Suppose the charge of a proton and an electron differ slightly. One of them is -e, the other is (e + e). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then e is of the order of [Given mass of hydrogen mh = 1.67 x 10-27 kg] Option 1) 10-20 C Option 2) 10-23 C Option 3) 10-37 C Option 4) 10-47 C

As we learnt in

Coulombic force -

$\dpi{100} F\propto Q_{1}Q_{2}=F\propto \frac{Q_{1}Q_{2}}{r^{2}}=F=\frac{KQ_{1}Q_{2}}{r^{2}}$

- wherein

K - proportionality Constant

Q1 and Q2 are two Point charge

Charge on one hydrogen atoms

$q_{e}+q_{p}=e+\left ( e+se \right )=se$

$\therefore F_{e} =\frac{1}{4\pi \varepsilon _{0}} \frac{\left ( se \right )^{2}}{d^{2}}$

gravitational force between two hydrogen

$F_{g} = \frac{Gm_{n}m_{n}}{d^{2}}$

2f force on the system is 2ro

$\therefore F_{e}=F_{g} \Rightarrow \frac{\left ( se \right )^{2}}{4\pi \varepsilon _{0}d^{2}} =\frac{Gm^{2}n}{d^{2}}$

$\left ( se \right )^{2} =4\pi \varepsilon _{0}Gm^{2}n$

$se = 10^{-37}c$

Option 1)

10-20 C

Incorrect

Option 2)

10-23 C

Incorrect

Option 3)

10-37 C

Correct

Option 4)

10-47 C

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