Q. A force of 2.25 N acts on a charge of $15 \times 10^{- 4}$C. Calculate the intensity of electric field at that point
A) 1500$\ NC^{- 1}$
B) 150$\ NC^{- 1}$
C) 1500$0\ NC^{- 1}$
D) None of these
Explanation
(a) $E = \frac{F}{q} = \frac{2.25}{15 \times 10^{- 4}} = 1500NC^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.