Question

Find the ratio of the electrical and gravitational forces between two protons.

Answer 1

The electrical and gravitational forces between two protons is $=1.238 \times 10^{36}$

Explanation:

Step 1:

We know that ,

$m=1.67 \times 10^{-27} \mathrm{kg}$

$q=1.6 \times 10^{-19} \mathrm{C}$

$\mathrm{G}=6.67 \times 10^{-11} \mathrm{N}-\mathrm{m}^{2} / \mathrm{kg}^{2}$

m is the  proton's mass,  

q is Charge on a proton,  

G is Gravitational constant,

Step 2:

Electrostatic force by Coulomb's Law,      

$F_{e}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r^{2}}$    

step 3:

Gravitational force by Coulomb's Law,    

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

$\frac{F_{e}}{F_{g}}=\frac{\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r^{2}}}{G \frac{m^{2}}{r^{2}}}$

$\frac{F_{e}}{F_{g}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{G m^{2}}$

$=9 \times 10^{9} \times \frac{1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{6.67 \times 10^{-11} \times\left(1.67 \times 10^{-27}\right)^{2}}$

$=9 \times 10^{9} \times \frac{2.56 \times 10^{-38}}{6.67 \times 10^{-11} \times\left(1.67 \times 10^{-27}\right)^{2}}$

$=\frac{23.04 \times 10^{-29}}{6.67 \times 10^{-11} \times 2.78 \times 10^{-54}}$

$=\frac{23.04 \times 10^{-29}}{6.67 \times 2.78 \times 10^{-65}}$

$=\frac{23.04 \times 10^{36}}{18.60}$

$=1.238 \times 10^{36}$