(a) Derive the expression for the energy stored in a parallel plate capacitor. Hence obtain the expression for the energy density of the electric field. (b) A fully charged parallel plate capacitor is connected across an uncharged identical capacitor. Show that the energy stored in the combination is less than that stored initially in the single capacitor.

(a)Power across capacitor is P = VI = VdQ/dt
By definition of capacitance, Q = CV
Pdt = Q/C dQ
Energy stored is U = ∫Pdt = ∫(Q/C)dQ = Q²/2C .. (i)
Electric field inside the capacitor is given by: E = σ/ε₀ = Q/Aε₀
Capacitance C = ε₀A/d
dQ = ε₀A/d * dV = ECdV (ii)
Substituting (ii) in (i), U = (ECdV)²/(2C) = (E²C²d²)/(2C) = E²(ε₀A/d)d²/2 = E²ε₀Ad/2
Energy density U/Ad = (1/2)ε₀E²
(b)Let charge on the capacitor before connecting be Q.
After connection, each capacitor has a charge Q/2.
Energy stored before connection, U₁ = Q²/2C
Energy stored after connection, U₂ = 2(Q/2)²/2C = Q²/4C
Hence, energy stored in the combination is less than that of the single capacitor.