Suppose that intensity of a laser is $$\left(\frac{315}{\pi}\right)\,\text{W m}^{-2}$$. The rms electric field, in units of $$\text{V m}^{-1}$$ associated with this source is close to the nearest integer is ___ ($$\varepsilon_0 = 8.86 \times 10^{-12}\,\text{C}^2\,\text{N m}^{-2}$$; $$c = 3 \times 10^8\,\text{m s}^{-1}$$)

The average intensity $$I$$ of a monochromatic electromagnetic (laser) beam is related to the root-mean-square (rms) value of its electric field $$E_{\text{rms}}$$ through the energy‐flux (Poynting-vector) formula for a plane wave in free space.

We first STATE the relation we shall use: for an electromagnetic wave,

$$I \;=\; \dfrac{1}{2}\,c\,\varepsilon_0\,E_{\text{rms}}^{\,2}.$$

Here $$c$$ is the speed of light and $$\varepsilon_0$$ is the permittivity of free space. Rearranging this equation to isolate $$E_{\text{rms}}$$, we obtain

$$E_{\text{rms}}^{\,2} \;=\; \dfrac{2\,I}{c\,\varepsilon_0}.$$ Hence

$$E_{\text{rms}} \;=\; \sqrt{\dfrac{2\,I}{c\,\varepsilon_0}}.$$

Now we SUBSTITUTe the numerical data given in the question:

$$I \;=\; \dfrac{315}{\pi}\;\text{W m}^{-2}, \qquad \varepsilon_0 \;=\; 8.86\times10^{-12}\;\text{C}^2\text{N}^{-1}\text{m}^{-2}, \qquad c \;=\; 3.00\times10^{8}\;\text{m s}^{-1}.$$

First we work out the denominator $$c\,\varepsilon_0$$ step by step:

$$c\,\varepsilon_0 \;=\; (3.00\times10^{8})\,(8.86\times10^{-12}) \;=\; (3.00\times8.86)\times10^{8-12} \;=\; 26.58\times10^{-4} \;=\; 2.658\times10^{-3}.$$

Next we evaluate the numerator $$2I$$:

$$2I \;=\; 2\left(\dfrac{315}{\pi}\right) \;=\; \dfrac{630}{\pi}. $$

Taking $$\pi\approx3.14$$ for numerical work,

$$\dfrac{630}{\pi} \;=\; \dfrac{630}{3.14} \;\approx\; 200.636.$$

Putting numerator and denominator together,

$$ E_{\text{rms}}^{\,2} \;=\; \dfrac{200.636}{2.658\times10^{-3}} \;=\; \dfrac{200.636}{0.002658} \;\approx\; 7.5498\times10^{4}. $$

We now take the square root to obtain the rms field:

$$ E_{\text{rms}} \;=\; \sqrt{7.5498\times10^{4}} \;\approx\; 2.75\times10^{2}\;\text{V m}^{-1} \;=\; 275\;\text{V m}^{-1}\;(\text{to the nearest integer}). $$

So, the answer is $$275$$.