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Q.

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

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a

6.25 x 107 N/C

b

9.28 x 107 N/C

c

13.1x1111 N/C

d

20.5 x 107 N/C

answer is A.

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Detailed Solution

We know, E=9×109·2prr2-l22 where p=500×10-6×10×10-2=5×10-5Cmr=25 cm=0.25 m,l=5 cm=0.05 mE=9×109×2×5×10-5×0.25(0.25)2-(0.05)22 =6.25×107 N/C
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