Two electric dipoles each of dipole moment $p = 6.2 \times 10^{^{- 30}} C - m$ are placed with their axis along the same line and their centres at a distance d $= 10^{- 8} c m$ . The force of attraction between dipoles is:

2.1 \times 10^{- 16} N

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2.1 \times 10^{- 12} N

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2.1 \times 10^{- 10} N

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2.1 \times 10^{- 8} N

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Solution

The correct option is C $2.1 \times 10^{- 8} N$
Electric field due to a dipole on its axis is given by $2 \to p / (4 π ϵ_{0} r^{3})$
Therefore using necessary approximation and $F = q E$ we get,
$F = 2 q p / (4 π ϵ_{0}) (1 / (d - l)^{3} - 1 / (d + l)^{3}) = 2 q p / (4 π ϵ_{0}) 6 d^{2} l / d^{6} = 6 p^{2} / 4 π ϵ_{0} d^{4}$
putting in the values,
$F = 6 \times 9 \times 10^{9} \times {6.2}^{2} \times 10^{- 60} \times 10^{40} = 2.1 \times 10^{- 8} N$

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Two electric dipoles each of dipole moment p=6.2×10−30C−m are placed with their axis along the same line and their centres at a distance d =10−8cm. The force of attraction between dipoles is:

Two electric dipoles each of dipole moment $p = 6.2 \times 10^{^{- 30}} C - m$ are placed with their axis along the same line and their centres at a distance d $= 10^{- 8} c m$ . The force of attraction between dipoles is: