# The plates of a p

Area of the plates of a parallel capacitor (A) = 90cm2 = 910-3m

Distance between the plates (d) = 2.5mm = 2.510-3m

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 = 8.8510-12C2N-1m-2

E = 2.5510-6J

The electrostatic energy stored by the capacitor = 2.5510-6J

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

= 910-3m2 × 2.510-3m

= 2.2510-5m3

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 :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

The plates of a pNCERT - Physics Part-I

Show that the forNCERT - Physics Part-I

A 4 μF capacitor NCERT - Physics Part-I