The Total solution for NCERT class 6-12
The plates of a parallel plate capacitorhave an area of 90 cm2 each and are separated by 2.5 mm. Thecapacitor is charged by connecting it to a 400 V supply.
(a) How much electrostatic energy is stored by thecapacitor?
(b) View this energy as stored in the electrostaticfield between the plates, and obtain the energy per unit volume u.Hence arrive at a relation between u and the magnitude ofelectric field E between the plates.
Area of the plates of a parallel platecapacitor, A = 90 cm2 = 90 × 10−4 m2
Distance between the plates, d =2.5 mm = 2.5 × 10−3 m
Potential difference across theplates, V = 400 V
Where,
= Permittivity of free space = 8.85 ×10−12 C2 N−1 m−2
Hence, the electrostaticenergy stored by the capacitor is
(b) Volume of the given capacitor,
Energy stored in the capacitor per unit volume is givenby,
= Electric intensity = E