Chapter 17, Problem 57GP

To determine

Energy stored in per cubic meter.

Answer to Problem 57GP

  $1.0\times {10}^{-7}J/m^3$

Explanation of Solution

Given data:

Electric field, $E=150V/m$

Permittivity of free space, ${\text{ε}}_0=8.85\times {10}^{-12}{C}^2/N\cdot m^2$

Formula used:

The electric density is $=\frac{PE}{Volume}=\frac{1}{2}{\text{ε}}_0{E}^2\mathrm{...}\left(1\right)$

Where,

  $PE$ is the potential energy or stored energy.

  ${\text{ε}}_{\text{0}}$ is the permittivity of free space.

  $E$ is the electric field.

Calculation:

Substitute the given data into equation $\left(1\right)$ to calculate the stored energy,

  $=\frac{1}{2}{\text{ε}}_0{E}^2$

  $=\frac{1}{2}\left(8.85\times {10}^{-12}{C}^2/N\cdot m^2\right){\left(150V/m\right)}^2$

  $=1.0\times {10}^{-7}J/m^3$

Conclusion:

The required value is $1.0\times {10}^{-7}J/m^3$ .