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
see full answer
High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET
🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya
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.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
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