Force between two unit pole strength placed at a distance of one metre is

1 N

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\frac{10^{- 7}}{4 π} N

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10^{- 7} N

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4 π \times 10^{- 7} N

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Solution

The correct option is C $10^{- 7} N$

$F = 10^{- 7} \times \frac{m^{2}}{r^{2}} = \frac{10^{- 7} (1)^{2}}{(1)^{2}} = 10^{- 7} N$

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Force between two unit pole strength placed at a distance of one metre is

The correct option is C 10−7N F=10−7×m2r2=10−7(1)2(1)2=10−7N

The correct option is C $10^{- 7} N$

$F = 10^{- 7} \times \frac{m^{2}}{r^{2}} = \frac{10^{- 7} (1)^{2}}{(1)^{2}} = 10^{- 7} N$