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 = dqÂ = | = ^{Â }= [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 = Â = ^{Â }= Â (where surface charge density Ïƒ = , and capacitance of the parallel plate capacitor C = )

Or, U = Â = = E^{2} [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 =Â E^{2}Â

The energy stored due to change in volume Adx is dw = UAdx = E^{2}Adx —–(2)

From equations (1) and (2) we get, Fdx = E^{2}Adx

Or, F = E^{2}A

Or, F = (EA)E

F = QE (from Gaussâ€™s law E.A = ).

Shearing of charge and lost energy: Â

Let us consider two capacitors each of capacitance C_{1} and C_{2} having charge Q_{1} and Q_{2} (Q_{1} > Q_{2}) respectively. The potentials of the capacitors are V_{1} and V_{2} respectively. Then Q_{1} = C_{1}V_{1} and Q_{2} = C_{2}V_{2}.

Now two capacitors are connected by a wire and charge is transferred from C_{1} to C_{2} until the potentials of each capacitor become same (say V). Then the charge of capacitors C_{1} and C_{2} are respectively Q_{1}^{/} = C_{1}V and Q_{2}^{/} = C_{2}V.

Â As the charge is conserved, then Q_{1} + Q_{2} = Q_{1}^{/} + Q_{2}^{/}

Or, C_{1}V_{1} + C_{2}V_{2} = C_{1}V + C_{2}VÂ

Or, V =

The shearing of charge from C_{1} to C_{2} is Q = Q_{1} – Q_{1}^{/}

Or, Q = C_{1}V_{1} – C_{1}V

Or, Q = C_{1}[V_{1} â€“ V]

Or, Q = C_{1}[V_{1} â€“ ]

Or, Q = .

Lost energy: Â Total potential energy of two capacitors before shearing of charge is U_{i} = (C_{1}V_{1}^{2} + C_{2}V_{2}^{2})

Total potential energy of two capacitors after shearing of charge is U_{f} = (C_{1}V^{2} + C_{2}V^{2}) = =

The lost energy U = U_{i} – U_{f} = [(C_{1}V_{1}^{2} + C_{2}V_{2}^{2}) – ]

Or, U =

U = .

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