Q. Two point charges of $1\mu$C and $- 1\mu$C are separated by a distance of 100$\ \mathring{\mathrm{A}}.$ A point P is at a distance of 10 cm from the mid-point and on the perpendicular bisector of the line joining the two charges. The electric field at P will be
A) 9 $NC^{- 1}$
B) 0.9 $Vm^{- 1}$
C) 90 $Vm^{- 1}$
D) 0.09 $NC^{- 1}$
Explanation
(d) The point lies on equatorial line of a short dipole.
$\therefore\ \ E = \frac{p}{4\pi\varepsilon_{0}r^{3}} = \frac{9 \times 10^{9}\left( 10^{- 6} \times 10^{- 8} \right)}{\left( 10^{- 1} \right)^{3}}\ $
$= 9 \times 10^{- 2} = 0.09\ NC^{- 1}$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.