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Q.

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

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a

6.25×107 N/C

b

9.28×107 N/C

c

13.1×1111 N/C

d

20.5×107 N/C

answer is A.

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Detailed Solution

By using E=9×109.2p r(r2−l2)2; wherep=500×10−6×10×10−2=5×10−5c×m,r=25cm=0.25m,l=5cm=0.05mE=9×109×2×5×10−5×0.25(0.25)2−(0.05)22=6.25×107N/C
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