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The plates of a parallel plate capacitor have an area of 90cm2 each and are separated by 2.5mm. The capacitance is charged by connecting it to a 400V 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.

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Area of the plate of a parallel plate capacitor, A=90cm2=90×104m2
Distnace betweent he plates , d=2.5 mm =2.5×103m
potential difference across the plates , V=400 v
(a) capacitance of the capacitor is given by the relation.
C=0Ad
Electrostatic energy stored in the capacitor is given by the relation , E1=12CV2
=120AdV2
where, 0=permitivity of free space =8.85×1012C2N1m2
E1=1×8.85×1012×90×(400)22×2.5×103=2.55×106J
Hence, the electrostatic energy stored by the capacitor is 2.55×106J
(B) volume of the given capacitor
V'=A×d
=90×104×25×103
=2.25×104m3
Energy stored in the capacitor per unit volume is given by
u=E1V'
=2.55×1062.25×104=0.11Jm3
Again u=E1v'
=12CV2Ad=0A2dV2Ad=120(Vd)2
where,
Vd=electric intensity =E
u=120E2

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Updated on:21/07/2023

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