Ratio of gravitational force and electrostatic repulsive force between two electrons is approximatel

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1. Electric Charges and Fields

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The ratio of gravitational force and electrostatic repulsive force between two electrons is approximately (gravitational constant $=6.7 \times 10^{-11} \,Nm ^2 / kg ^2$, mass of an electron $=9.1 \times 10^{-31} \,kg$, charge on an electron $=1.6 \times 10^{-19} C$ )

$24 \times 10^{-24}$

$24 \times 10^{-36}$

$24 \times 10^{-44}$

$24 \times 10^{-54}$

(KVPY-2020)

Solution

(c)

Given, gravitational constant $=6.7 \times 10^{-11} \,Nm ^{-2} / kg ^2$

Mass of an electron $=9.1 \times 10^{-31} \,kg$

Charge of an electron $=1.6 \times 10^{-19} \,C$

Gravitational force, $F_G=\frac{G n_1 m_2}{r^2}$

$=\frac{6.7 \times 10^{-11} \times\left(9.1 \times 10^{-31}\right)^2}{r^2}$

Electrostatic repulsive force,

$F_E =\frac{K q_1 q_2}{r^2}$

$=\frac{9 \times 10^9 \times\left(1.6 \times 10^{-19}\right)^2}{r^2}$

$\therefore \frac{F_G}{F_E}=\frac{6.7 \times 10^{-11} \times\left(9.1 \times 10^{-31}\right)^2}{9 \times 10^9 \times\left(1.6 \times 10^{-19}\right)^2}$

$=24 \times 10^{-44}$

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The ratio of gravitational force and electrostatic repulsive force between two electrons is approximately (gravitational constant $=6.7 \times 10^{-11} \,Nm ^2 / kg ^2$, mass of an electron $=9.1 \times 10^{-31} \,kg$, charge on an electron $=1.6 \times 10^{-19} C$ )

Solution

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