The electric field in an electromagnetic wave is given by
$$E = (50 \text{ N C}^{-1})\sin\omega\left(t - \frac{z}{c}\right)$$
The energy contained in a cylinder of volume $$V$$ is $$5.5 \times 10^{-12}$$ J. The value of $$V$$ is _________ cm$$^3$$.
(given $$\epsilon_0 = 8.8 \times 10^{-12}$$ C$$^2$$ N$$^{-1}$$ m$$^{-2}$$)

We start with the expression for the electric field of the plane electromagnetic wave

$$E = E_0 \sin\!\left[\omega\!\left(t - \frac{z}{c}\right)\right]$$

and from the question we read the amplitude as

$$E_0 = 50 \ \text{N C}^{-1}.$$

An electromagnetic wave stores energy in both its electric and magnetic fields. The instantaneous energy density is

$$u = \frac{1}{2}\varepsilon_0 E^2 + \frac{1}{2\mu_0}B^2.$$

For a wave in free space we have the relation $$B = \dfrac{E}{c},$$ so that the magnetic part equals the electric part. Therefore the total instantaneous energy density can be written solely with the electric field as

$$u = \varepsilon_0 E^2.$$

Because the field is oscillatory, what actually matters for a macroscopic amount of energy is the time-average of this density. For any sinusoidal term $$\sin^2(\theta)$$ the time average is

$$\langle \sin^2(\theta) \rangle = \frac{1}{2}.$$

Applying this to the present case we obtain the time-averaged energy density

$$\langle u \rangle \;=\; \varepsilon_0 \langle E^2 \rangle \;=\; \varepsilon_0 \, E_0^{\,2}\,\langle\sin^2(\dots)\rangle \;=\; \frac{\varepsilon_0 E_0^{\,2}}{2}.$$

Substituting the numerical values (with $$\varepsilon_0 = 8.8 \times 10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$$ and $$E_0 = 50 \ \text{N C}^{-1}$$) we get

$$\langle u \rangle = \frac{8.8 \times 10^{-12}\; (50)^2}{2} = \frac{8.8 \times 10^{-12}\; \times 2500}{2}.$$

Multiplying first:

$$8.8 \times 2500 = 22000,$$

so

$$8.8 \times 10^{-12}\; \times 2500 = 2.2 \times 10^{-8}.$$

Dividing by 2 gives

$$\langle u \rangle = 1.1 \times 10^{-8}\ \text{J m}^{-3}.$$

The total (time-averaged) energy contained in a volume $$V$$ is related by

$$U = \langle u \rangle\, V.$$

We are told this energy equals

$$U = 5.5 \times 10^{-12}\ \text{J},$$

so we solve for $$V$$:

$$V = \frac{U}{\langle u \rangle} = \frac{5.5 \times 10^{-12}}{1.1 \times 10^{-8}} = 5.0 \times 10^{-4}\ \text{m}^3.$$

To convert cubic metres to cubic centimetres we recall

$$1\ \text{m}^3 = 10^6\ \text{cm}^3.$$

Therefore

$$V = 5.0 \times 10^{-4}\ \text{m}^3 \times 10^6\ \frac{\text{cm}^3}{\text{m}^3} = 5.0 \times 10^{2}\ \text{cm}^3 = 500\ \text{cm}^3.$$

Hence, the correct answer is Option 500.