Question

Match the following two columns

Answer 1

Mass $=\left[M\right],$ length $=\left[L\right],$ time $=T$, current $=\left[A\right]$
$F=qE$ where $E=$ electric field.
$\left[ML{T}^{-2}\right]=\left[AT\right]\left[E\right]$ as $I=\frac{dq}{dt}$
$\left[E\right]=\left[ML{A}^{-1}{T}^{-3}\right]$

Thus $B\to 1$.

As $E=-\frac{dV}{dx}$

So $\left[V\right]=\left[E\right]\left[L\right]=\left[M{L}^2{A}^{-1}{T}^{-3}\right]$

Thus $A\to 3$.

Electric flux $\left(\varphi \right)=$ electric field $\times$ area.
$\therefore \left[\varphi \right]=\left[E\right]\left[L^2\right]=\left[M{L}^3{A}^{-1}{T}^{-3}\right]$

Thus $C\to 2$.

As $E=\frac{q}{4\pi {ϵ}_0}$

So $\left[{ϵ}_0\right]=\frac{\left[q\right]}{\left[E\right]}=\frac{\left[AT\right]}{\left[ML{A}^{-1}{T}^{-3}\right]}=\left[M^{-1}{L}^{-1}{A}^2{T}^4\right]$

Thus $D\to 4$.