Show that electrostatic force is about 36 orders of magnitude stronger

1.	Show that electrostatic force is about 36 orders of magnitude stronger than gravitational force
Answer» To SHOW: Electrostatic FORCE is about 36 orders of magnitude stronger than gravitational force. Calculation: LET's consider 2 protons: So, electrostatic force: $F_{e} = \dfrac{k {q}^{2} }{ {r}^{2} }$ $\implies F_{e} = \dfrac{9 \times {10}^{9} \times {(1.6 \times {10}^{ - 19} )}^{2} }{ {r}^{2} }$ Now, gravitational force between the protons: $F_{g} = \dfrac{G {m}^{2} }{ {r}^{2} }$ $F_{g} = \dfrac{6.67 \times {10}^{ - 11} \times {(1.67 \times {10}^{ - 27} )}^{2} }{ {r}^{2} }$ So, the RATIO: $\therefore \dfrac{F_{e}}{F_{g}} = \dfrac{9 \times {10}^{9} \times {(1.6 \times {10}^{ - 19} )}^{2} }{6.67 \times {10}^{ - 11} \times {(1.67 \times {10}^{ - 27} )}^{2} }$ $\implies\dfrac{F_{e}}{F_{g}} = 1.23 \times {10}^{36}$ $\boxed{ \bf \implies \: F_{e} = (1.23 \times {10}^{36} ) \times F_{g}}$ [Hence proved]

Show that electrostatic force is about 36 orders of magnitude stronger than gravitational force

Answer»

To SHOW:

Electrostatic FORCE is about 36 orders of magnitude stronger than gravitational force.

Calculation:

LET's consider 2 protons:

So, electrostatic force:

$F_{e} = \dfrac{k {q}^{2} }{ {r}^{2} }$

$\implies F_{e} = \dfrac{9 \times {10}^{9} \times {(1.6 \times {10}^{ - 19} )}^{2} }{ {r}^{2} }$

Now, gravitational force between the protons:

$F_{g} = \dfrac{G {m}^{2} }{ {r}^{2} }$

$F_{g} = \dfrac{6.67 \times {10}^{ - 11} \times {(1.67 \times {10}^{ - 27} )}^{2} }{ {r}^{2} }$

So, the RATIO:

$\therefore \dfrac{F_{e}}{F_{g}} = \dfrac{9 \times {10}^{9} \times {(1.6 \times {10}^{ - 19} )}^{2} }{6.67 \times {10}^{ - 11} \times {(1.67 \times {10}^{ - 27} )}^{2} }$