Question

The electric field ( in NC^-1) in an electromagnetic wave is given by $E=50\sin \omega (t-\frac{x}{c})$.The energy stored in a cylinder of cross-section $10c{m}^2$ and length $100cm$along the x-axis will be

• A. $5.5\times {10}^{-12}J$
• B. $1·1\times {10}^{-11}J$
• C. $2·2\times {10}^{-11}J$
• D. $1·65\times {10}^{-11}J$

Answer 1

The correct option is B

$1·1\times {10}^{-11}J$

Step 1: Given data

Electric field $E=50\sin \omega (t-\frac{x}{c})$

Area of the cylinder $\left(A\right)=10c{m}^2=10\times {10}^{-4}{m}^2$

Length $\left(l\right)=100cm=100\times {10}^{-2}m=1m$

Step 2: Formula used

The energy density can be computed using the following formula:

$U=\frac{1}{2}{\epsilon }_{\circ }{E}_{\circ }^2\dots \dots \dots \dots \dots \left(1\right)$

Where $U=$energy density, ${\epsilon }_{\circ }=$permittivity and $E_{\circ }=$electric field

$\begin{array}{rcl}Energy& =& Energydensity\times volume\dots \dots \dots \dots \dots \left(2\right)\end{array}$

$Volume\left(V\right)=Area\left(A\right)\times Length\left(l\right)\dots \dots \dots \dots \dots \dots \left(3\right)$

Step 3: Compute the Volume:

Substitute known values in the equation $\left(3\right)$ to obtain the volume.

$$\begin{gathered} V=10\times {10}^{-4}\times 1 \\ V={10}^{-3}{m}^3 \end{gathered}$$

Step 3: Compute energy density

From the equation $E=50\sin \omega (t-\frac{x}{c})$, it is clear that $E_o=50$

Substitute the known values in the equation $\left(1\right)$.

$\begin{array}{l}U=\frac{1}{2}\times 8.85\times {10}^{-12}\times {50}^2\\ U=1.1\times {10}^{-8}\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$m^3$}\right.\end{array}$

Step 4: Compute stored energy

Substitute the value of energy density and volume in equation $\left(2\right)$.

$\begin{array}{rcl}& =& 1.1\times {10}^{-8}\times {10}^{-3}\\ & =& 1.1\times {10}^{-11}J\end{array}$

Thus, the energy stored in a cylinder is $1.1\times {10}^{-11}J$.

Hence, option B is the correct answer.