The unit of √2I / ϵ₀c is: (I = intensity of an electromagnetic wave, c: speed of light)

 

❓ Question

The unit of

$$\frac{\sqrt{2I}}{{\epsilon }_0\mathrm{c}}$$

is to be found, where

• $I$ = intensity of an electromagnetic wave,

• $\varepsilon_0$ = permittivity of free space, and

• $c$ = speed of light.

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🖼️ Question Image

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✍️ Short Solution

Step 1: Expressing $E_0$

$$I=\frac{1}{2}{\epsilon }_{0}c{E}_0^2\Rightarrow E_0=\sqrt{\frac{2I}{{\epsilon }_{0}c}}$$

So,

$$\frac{\sqrt{2I}}{{\epsilon }_{0}c}=\frac{E_0}{{\epsilon }_{0}c}\mathrm{.}$$

Step 2: Dimensional Analysis

We know the SI units:

• $E_0$: electric field intensity → $\text{V/m}=\text{N/C}={\text{kg}\text{<span style="color: black;">m</span>}\text{<span style="color: black;">s</span>}}^{-3}{\text{<br>}\text{A}}^{-1}$

• $\varepsilon_0$: permittivity → ${\text{C}}^2{\text{N}}^{-1}{\text{<br>}\text{m}}^{-2}={\text{A}}^2{\text{<span style="color: black;">s</span>}}^4{\text{<span style="color: black;">kg</span>}}^{-1}{\text{<span style="color: black;">m</span>}}^{-3}$

• $c$: speed of light → $\text{m/s}$
  

Now substitute:

$$\text{Unit of } \frac{E_0}{\varepsilon_0 c} = \frac{(\text{kg·m·s}^{-3}\text{·A}^{-1})} {(\text{A}^2\text{·s}^4\text{·kg}^{-1}\text{·m}^{-3})(\text{m·s}^{-1})}$$

Simplify powers of base units step-by-step:

• kg: $1 - (-1) = 2$
  

• m: $1 - (-3 + 1) = 1 + 2 = 3$
  

• s: $-3 - (4 - 1) = -3 - 3 = -6$
  

• A: $-1 - 2 = -3$
  

$$\Rightarrow [\text{Unit}]={\text{kg}}^2{\text{m}}^3{\text{s}}^{-6}{\text{A}}^{-3}$$

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🧮 Image Solution

✅ Conclusion & Video Solution

✅ Final Answer:

$$\boxed{{\displaystyle \text{Unit of }\frac{\sqrt{2I}}{{\epsilon }_{0}c}={\text{kg}}^2{\text{m}}^3{\text{s}}^{-6}{\text{A}}^{-\mathrm{3}}}}$$

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📘 Concept Recap:

• Intensity of an EM wave relates directly to the square of its electric field amplitude.

• $I = \frac{1}{2}\varepsilon_0 c E_0^2$ → gives the energy flow per unit area per unit time (Poynting vector).

• By rearranging and dimensional reasoning, we can derive consistent SI units for any physical expression.