Fg and Fe represents gravitational and electrostatic force respectively between electron and proton at a distance of 10 cm. The ratio of Fg / Fe is of the order of

Question

Fg and Fe represents gravitational and electrostatic force respectively between electron and proton at a distance of 10 cm. The ratio of Fg / Fe is of the order of (A) 1042 (B) 10-39 (C) 1 (D) 10-43

Tardigrade Admin · Accepted Answer

The electric force between an electron and proton at a distance

r

apart is

F_{e} = \frac{- k e ^{2}}{r ^{2}} ... (i)

where, the negative sign indicates that the force is attractive. The corresponding gravitational force (always attractive) is

F_{G} = - G \frac{m _{p} m _{e}}{r ^{2}} ... (ii)

where,

m_{p}

and

m_{e}

are the masses of the proton and electron.
On comparing Eqs. (i) and (ii), we get

∣ ∣ \frac{F _{e}}{F _{G}} ∣ ∣ = \frac{k e ^{2}}{G m _{p} m _{e}} = 2.4 \times 1 0^{39}

\Rightarrow F_{G} / F_{e} \approx 1 0^{- 39}

Q. $F_{g}$ and $F_{e}$ represents gravitational and electrostatic force respectively between electron and proton at a distance of $10 c m$ . The ratio of $F_{g} / F_{e}$ is of the order of

$1 0^{42}$

$1 0^{- 39}$

$1$

$1 0^{- 43}$

Q. Fg​ and Fe​ represents gravitational and electrostatic force respectively between electron and proton at a distance of 10cm. The ratio of Fg​/Fe​ is of the order of

1042

10−39

1

10−43

Q. $F_{g}$ and $F_{e}$ represents gravitational and electrostatic force respectively between electron and proton at a distance of $10 c m$ . The ratio of $F_{g} / F_{e}$ is of the order of

$1 0^{42}$

$1 0^{- 39}$

$1$

$1 0^{- 43}$