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The electric intensity due to a dipole of length 10 cm and having a charge of 500µC, at a point on the axis at a distance 20 cm from one of the charges in air, is

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6.25 × 10⁷N/C

9.28 × 10⁷N/C

13.1 × 11¹¹N/C

20.5 × 10⁷N/C

answer is A.

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Detailed Solution

$E_{axial} = \frac{1}{4 {πε}_{0}} \times \frac{2 pr}{{(r^{2} - a^{2})}^{2}} = \frac{9 \times 10^{9} \times 2 \times (500 \times 10^{- 6} \times 10 \times 10^{- 2})}{{(0 {.2}^{2} - 0 {.05}^{2})}^{2}}$
$= 6.25 \times 10^{7} N / C$

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