Electrostatics Question 365
Question: 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 [MP PET 1996]
Options:
A) $ 2.304\times {{10}^{-10}}N/C $
B) $ 14.4V/m $
C) $ 16V/m $
D) $ 1.44\times 10^{11}N/C $
Show Answer
Answer:
Correct Answer: D
Solution:
$ E=\frac{q}{4\pi {\varepsilon _{0}}r^{2}} $
$ =9\times 10^{9}\times \frac{1.6\times {{10}^{-19}}}{{{({{10}^{-10}})}^{2}}}=1.44\times 10^{11}N/C $
BETA
