The unit of permittivity of free space $\varepsilon_0$, is:

The unit of permittivity of free space $\varepsilon_0$, is:
  1. coulomb/newton-metre
  2. newton-metre ${ }^2 /$ coulomb $^2$
  3. coulomb ${ }^2 /$ newton-metre $^2$
  4. coulomb $^2 /$ (newton-metre) ${ }^2$

Solution

We know that $\begin{aligned} & F=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r^2} \\ & \Rightarrow \quad \varepsilon_0=\frac{1}{4 \pi} \frac{a^2}{\mathrm{Fr} r^2}=\frac{\mathrm{C}^2}{\mathrm{~N}-\mathrm{m}^2} \\ & \end{aligned}$

Asked in: NEET 2004

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