91Ó°ÊÓ

The given data is listed below as-

• The Energy density is,$\mathrm{η}=10\frac{J}{m^3}$
• Electric field is,$E=3×{10}^6\frac{V}{m}$

The energy density is stored in a region where there is an electric field of magnitude E.

The concept of Energy density gives the magnitude of the electric field inside the capacitor.

The equation of the magnitude of the electric field is expressed as,

$\mathrm{η}=\frac{1}{2}{\mathrm{Î\mu }}_0{E}^2$

Here, is permittivity of free space, E is Electric field.

For, $\mathrm{η}=10\frac{J}{m^3}$and ${\mathrm{Î\mu }}_0=8.85×{10}^{-12}\frac{F}{m}$

$$\begin{gathered} \mathrm{η}=\frac{1}{2}{\mathrm{Î\mu }}_0{E}^2 \\ \left(10 \frac{J}{m^3}\right)=\frac{1}{2}×\left(8.85×{10}^{-12} \frac{F}{m}\right)×E^2 \\ {E}^2=\frac{\left(10 \frac{J}{m^3}\right)×2}{\left(8.85×{10}^{-12} \frac{F}{m}\right)} \\ {E}^2=2.25×{10}^{12} \end{gathered}$$

Thus, the magnitude of the electric field is. 1.5 X 10⁶V/m

The equation of the magnitude of the electric field is expressed as,

Here, is permittivity of free space, is Electric field.

Thus, the magnitude of the energy density is 39.825.