Question

Determine the electrical force that two protons in the nucleus of a helium atom exert on each other when separated by 2.8×10−15m. Express your answer with the appropriate units.

Answer 1

Answer:

29.4N

Explanation:

By Coulomb's law, the electrical force of attraction/repulsion, F,  between two charges of magnitudes Q₁ and Q₂, is directly proportional to the product of the magnitudes of these charges and inversely proportional to the square of the distance, r, between these charges. i.e;

F ∝ Q₁ x Q₂ / r²

F = k x Q₁ x Q₂ / r²    -------------------(i)

Where;

k = proportionality constant = 8.99 x 10⁹Nm²/C²

From the question;

Two protons are given;

Let the charge of first proton be Q₁

Let the charge of second proton be Q₂

The charge of a proton  = 1.602 x 10⁻¹⁹C

=> Q₁ = Q₂ =  1.602 x 10⁻¹⁹C

Also;

r = 2.8 x 10⁻¹⁵m

Substitute these values into equation (i) s follows;

F = 8.99 x 10⁹ x 1.602 x 10⁻¹⁹ x  1.602 x 10⁻¹⁹ / (2.8 x 10⁻¹⁵)²

F = 8.99 x 10⁹ x 1.602 x 10⁻¹⁹ x  1.602 x 10⁻¹⁹ / (7.84 x 10⁻³⁰)

F = 23.07 x 10⁻²⁹ / (7.84 x 10⁻³⁰)

F = 29.4N

Therefore, the electrical force that the two protons exert on each other is 29.4N

Answer 2

Answer:

6.27 x 10^10 N  

Explanation:

Since the mass of the proton is 1.602 x 10-19 C, therefore the electric force between two protons separated by a distance:  

r = 2.8×10^−15m  

is given as following:  

F=k*║q_1║*║q_2║/r^2

  =k*║q║*║q║/r^2

  =k*║q║^2/r^2

  =k*q^2/r^2

  = 8.99 x 10^9 Nm^2c^2*[(1.67 x 10^-19 c)^2/ (2.8 x 10^-15 m)^2  

  = 6.27 x 10^10 N