Question

Energy density in a parallel plate capacitor is given ad 2.1×10^-9.the value of the electric field in the region between the plates is

Answer 1

A capacitor of capacitance C is applied in electric circuit of electromotive force (voltage) V supply through it.

then, energy stored in capacitor is given by, E = 1/2 CV²

if A is cross sectional area and d is seperation between two parallel plates then, C = $\frac{\epsilon_0A}{d}$

and then, E = $\frac{\epsilon_0A}{2d}V^2$

now, energy density in parallel plate capacitor, $\mu=\frac{E}{A.d}$ [ volume = cross sectional × distance between plate ]

so, $\mu=\frac{1}{2}\epsilon_0\left(\frac{V}{d}\right)^2$

but we know, V/d = Electric field.

so, energy density = 1/2 × $\epsilon_0$ × electric field²

given, energy density = 2.1 × 10^-9 J/m³

so, electric field² = 2 × 2.1 × 10^-9/8.85 × 10^-12

or, electric field = √{(4.2/8.85) × 10³}

= 21.78 N/C