Q. The electric intensity due to a dipole of length $10\ cm$ and having a charge of $500\mu C,$ at a point on the axis at a distance $20\ cm$ from one of the charges in air, is
A) $6.25 \times 10^{7}\ N/C$
B) $9.28 \times 10^{7}\ N/C$
C) $13.1 \times 11^{11}\ N/C$
D) $20.5 \times 10^{7}\ N/C$
Explanation
(a) By using $E = 9 \times 10^{9}.\frac{2pr}{\left( r^{2} - l^{2} \right)^{2}};$ where
$p = \left( 500 \times 10^{- 6} \right) \times \left( 10 \times 10^{- 2} \right) = 5 \times 10^{- 5}c \times m,$
$r = 25\ cm = 0.25\ m,\ l = 5\ cm = 0.05\ m$
$E = \frac{9 \times 10^{9} \times 2 \times 5 \times 10^{- 5} \times 0.25}{\left\{ (0.25)^{2} - (0.05)^{2} \right\}^{2}} = 6.25 \times 10^{7}N/C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.