Question
Find the dimensional formula of $\in_0.$
✓
Answer
By Coulomb's Law,
$\text{F}=\frac{1}{4\pi\in_0}\frac{\text{q}_1\text{q}_2}{\text{r}^2}$
$\Rightarrow\in_0=\frac{1}{4\pi\text{F}}\frac{\text{q}_1\text{q}_2}{\text{r}^2}$
Using $[\text{F}]=\big[\text{MLT}^{-2}\big]$
$[\text{r}]=\big[\text{M}^0\text{L}^1\text{T}^0\big]$
$[\text{q}]=\big[\text{M}^2\text{L}^0\text{T}^1\text{A}^1\big],$ we get
$[\in_0]=\big[\text{M}^{-1}\text{L}^{-3}\text{T}^4\text{A}^2\big]$
$\text{F}=\frac{1}{4\pi\in_0}\frac{\text{q}_1\text{q}_2}{\text{r}^2}$
$\Rightarrow\in_0=\frac{1}{4\pi\text{F}}\frac{\text{q}_1\text{q}_2}{\text{r}^2}$
Using $[\text{F}]=\big[\text{MLT}^{-2}\big]$
$[\text{r}]=\big[\text{M}^0\text{L}^1\text{T}^0\big]$
$[\text{q}]=\big[\text{M}^2\text{L}^0\text{T}^1\text{A}^1\big],$ we get
$[\in_0]=\big[\text{M}^{-1}\text{L}^{-3}\text{T}^4\text{A}^2\big]$
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