Value of permittivity in free space is

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8 {.85×10}^{-10} C^{2} / {Nm}^{2}

88 {.5×10}^{-12} C^{2} / {Nm}^{2}

8 {.85×10}^{-12} C^{2} / {Nm}^{2}

0 {.885×10}^{-12} C^{2} / {Nm}^{2}

answer is A.

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Detailed Solution

F = \frac{1}{4 π ε_{0}} \times \frac{q_{1} q_{2}}{r^{2}}

$r = 1 m, F = 1 {N, q}_{1} = 1 {C, q}_{2} = 1 C$

$\frac{1}{4 π ε_{0}} = 9 \times 10^{9} \frac{N - m^{2}}{C^{2}}$

$ε_{0} = \frac{1}{4 π \times 9 \times 10^{9}}$

$= \frac{1 \times 10^{- 9}}{4 \times 3.14 \times 9}$

$= 0.008846 \times 10^{- 9}$

$= 8.85 \times 10^{- 12} \frac{C^{2}}{N - m^{2}}$

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