Question

Answer the question which is given in the attachment.

Answer 1

We are asked to find the dimensions of:

$\frac{1}{2} \varepsilon_{\circ}E^2$

Here, $\varepsilon_{\circ}$ is Relative Permittivity, and $E$ is Electric Field.

We can proceed in two ways:

METHOD 1:


We find the dimensions of $\varepsilon_{\circ}$ and $E$.

Coulomb's Law states that:


$F = \frac{1}{4\pi \varepsilon_{\circ}} \frac{q_1 q_2}{r^2} \\ \\ \\ \implies \varepsilon_{\circ} = \frac{1}{4 \pi F} \frac{q_1 q_2}{r^2} \\ \\ \\ \implies [ \, \varepsilon_{\circ} \, ] = \left[ \frac{Q^2}{F R^2} \right]$


Now, E is Electric Field. 

Electric Field is Just Electromagnetic Force per unit Charge. 

In mathematical form:

$E = \frac{F}{Q} \\ \\ \\ \implies [ \, E \, ] = \left[ \frac{F}{Q} \right]$

So,now we can find dimension of $\frac{1}{2} \varepsilon_{\circ}E^2$

$\left[ \frac{1}{2} \varepsilon_{\circ} E^2 \right] \\ \\ \\ = \left[ \varepsilon_{\circ} E^2 \right] \\ \\ \\ = \left[ \frac{1}{F} \, \frac{Q^2}{R^2} \, \times \frac{F^2}{Q^2} \right] \\ \\ \\ = \left[ \frac{F}{R^2} \right] \\ \\ \\ = \frac{[ \, M \, L^1 \, T^{-2} \, ]}{[ \, L^2 \, ]}\\ \\ \\ = [ \, M \, L^{-1} \, T^{-2} \, ] \\ \\ \\ \\ \implies \boxed{\left[ \frac{1}{2}\varepsilon_{\circ}E^2 \right] = [ \, M \, L^{-1} \, T^{-2}\, ]}$


METHOD 2:


The expression $\frac{1}{2}\varepsilon_{\circ}E^2$ is known as the Electric Field Energy Density. It comes under the topic of Electromagnetic Waves.


Now, Electric Energy Density = Energy / Volume

$\frac{\text{Energy in Electric Field}}{\text{Volume}} = \frac{1}{2}\varepsilon_{\circ}E^2$

So, in a way, we only need to find the dimensions of Energy / Volume.

Now, Dimension of Volume is simply $[ \, L^3 \, ]$

Then, the unit of Energy is same as that of Work, and Work is defined as :

W = Force $\times$ Displacement
So, [ W ] = [ F ] $\times$ [ L ]
So, [ Energy ] = [ F ] $\times$ [ L ]

We know unit of Force is $kg \, m \, / \, s^2$.So its unit is:

$F = [ \, M \, L \, T^{-2} \, ]$

So,
$[ \, Energy \, ] = [ \, M \, L^2 \, T^{-2} \, ]$



Now, we can find our answer:


$\left[ \frac{1}{2}\varepsilon_{\circ}E^2 \right] \\ \\ \\ = \left[ \frac{Energy}{Volume} \right] \\ \\ \\ = \frac{[ \, M \, L^2 \, T^{-2} \, ]}{[ \, L^3 \, ]} \\ \\ \\ = [ \, M \, L^{-1} \, T^{-2} \, ] \\ \\ \\ \\ \\ \implies \boxed{\left[ \frac{1}{2}\varepsilon_{\circ}E^2 \right] = [ \, M \, L^{-1} \, T^{-2} \, ]}$


Thus, the  answer is Option 2.




Hope it helps
Purva
Brainly Community