The ratio of electric force to the gravitational force between two protons separated by a finite distance is of the order of:

Question

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Answer 1

$G r a v i t a t i o n a l f o r c e = \frac{G ( m _{p} ) ^{2}}{r ^{2}}$
$E l e c t r i c f o r c e = \frac{K ( e ) ( e )}{r ^{2}}$
$R a t i o = \frac{K ( 1 . 6 \times 1 0 ^{- 19} ) ^{2}}{r ^{2}} \times \frac{r ^{2}}{G \times ( 1 . 6 7 \times 1 0 ^{- 27} ) ^{2}}$
$= \frac{( 9 \times 1 0 ^{9} ) \times ( 1 . 6 \times 1 0 ^{- 19} ) ^{2}}{( 6 . 6 7 \times 1 0 ^{- 11} ) ( 1 . 6 7 \times 1 0 ^{- 27} ) ^{2}}$
$= 1.23 \times 1 0^{20} \times 1 0^{16}$
$= 1 0^{36}$ (order of magnitude)

Answer 2

G r a v i t a t i o n a l f o r c e = \frac{G ( m _{p} ) ^{2}}{r ^{2}}

E l e c t r i c f o r c e = \frac{K ( e ) ( e )}{r ^{2}}

R a t i o = \frac{K ( 1 . 6 \times 1 0 ^{- 19} ) ^{2}}{r ^{2}} \times \frac{r ^{2}}{G \times ( 1 . 6 7 \times 1 0 ^{- 27} ) ^{2}}

= \frac{( 9 \times 1 0 ^{9} ) \times ( 1 . 6 \times 1 0 ^{- 19} ) ^{2}}{( 6 . 6 7 \times 1 0 ^{- 11} ) ( 1 . 6 7 \times 1 0 ^{- 27} ) ^{2}}

= 1.23 \times 1 0^{20} \times 1 0^{16}

= 1 0^{36}

(order of magnitude)