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The electric field intensity at a point 6 μm away from the centre of an electric dipole having charges ± $q = 1 .6 \times 10^{- 19} C$ and length $(2 L) = 5 Å$ on its axis is $0 .67 \times 10^{- Y}$ $N / C$ . Then find Y.

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Detailed Solution

$p = q (2 L) = 8 \times 10^{- 29} Cm$

$E = \frac{1}{4 π \in_{o}} \frac{2 p}{r^{3}}$

$E = 9 \times 10^{9} \times \frac{2 \times 8 \times 10^{- 29}}{{(6 \times 10^{- 6})}^{3}} = 0 .67 \times 10^{- 2} {NC}^{- 1}$

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