Question

a) Two protons in a molecule are 3.40 10-10 m apart. Find the electric force exerted by one proton on the other. magnitude 1.99e-9 Correct: Your answer is correct. N direction repulsive Correct: Your answer is correct. (b) State how the magnitude of this force compares with the magnitude of the gravitational force exerted by one proton on the other. Fe Fg

Answer 1

Answer: a) $F_{e}=$ 1.99 x 10⁻⁹N

              b) Electric force is 1.24 x 10³⁶ times larger than gravitational force

Explanation: Electric Force is an interaction between two charges at a distance. It is calculated as

$F_{e}=k\frac{|q||Q|}{r^{2}}$

k is electric constant that values 9 x 10⁹N.m²/C²

q and Q are the quantity of charges

r is distance between the charges

a) For the two protons, charge = 1.602 x 10⁻¹⁹C:

$F_{e}=9.10^{9}\frac{(1.602.10^{-19})(1.602.10^{-19})}{(3.4.10^{-10})^{2}}$

$F_{e}=$ 1.99 x 10⁻⁹ N

Magnitude of electric force between 2 protons is 1.99 x 10⁻⁹ N and it is repulsive because they are both positive

Gravitational Force is the force of attraction that acts between all materials.

It is proportional to the object's masses and inversely proportional to their squared distance:

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

G is gravitational constant that values G = 6.67 x 10⁻¹¹ N.m²/kg²

b) For the two protons, m = 1.67 x 10⁻²⁷kg

$F_{g}=6.67.10^{-11}\frac{(1.67.10^{-27})(1.67.10^{-27})}{(3.4.10^{-10})^{2}}$

$F_{g}=$ 1.61 x 10⁻⁴⁵N

Comparing electrical and gravitational forces:

$\frac{F_{e}}{F_{g}} =\frac{1.99.10^{-9}}{1.61.10^{-45}}$

$\frac{F_{e}}{F_{g}}=$ 1.24 x 10³⁶

This shows that electrical force is 1.24 x 10³⁶ times greater than gravitational force.