Capacitor Part – 4

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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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