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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 $=$ $1.6 \times 10^{-19}$ $C$, mass of electron = $ 9.1 \times 10^{-31}$ $kg$, mass of proton = $1.6 \times {10^{ - 27}}\,kg,$ $\,G = 6.7 \times {10^{ - 11}}\,N{m^2}/k{g^2})$
$2.36 \times 10^{39}$
$2.36 \times 10^{40}$
$2.34 \times 10^{41}$
$2.34 \times 10^{42}$
Solution
(a) Gravitational force ${F_G} = \frac{{G{m_e}{m_p}}}{{{r^2}}}$
${F_G} = \frac{{6.7 \times {{10}^{ – 11}} \times 9.1 \times {{10}^{ – 31}} \times 1.6 \times {{10}^{ – 27}}}}{{{{(5 \times {{10}^{ – 11}})}^2}}}= 3.9 \times 10^{-47} \,N$
Electrostatic force ${F_e} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{e^2}}}{{{r^2}}}$
${F_e} = \frac{{9 \times {{10}^9} \times 1.6 \times {{10}^{ – 19}} \times 1.6 \times {{10}^{ – 19}}}}{{{{(5 \times {{10}^{ – 11}})}^2}}} = 9.22 \times 10^{-8}\, N$
So, $\frac{{{F_e}}}{{{F_G}}} = \frac{{9.22 \times {{10}^{ – 8}}}}{{3.9 \times {{10}^{ – 47}}}} = 2.36 \times {10^{39}}$
