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Electrostatics — Physics STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 SciencePhysicsElectrostatics4 Marks
Question
Calculate and compare the electrostatic and gravitational forces between two protons which are $10^{-15} m$ apart. Value of $G = 6.674 \times 10^{-11} m^3 \ kg^{-1} s^{-2}$ and mass of the porton is $1.67 \times 10^{-27} \ kg.$
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Answer

Given $: G = 6.674 \times 10^{-11} m^3 \ kg^{-1} s^{-2}$
$m_p = 1.67 \times 10^{-27} \ kg.$
$q_p = 1.67 \times 10^{-19} C, r = 10^{-15}$
To find:
$i.$ Electrostatic Force $(F_E)$
$ii.$ Gravitational Force $(F_G)$
Formula: $i. F _{ E }=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}$
$ii. F _{ E }=\frac{G m_1 m_2}{r^2}$
Calculation:
From formula $(i),$
$F _{ E }=9 \times 10^9 \times \frac{1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{\left(10^{-15}\right)^2}$
$= 9 \times 1.6 \times 1.6 \times 10$
$= 90 \times 1.6 \times 1.6$
$= \text {antilog} [\log 90 + \log 1.6 + \log 1.6]$
$= \text {antilog} [1.9542 + 0.2041 + 0.2041]$
$= \text {antilog} [2.3624]$
$= 2.303 \times 10^2 N$
From formula $(ii),$
$F _{ G }=6.674 \times 10^{-11} \times \frac{1.67 \times 10^{-27} \times 1.67 \times 10^{-27}}{\left(10^{-15}\right)^2}$
$= 6.674 \times 1.67 \times 1.67 \times 10^{-35}$
$= {\text {antilog} [\log 6.674 + \log 1.67 + \log 1.67]} \times 10^{-35}$
$= {\text {antilog} [0.8244 + 0.2227 + 0.2227]} \times 10^{-35}$
$= {\text {antilog} [1.2698]} \times 10^{-35}$
$= 1.861 \times 10^1 \times 10^{-35}$
$= 1.861 \times 10^{-34} N$
Now,
$\frac{ F _{ E }}{ F _{ G }}=\frac{2.303 \times 10^2}{1.861 \times 10^{-34}}$
$= {\text {antilog} [\log 2.303 – \log 1.861]} \times 10^{36}$
$= {\text {antilog} [0.3623 – 0.2697]} \times 10^{36}$
$= {\text {antilog} [0.0926]}$
$= 1.238 \times 10^{36}$
$\therefore F_E ≈ 10^{36} \times F_G$

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