First slide

Electrostatic force

Question

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 = $1 .6 \times 10^{- 27} kg$ , $G = 6 .7 \times 10^{- 11} {Nm}^{2} / {kg}^{2})$

Easy

A

$2 .36 \times 10^{39}$
B

$2 .36 \times 10^{40}$
C

$2 .34 \times 10^{41}$
D

$2 .34 \times 10^{42}$

Solution

Gravitational force $F_{G} = \frac{{Gm}_{e} m_{p}}{r^{2}}$
$F_{G} = \frac{6 .7 \times 10^{- 11} \times 9 .1 \times 10^{- 31} \times 1 .6 \times 10^{- 27}}{{(5 \times 10^{- 11})}^{2}} = 3 .9 \times 10^{- 47} N$
Electrostatic force $F_{e} = \frac{1}{4 {πε}_{0}} \frac{e^{2}}{r^{2}}$
$F_{e} = \frac{9 \times 10^{9} \times 1 .6 \times 10^{- 19} \times 1 .6 \times 10^{- 19}}{{(5 \times 10^{- 11})}^{2}} = 9 .22 \times 10^{- 8} N$
So, $\frac{F_{e}}{F_{G}} = \frac{9 .22 \times 10^{- 8}}{3 .9 \times 10^{- 47}} = 2 .36 \times 10^{39}$

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