NEET Solved Paper 2017 Question 36
Question: Suppose the charge of a proton and an electron differ slightly. One of them is ?e, the other is $ (e+\Delta e). $ If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then $ \Delta e $ is of the order of [Given mass of hydrogen $ m _{h}=1.67\times {10^{-27}}kg] $
Options:
A) $ {10^{-47}}C $
B) $ {10^{-20}}C $
C) $ {10^{-23}}C $
D) $ {10^{-37}}C $
Show Answer
Answer:
Correct Answer: D
Solution:
$ F _{e}=F _{g} $ $ \frac{1}{4\pi {\varepsilon_0}}\frac{\Delta e^{2}}{d^{2}}=\frac{Gm^{2}}{d^{2}} $ $ 9\times 10^{9}(\Delta e^{2})=6.67\times {10^{-11}}\times 1.67 $ $ \times {10^{-27}}\times 1.67\times {10^{-27}} $ $ \Delta e^{2}=\frac{6.67\times 1.67\times 1.67}{9}\times {10^{-74}} $ $ \Delta \epsilon \approx {10^{-37}} $
BETA
