Q. The distance between a proton and electron both having a charge $1.6 \times 10^{- 19}\ coulomb,$ of a hydrogen atom is $10^{- 10}metre.$ The value of intensity of electric field produced on electron due to proton will be
A) $2.304 \times 10^{- 10}N/C$
B) $14.4\ V/m$
C) $16V/m$
D) $1.44 \times 10^{11}N/C$
Explanation
(d) $E = \frac{q}{4\pi\varepsilon_{0}r^{2}} = 9 \times 10^{9} \times \frac{1.6 \times 10^{- 19}}{\left( 10^{- 10} \right)^{2}} = 1.44 \times 10^{11}N/C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.