The dimension of $\frac{1}{2} \varepsilon_0 \mathrm{E}^2$, where $\varepsilon_0$ is permittivity of free…
The dimension of $\frac{1}{2} \varepsilon_0 \mathrm{E}^2$, where $\varepsilon_0$ is permittivity of free space and $E$ is electric field, is
$\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$
$\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$
$\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]$
$\left[\mathrm{MLT}^{-1}\right]$
Solution
Step 1: Given :
Formula of electric field \((1 / 2) \varepsilon_0 E^2, \varepsilon_0\) is the permittivity of free space
Step 2: Find the dimension of the \(\varepsilon_0\) :
For the dimension of \(\varepsilon_0\)
We know that, \(F=\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}\), where \(F=\) Force, \(q=\) charge, and \(r=\) distance between the charges
So, \(\in_0=\frac{1}{4 \pi \mathrm{~F}} \frac{q_1 q_2}{r^2}\)
Dimensional Formula will be \(\in_0=\frac{[A T][A T]}{\left[M L T^{-2}\right]\left[L^2\right]}=\left[M^{-1} L^{-3} T^4 A^2\right]\)
Step 3: Dimension of electric field:
And the dimension of electric field \(E\),
We know that \(E=\frac{F}{Q}\). (1), where \(E=\) Electric Field
Also, \(F=m \times a\), where \(m=\) mass, and \(a=\) acceleration
\(\operatorname{Char}(q)=i \times t\)
Dimension of Electric Field, using (1),
$\begin{aligned}
E & =\frac{F}{q} \\
& =\frac{\left[M L T^{-2}\right]}{[A][T]} \\
& =\left[M L T^{-3} A^{-1}\right]
\end{aligned}$
Dimension of \((\frac{1}{2}) \varepsilon_0 E^2\)
$\begin{aligned}
& = \frac{\varepsilon_0 E^2}{2}\\
& =\left[M^{-1} L^{-3} T^4 A^2\right]\left[M L T^{-3} A^{-1}\right]^2 \\
& =\left[M^{-1} L^{-3} T^4 A^2\right]\left[M^2 L^2 T^{-6} A^{-2}\right] \\
& =\left[M L^{-1} T^{-2}\right]
\end{aligned}$