Electrostatics Question 219
Question: The ratio of electrostatic and gravitational forces acting between electron and proton separated by a distance $ 5\times {{10}^{-11}}m, $ will be (Charge on electron $ =\text{ 1}.\text{6}\times \text{1}{{0}^{\text{19}}}C $ , mass of electron $ =\text{ 9}.\text{1}\times \text{1}{{0}^{\text{31}}}kg $ , mass of proton = $ 1.6\times {{10}^{-27}}kg, $
$ G=6.7\times {{10}^{-11}}Nm^{2}/kg^{2}) $ [RPET 1997; Pb PMT 2003]
Options:
A) $ \text{2}.\text{36}\times \text{1}{{0}^{\text{39}}} $
B) $ \text{2}.\text{36}\times \text{1}0^{40} $
C) $ \text{2}.\text{34}\times \text{1}0^{41} $
D) $ \text{2}.\text{34}\times \text{1}0^{42} $
Show Answer
Answer:
Correct Answer: A
Solution:
Gravitational force $ F _{G}=\frac{Gm _{e}m _{p}}{r^{2}} $
$ F _{G}=\frac{6.7\times {{10}^{-11}}\times 9.1\times {{10}^{-31}}\times 1.6\times {{10}^{-27}}}{{{(5\times {{10}^{-11}})}^{2}}} $ = 3.9x 10?47 N Electrostatic force $ F _{e}=\frac{1}{4\pi {\varepsilon _{0}}}\frac{e^{2}}{r^{2}} $
$ F _{e}=\frac{9\times 10^{9}\times 1.6\times {{10}^{-19}}\times 1.6\times {{10}^{-19}}}{{{(5\times {{10}^{-11}})}^{2}}} $ = 9.22x 10?8 N So, $ \frac{F _{e}}{F _{G}}=\frac{9.22\times {{10}^{-8}}}{3.9\times {{10}^{-47}}}=2.36\times 10^{39} $
BETA
