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The ratio of coulomb's electrostatic force to the gravitational force between an electron and a proton separated by some distance is $2.4 \times 10^{39}$. The ratio of the proportionality constant, $K=\frac{1}{4 \pi \varepsilon_0}$ to the Gravitational constant $G$ is nearly (Given that the charge of the proton and electron each $=1.6 \times 10^{-19}\; C$, the mass of the electron $=9.11 \times 10^{-31}\; kg$, the mass of the proton $=1.67 \times 10^{-27}\,kg$ ):
$10^{20}$
$10^{30}$
$10^{40}$
$10$
Solution
$\frac{ F _e}{ F _{ G }}=\frac{\frac{ Kq _1 q _2}{ r ^2}}{\frac{ Gm _1 m _2}{ r ^2}}$
$2.4 \times 10^{39}=\frac{ K }{ G } \times \frac{\left(1.6 \times 10^{-19}\right)^2}{\left(9.11 \times 10^{-31} \times 1.67 \times 10^{-27}\right)}$
$\frac{ K }{ G }=\frac{2.4 \times 10^{39} \times 15.2137 \times 10^{-58}}{2.56 \times 10^{-38}}$
$=14.26 \times 10^{19}$
$=1.426 \times 10^{20}$
$\approx 10^{20}$
Similar Questions
Four charge $Q _1, Q _2, Q _3$, and $Q _4$, of same magnitude are fixed along the $x$ axis at $x =-2 a – a ,+ a$ and $+2 a$, respectively. A positive charge $q$ is placed on the positive $y$ axis at a distance $b > 0$. Four options of the signs of these charges are given in List-$I$ . The direction of the forces on the charge q is given in List-$II$ Match List-$1$ with List-$II$ and select the correct answer using the code given below the lists.$Image$
| List-$I$ | List-$II$ |
| $P.$ $\quad Q _1, Q _2, Q _3, Q _4$, all positive | $1.\quad$ $+ x$ |
| $Q.$ $\quad Q_1, Q_2$ positive $Q_3, Q_4$ negative | $2.\quad$ $-x$ |
| $R.$ $\quad Q_1, Q_4$ positive $Q_2, Q_3$ negative | $3.\quad$ $+ y$ |
| $S.$ $\quad Q_1, Q_3$ positive $Q_2, Q_4$ negative | $4.\quad$ $-y$ |
