Question

What electric field strength would store 12.5 JJ of energy in every 6.00 mm3mm3 of space?

Answer 1

To develop this problem we will apply the concepts related to the potential energy per unit volume for which we will obtain an energy density relationship that can be related to the electric field. From this formula it will be possible to find the electric field required in the problem. Our values are given as

The potential energy,  $U = 13.0 J$

The volume,  $V = 6.00 mm^3 = 6.00*10^{-9}m^3$

The potential energy per unit volume is defined as the energy density.

$u = \frac{U}{V}$

$u= \frac{(13.0 J)}{(6.00*10^{-9} m^3)}$

$u= 2.167109 J/m^3$

The energy density related with electric field is given by

$u = \frac{1}{2} \epsilon_0 E^2$

Here, the permitivity of the free space is

$\epsilon_0 = 8.85*10^-{12} C^2/N \cdot m^2$

Therefore, rerranging to find the electric field strength we have,

$E = \sqrt{\frac{2u}{\epsilon_0}}$

$E = \sqrt{\frac{2(2.167109)}{8.85*10^{-12}}}$

$E = 2.211010 V/m$

Therefore the electric field is 2.21V/m