Question

The unit of permittivity of free space, \(\varepsilon_{0}\), is
(a) coulomb/newton-metre
(b) newton-metre \(^{2} /\) coulomb \(^{2}\)
(c) coulomb \(^{2}\) /newton-metre \(^{2}\)
(d) coulomb \(^{2} /\) (newton-metre) \(^{2}\)

Answer 1

Correct Option (c) coulomb \(^{2} /\) newton-metre \(^{2}\)
Explanation:
Force between two charges
$$
\begin{aligned}
F &=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r^{2}} \\
\varepsilon_{0} &=\frac{1}{4 \pi} \frac{q^{2}}{F r^{2}} \\
&=\mathrm{C}^{2} / \mathrm{N}-\mathrm{m}^{2}
\end{aligned}
$$

Answer 2

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Answer 3

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