Question

Question: The energy density inside a certain capacitor is . What is the magnitude of electric field inside the capacitor? What is the energy density associated with an electric field of (large enough to initiate a spark)?

Short answer

Answer

Electric field inside the capacitor =$1.5\times {10}^6\frac{N}{C}$

Energy density associated with an electric field =$39.825 \frac{\text{J}}{{\text{m}}^{\text{3}}}$

Step-by-step solution

 Step 1: Identification of the given data

The given data is listed below as-

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

 Step 2: Significance of the Energy density

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.

Determination of the magnitude of the electric field

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

$\eta =\frac{1}{2}{\epsilon }_0{E}^2$

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

For, $\eta =10\frac{J}{m^3}$and ${\epsilon }_0=8.85\times {10}^{-12}\frac{F}{m}$

$$\begin{gathered} \eta =\frac{1}{2}{\epsilon }_0{E}^2 \\ \left(10 \frac{J}{m^3}\right)=\frac{1}{2}\times \left(8.85\times {10}^{-12} \frac{F}{m}\right)\times E^2 \\ {E}^2=\frac{\left(10 \frac{J}{m^3}\right)\times 2}{\left(8.85\times {10}^{-12} \frac{F}{m}\right)} \\ {E}^2=2.25\times {10}^{12} \end{gathered}$$

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

Determination of the magnitude of energy density associated with the electric field

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.