitor:
Energy density of a capacitor is the energy stored per unit volume of the capacitor. If the area of each plate is A and the separation between the plates is d then volume of the capacitor is V = Ad.
The energy density of plate capacitor is U = 
= 
= 
(where surface charge density σ = 
, and capacitance of the parallel plate capacitor C = 
)
Or, U = 
= 
= 
E2 [E = 
is the electric field between the plates].
Force acting between the plates of the capacitor:
Let us consider a parallel plate capacitor consists of two parallel plates each of plate area A are separated by a distance d. If F is force acting between two plates during small displacement of one plate by dx, the work done stored as dw = Fdx —–(1).
The change in volume in this case is Adx. The energy density of the capacitor is U = E2
The energy stored due to change in volume Adx is dw = UAdx = E2Adx —–(2)
From equations (1) and (2) we get, Fdx = E2Adx
Or, F = E2A
Or, F = (EA)E
F = QE (from Gauss’s law E.A = 
).
Shearing of charge and lost energy:
Let us consider two capacitors each of capacitance C1 and C2 having charge Q1 and Q2 (Q1 > Q2) respectively. The potentials of the capacitors are V1 and V2 respectively. Then Q1 = C1V1 and Q2 = C2V2.
Now two capacitors are connected by a wire and charge is transferred from C1 to C2 until the potentials of each capacitor become same (say V). Then the charge of capacitors C1 and C2 are respectively Q1/ = C1V and Q2/ = C2V.
As the charge is conserved, then Q1 + Q2 = Q1/ + Q2/
Or, C1V1 + C2V2 = C1V + C2V
Or, V = 
The shearing of charge from C1 to C2 is 
Q = Q1 – Q1/
Or, Q = C1V1 – C1V
Or, Q = C1[V1 – V]
Or, Q = C1[V1 – ]
Or, Q = 
.
Lost energy: Total potential energy of two capacitors before shearing of charge is Ui = (C1V12 + C2V22)
Total potential energy of two capacitors after shearing of charge is Uf = (C1V2 + C2V2) = 
= 
The lost energy U = Ui – Uf = [(C1V12 + C2V22) – 
]
Or, U = 
U = 
.
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