The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply. (a) How much electrostatic energy is stored by the capacitor? (b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates
(a) Area of the plates of a parallel plate capacitor, A=90cm2=90×10−4m2 Distance between the plates, d=2.5mm=2.5×10−3m Potential difference across the plates, V=400V Capacitance of the capacitor is given by the relation, C=d∈0A Electrostatic energy stored in the capacitor is given by the relation, E1=21CV2 =21d∈0AV2 Where, ∈0=Permittivity of free space =8.85×10−12C2N−1m−2 ∴E1=2×2.5×10−31×8.85×106−12×90×10−4×(400)2=2.55×10−6J Hence, the electrostatic energy stored by the capacitor is 2.55×10−6J. (b) Volume of the given capacitor, V′=A×d =90×10−4×25×10−3 =2.25×10−4m3 Energy stored in the capacitor per unit volume is given by, u=V′E1 =2.25×10−42.55×10−6=0.113Jm−3 Again, u=V′E1 =Ad21CV2=Ad2d∈0AV2=21∈0(dV)2 Where, dV= Electric intensity =E ∴u=21∈0E2
