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 $



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