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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, $\mathrm{K}=$ $\frac{1}{4 \pi \varepsilon_0}$ to the Gravitational constant G is nearly (Given that the charge of the proton and electron each $=1.6 \times 10^{-19} \mathrm{C}$, the mass of the electron $=9.11 \times 10^{-31} \mathrm{~kg}$, the mass of the proton $=$ $\left.1.67 \times 10^{-27} \mathrm{~kg}\right)$
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1719 Upvotes
Verified Answer
The correct answer is:
$10^{20}$
A.T.Q, we have,
$\begin{aligned}
\frac{\mathrm{F}_e}{\mathrm{~F}_{\mathrm{G}}} & =\frac{\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}}{\frac{\mathrm{G} m_1 m_2}{r^2}} \\
2.4 \times 10^{39} & =\frac{\mathrm{K}}{\mathrm{G}} \times \frac{\left(1.6 \times 10^{-19}\right)^2}{\left(9.1 \times 10^{-31} \times 1.67 \times 10^{-22}\right)} \\
& {\left[\because \mathrm{K}=\frac{1}{4 \pi \varepsilon_0}\right] }
\end{aligned}$
Hence,
$\begin{aligned}
\frac{\mathrm{K}}{\mathrm{G}} & =\frac{2.4 \times 10^{39} \times 9.1 \times 10^{-31} \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 1.6 \times 10^{-19}} \\
& =14.247 \times 10^{19} \\
& =1.4247 \times 10^{20} \approx 10^{20}
\end{aligned}$
$\begin{aligned}
\frac{\mathrm{F}_e}{\mathrm{~F}_{\mathrm{G}}} & =\frac{\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}}{\frac{\mathrm{G} m_1 m_2}{r^2}} \\
2.4 \times 10^{39} & =\frac{\mathrm{K}}{\mathrm{G}} \times \frac{\left(1.6 \times 10^{-19}\right)^2}{\left(9.1 \times 10^{-31} \times 1.67 \times 10^{-22}\right)} \\
& {\left[\because \mathrm{K}=\frac{1}{4 \pi \varepsilon_0}\right] }
\end{aligned}$
Hence,
$\begin{aligned}
\frac{\mathrm{K}}{\mathrm{G}} & =\frac{2.4 \times 10^{39} \times 9.1 \times 10^{-31} \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 1.6 \times 10^{-19}} \\
& =14.247 \times 10^{19} \\
& =1.4247 \times 10^{20} \approx 10^{20}
\end{aligned}$
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