Two charges $+ 20 μ C$ and $- 20 μ C$ are placed 10 mm apart. The electric field at point $P$ , on the axis of the dipole $10 c m$ away from its centre $O$ on the side of the positive charge is

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$4.6 \times 10^{5} N C^{- 1}$

$8.6 \times 10^{9} N C^{- 1}$

$3.6 \times 10^{6} N C^{- 1}$

$4.1 \times 10^{6} N C^{- 1}$

answer is C.

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Detailed Solution

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$|\vec{p}| = q \times 2 a = 20 \times 10^{- 6} \times 10 \times 10^{- 3} = 2 \times 10^{- 7} C . m$

As $a < < r,$

$\vec{E} = \frac{2 |\vec{p}|}{4 π ε_{0} r^{3}} = \frac{2 \times 2 \times 10^{- 7} \times 9 \times 10^{9}}{{(10 \times 10^{- 2})}^{3}} = 3.6 \times 10^{6} N C^{- 1}$

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