Q. 264.3( 9 Votes )

# The plates of a p

Answer :

Area of the plates of a parallel capacitor (A) = 90cm2 = 9 10-3m

Distance between the plates (d) = 2.5mm = 2.5 10-3m

Potential difference across the plates (V) = 400V

(a) Electrostatic energy stored in the capacitor(E) = CV2

Since, Capacitance (C) =  Where, = Absolute Permittivity of free space = 8.85 10-12C2N-1m-2 E = 2.55 10-6J

The electrostatic energy stored by the capacitor = 2.55 10-6J

(b) Volume between plates of capacitor (v) = A × d

= 9 10-3m2 × 2.5 10-3m

= 2.25 10-5m3

Therefore, Energy per unit volume (u) = E/v  The energy per unit volume (u) = 0.113 Jm-3

Relation between u and the magnitude of electric field E between the plates

Since,

Energy per unit volume (u) = E/v

= CV2 /v

Here, V = voltage across capacitor

v = volume between plates of capacitor,  Here, V/d is the Electric intensity or Electric field so, Rate this question :

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