Question

Proton is placed 100 micrometers from a helium nucleus. Gravity pulls the proton and nucleus together while the electric force pushes them apart. Which is stronger and by how much ?

Answer 1

Answer:

Electric force is $6.2\cdot 10^{35}$ times stronger than gravitational force

Explanation:

The gravitational force between two objects is given by:

$F_G = G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

Here we have:

$m_1 = 1.67\cdot 10^{-27}kg$ (mass of the proton)

$m_2=4\cdot 1.67\cdot 10^{-27} =6.68\cdot 10^{-27} kg$ (mass of the helium nucleus is equal to 4 times the mass of a proton)

$r=100 \mu m = 100\cdot 10^{-6} m$

So,

$F_G = (6.67\cdot 10^{-11}) \frac{(1.67\cdot 10^{-27})(6.68\cdot 10^{-27})}{(100\cdot 10^{-6})^2}=7.44\cdot 10^{-56} N$

The electric force between two charged object is given by

$F_E=k\frac{q_1 q_2}{r^2}$

where

k is the  Coulomb constant

q1, q2 are the two charges

r is the separation

Here we have

$q_1 = 1.6\cdot 10^{-19}C$ (charge of the proton)

$q_2 = 2\cdot (1.6\cdot 10^{-19})=3.2\cdot 10^{-19}C$ (charge of the helium nucleus is twice that of the proton)

$r=100 \mu m = 100\cdot 10^{-6} m$

So,

$F_E=(9\cdot 10^9) \frac{(1.6\cdot 10^{-19})(3.2\cdot 10^{-19})}{(100\cdot 10^{-6})^2}=4.6\cdot 10^{-20}N$

Therefore, we see that the electric force is much stronger than the gravitational force, by a factor of:

$\frac{F_E}{F_G}=\frac{4.6\cdot 10^{-20}}{7.44\cdot 10^{-56}}=6.2\cdot 10^{35}$