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1. Electric Charges and Fields
hard
In hydrogen like system the ratio of coulombian force and gravitational force between an electron and a proton is in the order of:
A
$10^{39}$
B
$10^{19}$
C
$10^{29}$
D
$10^{36}$
(JEE MAIN-2024)
Solution
$\mathrm{F}_{\mathrm{e}}=\frac{\mathrm{kQ}_1 \mathrm{Q}_2}{\mathrm{r}^2}=\frac{9 \times 10^9 \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{\mathrm{r}^2}$
$\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{r}^2}=\frac{6.67 \times 10^{-11} \times 9.1 \times 10^{-51} \times 1.6 \times 10^{-2 /}}{\mathrm{r}^2}$
$\frac{\mathrm{F}_{\mathrm{e}}}{\mathrm{F}_{\mathrm{g}}} \cong 0.23 \times 10^{40} \cong 2.3 \times 10^{39}$
Option $(1)$
Standard 12
Physics
